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n, inputs = [int(n) for n in input().split(" ")]
list = [0]*(n+1)
for _ in range(inputs):
x, y, incr = [int(n) for n in input().split(" ")]
list[x-1] += incr
if((y)<=len(list)): list[y] -= incr;
max = x = 0
for i in list:
x=x+i;
if(max<x): max=x;
print(max)
And the same approach in ruby -- I claim no credit for working this out -- I wrote this after reading the comments and code posted here.
Just thought I'd add a ruby solution for anyone looking for one.
N,M=gets.chomp.split(' ').map(&:to_i)# create array of zeros of length N + 1arr=Array.new(N+1,0)M.timesdo# cycle through and get the inputsstart,finish,value=gets.chomp.split(' ').map(&:to_i)# increment value at start of sequencearr[start-1]+=value# decrement value at first position after sequencearr[finish]-=valueendtmp=0max=0arr.eachdo|value|# step through summing arraytmp+=value# capture the max value of tmpmax=tmpifmax<tmpendputsmax
intmain(){longlongintn,k,i,max=0,x=0;scanf("%lld %lld",&n,&k);int*a=(int*)malloc(sizeof(int)*(n+1));for(i=0;i<n;i++){*(a+i)=0;}for(i=0;i<k;i++){longlongintc,d,g;scanf("%lld %lld %lld",&c,&d,&g);*(a+c)+=g;if(d+1<=n){*(a+d+1)-=g;}}for(i=1;i<=n;i++){x+=*(a+i);if(max<x){max=x;}}printf("%lld",max);/* Enter your code here. Read input from STDIN. Print output to STDOUT */return0;}
Edit 1: I had used pointers so much because I was learning about pointers at that time
Edit 2: Many were asking about the logic, I will try to give you the intuition
The basic idea is doing lazy update,i.e., not doing actual update of range every time and instead store the net update somwhere so that it can be easily done in one go
Suppose we have an array 3,2,4,5,7,9,4 and we want to make update, from 0 to 2 increase value by 1
We create update arr of equal size filled with 0 . For first update, update[0]+=1 and update[3]-=1. We add where the interval begins, we subtract just after the interval ends which just helps in excluding all indices from 3 and after that from being incremented. How so? When we have to do final update, we actually sum values from left. After suming values from left(update[i]+=update[i-1]) and filling update array you would get, update = [1,1,1,0,0,0,0]. Notice that we succesfully excluded all indices from 3 and after. If we hadn't subtracted at position 3, the whole update array would have been filled with 1. Similarly we can make more range updates using same logic. Sum at each index gives the net value to be updated. Hope this helps!
Scannerscan=newScanner(System.in);intn=scan.nextInt();intm=scan.nextInt();//This will be the "difference array". The entry arr[i]=k indicates that arr[i] is exactly k units larger than arr[i-1]long[]arr=newlong[n];intlower;intupper;longsum;for(inti=0;i<n;i++)arr[i]=0;for(inti=0;i<m;i++){lower=scan.nextInt();upper=scan.nextInt();sum=scan.nextInt();arr[lower-1]+=sum;if(upper<n)arr[upper]-=sum;}longmax=0;longtemp=0;for(inti=0;i<n;i++){temp+=arr[i];if(temp>max)max=temp;}System.out.println(max);
I still havent understood this logic.Even though i implemented this logic in java with ease,i dont understand how this logic helps us arrive at the solution.
After thinking like that i also understood the logic the solution.
Let's think our summing part input like that
{A B S} =
{1 3 100}
{2 5 150}
{3 4 110}
{2 4 160}
Instead of writing all elements of array we can write maximum value at just starting and ending indexes to have less writing operation. So, after first input row, array can be something like that.
0 100 0 100 0 0 0 0 0
But the problem is here that even we didn't write anything, value of index 2 is also 100. When we wanted to continue with second step we have to check whether index 2 is between indexes of first row operation or not.
Instead of doing like that we can write S value to index A and -S value to B+1, so it is still similar logic. Starting from A to B all indexes have S value and rest of them have less than these indexes as S as. Now the array is like that:
0 100 0 0 -100 0 0 0 0
While calculating second row, we are writing 150 to index 2 and -150 to index 6. It will be like that: 0 100 150 0 -100 0 -150 0 0
If we write array with old method, which means that all numbers calculated one, it will be:
0 100 250 250 150 150 0 0 0
It shows that value of index 2 is : 100+150 = 250. Value of index 5: 100 + 150 + (-100) = 150. So by calculating with the solution written above, instead of writing all numbers, we are writing changes at edge indexes.
vararr=[];varmax=0;// init each element of arr to 0for(letl=0;l<n;l++){arr[l]=0;}// for each sum operation in queriesfor(leti=0;i<queries.length;i++){// update arr with number to add at index=queries[i][0] and number to remove at index=queries[i][0]+1 => this will allow us to build each element of the final array by summing all elements before it. The aim of this trick is to lower time complexityarr[queries[i][0]-1]+=queries[i][2];if(queries[i][1]<arr.length){arr[queries[i][1]]-=queries[i][2];}}for(letj=1;j<n;j++){arr[j]+=arr[j-1];}for(letk=0;k<arr.length;k++){max=Math.max(max,arr[k]);}//max = Math.max(...arr); // not working for big arraysreturnmax;
Hey,
I did the code in Java8 and my code is getting failed for input type - where only single value is present in a row of array. meaning only left index value is provided and right and k value is missing from array.
So can you help me how to solve this issue?
but it would be great if you can please provide your comments, like, dislike on my video how i did it..It motivates me to create better content for my audience.
I appologize for my bad sound quality but i am trying to improve it.
but it will be very difficult to add subtitle for this video because its around half an hour which explains all the concepts in deep.
Making this long video took lot of effort and time now adding a subtitle will be very tedious without any support.
Will suggest you to try automatic transcribe feature from youtube to translate it.
I had the same issue. my mistake was decrementing lower and upper. you don't decrement upper, the difference array needs to show it went down AFTER then last index, not within.
You could simplify your code a smidge, and save a little processing power, by removing your final for loop, and putting in an if check in the second to last loop, like this:
My js solution ran out of time for a few test cases
function arrayManipulation(n, queries) {
let a = new Array(n).fill(0);
for(let j = 0; j < queries.length; j++) {
let temp = queries[j]
let k = temp[0];
while(k <= temp[1]) {
a[k-1] += temp[2];
k++;
}
this one is the one that finally made it click for me...crazy how many different explanations are "whoosh" until the right person finds the "right" explanations.
But let's keep doing standardized testing in schools /s
can you expalin this:
But the problem is here that even we didn't write anything, value of index 2 is also 100. When we wanted to continue with second step we have to check whether index 2 is between indexes of first row operation or not.
but if you dont mind can you please leave the same comment along with like, dislike on my video. it motivates me and help others too find the solution over internet.
Most of the peope solved this problem but time complexity of solution is O(n*m) (due to two nested for loops)which can not be used to solve this problem for given time constraint, so you need better approach which beats O(n*m)
I have created a video tutorial for you and uploaded the same on youtube with complete explanation along with code complexity analysis.
Here is the video tutorial for my solution O(n+m) complexity passed all test cases.
Most of the peope solved this problem but time complexity of solution is O(n*m) (due to two nested for loops)which can not be used to solve this problem for given time constraint, so you need better approach which beats O(n*m)
I have created a video tutorial for you and uploaded the same on youtube with complete explanation along with code complexity analysis.
Here is the video tutorial for my solution O(n+m) complexity passed all test cases.
We are creating a "difference array" Which shows how many steps up or down have occurred (the difference between 0 and result of each operation) and where in the array they have occurred. This way, you can see how high the max ends up and return that for the solution.
I'm still trying to figure it out myself. But if you graph result after doing the operations, you would see some rise and fall in the graph.
It looks like his solution tracks the differences between each data point. It went up by x, down by y, remained the same...etc. And his solutions finds the highest increase.
Example:
5 3
1 2 100
2 5 100
3 4 100
After doing the operations you get [100, 200, 200, 200, 100]
His solutions final array is [0, 100, 100, 0, 0, -100]
Meaning starting at 0 the graph went up by 100, went up by 100 again, remained the same, then went back down by 100.
One insight that might help is that we're keeping track of the change in values rather than the values themselves at each index. Therefore, by adding the k at a and subtracting it after b, we are saying that at a the total value increases by k compared to the previous element, and after b the total value decreases by k compared to b. Hope that helps!
Here's the same solution in swift in case anyone needs it :).
func arrayManipulation(n: Int, queries: [[Int]]) -> Int {
var sums = Int: Int
for query in queries {
sums[query[0]] = (sums[query[0]] ?? 0) + query[2]
sums[query[1] + 1] = (sums[query[1] + 1] ?? 0) - query[2]
}
var currentmax = 0
var sum = 0
sums.sorted{ `$0.0 < $`1.0 }.
compactMap{ `$0.1; sum += $`0.1; currentmax = sum > currentmax ? sum : currentmax}
return currentmax
Here the indices are starting from 1. So, we should be subtracting 1 from both lower index and upper index. Here you have done so for lower index, but haven't done for upper index.
It's because it doesn't go back down until the element after the section ends.
eg: n = 4, a = 1, b = 2 k = 3.
So we have 3 3 0 0 after reading in that line.
In his array he represents this as 3 0 -3 0
ie the subtraction is the element after the last element in the section.
The reason the lower value has a "-1" is because java uses 0-indexed arrays, ie they start at 0. But the input for this question only starts at 1. So he puts the values one index lower in the array.
The upper value has no "-1" for the reason in the above paragraph about subtracting after the last element in the section
It's a difference array. He is storing the difference/the changes that should be made in each index and then runs a loop to add up all these changes. You increment the lower bound because everything inbetween the lower and upper bound should be incremented but you dont want this change to continue for the rest of the array so you decrement at (upper+1).
WTF!!! It's true! I had a time out problem with test case 7 up to 12 I think and my code is good enough and didn't know what to do...until I read your comment, I removed all the empty spaces and the debugging prints (System.out.println for Java 8) and it worked :O
LOL thanks.
In java, we don't need to loop explicitly to assign zero to all the arr locations. When we create it, it has the zero as default values. I like your solution, thanks.
Is there a reason all the solutions posted above are written inside main() and not the provided function arrayManipulation() ? Or did hackerrank just change this over the past few years for readability?
// Complete the arrayManipulation function below.staticlongarrayManipulation(intn,int[][]queries){// initialize array with 0's of size nlongarr[]=newlong[n];// each successive element contains the difference between itself and previous elementfor(inti=0;i<queries.length;i++){// when checking query, subtract 1 from both a and b since 0 indexed arrayinta=queries[i][0]-1;intb=queries[i][1]-1;intk=queries[i][2];arr[a]+=k;if(b+1<n){arr[b+1]-=k;}}// track highest val seen so far as we golongmax=Long.MIN_VALUE;for(inti=1;i<arr.length;i++){arr[i]+=arr[i-1];max=Math.max(arr[i],max);}returnmax;}
I was wondering the same thing. The instructions say to complete the manipulation method, not to rewrite the main method. I assumed that it should work without timing out if I just get the manipulation method to the point where it is efficient enough.
Your solution is good but time complexity of your solution is O(n*m) which can not be used to solve this problem for given time constraint, so you need better approach which beats O(n*m)
Here is the video tutorial for my solution O(n+m) complexity.
thanks.
But if would be great ,if you can provide your comments, like, dislike on my video how i did it..It motivates me to create better content for my audience.
thanks. But if would be great ,if you can provide your comments, like, dislike on my video how i did it..It motivates me to create better content for my audience.
Thanks @Kanahaiya for sharing. Your video is very helpful.
I really liked the way you explained by dry running the solution. One of the best programming tutroial ever seen.
Surely, will look into other videos as well.. Thanks.
process.stdin.resume();process.stdin.setEncoding('ascii');varinput_stdin="";varinput_stdin_array="";varinput_currentline=0;process.stdin.on('data',function(data){input_stdin+=data;});process.stdin.on('end',function(){input_stdin_array=input_stdin.split("\n");main();});functionreadLine(){returninput_stdin_array[input_currentline++];}/////////////// ignore above this line ////////////////////functionmain(){varn_temp=readLine().split(' ');varlistSize=parseInt(n_temp[0]);varlineNumber=parseInt(n_temp[1]);letslopList=newArray();for(leti=0;i<listSize;i++){slopList.push(0);}for(vara0=0;a0<lineNumber;a0++){vara_temp=readLine().split(' ');constbeginElementPos=parseInt(a_temp[0]);constendElementPos=parseInt(a_temp[1]);constaddValue=parseInt(a_temp[2]);slopList[beginElementPos-1]+=addValue;if(endElementPos<listSize){slopList[endElementPos]-=addValue;}}letactualList=newArray();letsum=0;for(leti=0;i<listSize;i++){sum+=slopList[i];actualList.push(sum);}letmax=actualList.reduce((acc,val)=>{return(acc>val)?acc:val;},0);console.log(max);}
usingSystem;usingSystem.Collections.Generic;usingSystem.IO;usingSystem.Linq;usingSystem.Numerics;classSolution{staticvoidMain(String[]args){string[]tokens_n=Console.ReadLine().Split(' ');intn=Convert.ToInt32(tokens_n[0]);intm=Convert.ToInt32(tokens_n[1]);// Instantiate and populate an array of size NBigInteger[]arr=newBigInteger[n];for(inti=0;i<n;i++)arr[i]=0;for(inta0=0;a0<m;a0++){string[]tokens_a=Console.ReadLine().Split(' ');inta=Convert.ToInt32(tokens_a[0]);intb=Convert.ToInt32(tokens_a[1]);intk=Convert.ToInt32(tokens_a[2]);// Apply operationfor(intj=a-1;j<b;j++)arr[j]+=k;}Console.WriteLine(arr.Max(i=>i));}}
long tempMax = 0;
long max = 0;
for(int i=1; i<=n; i++)
{
tempMax += numList[i];
if(tempMax > max) max = tempMax;
}
What is this code doing , why we cant use numList.Sum()
Used the idea o modifiers, but without the n-sized array. Also in C#. Complexity is O(m log m), that can be less than O(n).
static long arrayManipulation(int n, int m, int[][] queries) {
long max = 0;
SortedDictionary<int,long> fakeArray = new SortedDictionary<int,long>();
// 'm' times
for (int i=0; i<m; i++) {
int a = queries[i][0];
int b = queries[i][1];
int k = queries[i][2];
// bit optimization: a '0' valued query will make no difference on resultant ones
if (queries[i][2] <= 0) continue;
if (!fakeArray.ContainsKey(a)) { // O( log m )
fakeArray.Add(a, k); // O( log m )
}
else {
fakeArray[a] += k; // O( log m )
}
if (b < n) {
b++;
if (!fakeArray.ContainsKey(b)) { // O( log m )
fakeArray.Add(b, -1 * k); // O( log m )
}
else {
fakeArray[b] -= k; // O( log m )
}
}
}
long current = 0;
foreach(long modifier in fakeArray.Values){ // O( 2*m )
current += modifier;
if (current > max) max = current;
}
return max;
}
Most of the peope solved this problem but time complexity of your solution is O(n*m) (due to two nested for loops)which can not be used to solve this problem for given time constraint, so you need better approach which beats O(n*m)
I have created a video tutorial for you and uploaded the same on youtube.
Here is the video tutorial for my solution O(n+m) complexity passed all test cases.
Most of the peope solved this problem but time complexity of solution is O(n*m) (due to two nested for loops)which can not be used to solve this problem for given time constraint, so you need better approach which beats O(n*m)
I have created a video tutorial for you and uploaded the same on youtube with complete explanation along with code complexity analysis.
Here is the video tutorial for my solution O(n+m) complexity passed all test cases.
i cant pass some test cases, those with very long inputs, i dont know what to do!
here is my code: -
long arrayManipulation(long long int n, vector> queries) {
vector<long long int> v;
vector<long long int>v2;
long long int max;
for(long long int i=0;i<n;i++){
v.push_back(0);
}
for(long long int i=0;i<queries.size();i++){
for(long long int j=0;j<queries[i].size();j++){
v2.push_back(queries[i][j]);
}
for(long long int k=(v2[0]-1);k<v2[1];k++){
if(v[k]!=0){
v[k]+=v2[2];
}
else{
v[k]=v2[2];
}
}
v2.pop_back();
v2.pop_back();
v2.pop_back();
}
max=*max_element(v.begin(), v.end());
return max;
in the code there are two steps to acquire a new array filled with zeroes:
At first step the allocation itself: int *a=(int )malloc(sizeof(int)(n+1));
Second step is to propagate zero values using for-loop upon newly created array.
In my opinion, we can do the same with single call of "calloc". It supposed to be faster in theory, cause calloc is hardware dependent and probably we don't have to loop over N items again just for get them zero valued, since calloc could know is it actually required or not, so on some harware it could be already zeroes or hardware knows better how to zero during allocation faster
This does not work for the current version of this problem. The prompt is asking for the solution of a method with two args, one n for the size ofthe array and a queries array of arrays with start index, end index, and value. There should be no need for gets.chomp().
Here, an easiest way to understand in ruby, at least for me.
def arrayManipulation(n, queries)
arr = Array.new(n + 1, 0)
queries.each do |q|
arr[q.first - 1] += q.last
arr[q[1]] -= q.last
end
tmp = 0
max = 0
arr.each do |n|
tmp += n
max = tmp if tmp > max
end
max
end
I believe what he's trying to say is this:
There are two approaches here -
1. The difference array approach (the one every one is praising as being more efficient)
2. The intuitive approach - i.e., just going through each group of a,b,k values and incrementing elements located between a and b in the array, then finally searching through for a max)
The reason approach 1 is more efficient is because the operation for the difference array only requires a maximum 2 adjustments per transformation (you increment the value at the start index by k, and decrement the value after the end index by k).
Contrast this with approach 2, where actually going through the array and incrementing every value within the specified 'a-b' range could result in N operations.
So approach 2 could take a max of O(N * M) time- where 'M' is the number of operations, and N is the size of the array
And approach 1 could take a max of O(2 * M) time, which is considered equivalent to O(M)
Does that make sense? Someone correct me if I'm wrong! Cheers :)
the best way to understand it is form a simple example.
say there are 4 of us in a line: 1. me 2. you 3. bob 4. sue
and we all start out with zero points.
Thcan be represented as (where my points are in the first index, your in the second, bob's in the third, sue's in fourth):
0 0 0 0
Furthermore, we go through rounds, where in each round a contiguous block of us can receive some set amount of points.
So in round one, say 2 points are awarded to anything in the range of start index = 1, and end index = 2. This means that you and bob, who are in the range, get 2 points.
But rather than write the current score as:
0 2 2 0
We instead want to write the score as:
0 2 0 -2
Because we want each value to represent how much greater/less than it is from the previous-index value. Doing this allows us to only ever need to change two elements in the list for a given round.
Now say we play one more round, and 1 point is awarded to all people in range of index = 0 to index = 1. This gives you
1 2 -1 -2
All I did was add the point value to the start index, and subtract it from the "end index + 1".
Then calculating the max is simply a matter of going through that list, adding each element to an accumulator variable (as this summing process reveals the actual value of an element at any given point - e.g., you would have 3 points at the end because your score is the result of 1 + 2), and having a variable which keeps track of the running max.
Most of the peope solved this problem but time complexity of solution is O(n*m) (due to two nested for loops)which can not be used to solve this problem for given time constraint, so you need better approach which beats O(n*m)
I have created a video tutorial for you and uploaded the same on youtube with complete explanation along with code complexity analysis.
Here is the video tutorial for my solution O(n+m) complexity passed all test cases.
Can some one please help as to why my solution gives a segmentation fault?
include
include
include
include
int main() {
unsigned long long int n,m,l,b,k,i,val=0;
scanf("%llu%llu",&n,&m);
unsigned long long int a[n+1];
for(i=1;i<=n;i++)
{
a[i]=0;
}
while(m--)
{
scanf("%llu%llu%llu",&l,&b,&k);
for(i=l;i<=b;i++)
{
a[i]+=k;
if(a[i]>val)
{
val=a[i];
}
}
}
printf("%llu",val);
return 0;
Maximum array size has to be 10^7 which is not possible in case of C. I tried Dynamic memory allocation (malloc) which worked but got TLE for bigger test cases
Most of the peope solved this problem but time complexity of solution is O(n*m) (due to two nested for loops)which can not be used to solve this problem for given time constraint, so you need better approach which beats O(n*m)
I have created a video tutorial for you and uploaded the same on youtube with complete explanation along with code complexity analysis.
Here is the video tutorial for my solution O(n+m) complexity passed all test cases.
fromitertoolsimportaccumulaten,m=map(int,input().split(' '))dx=[0]*(n+1)# allow run past endfor_inrange(m):a,b,c=map(int,input().split(' '))dx[a-1]+=cdx[b]-=cprint(max(accumulate(dx)))
The question asks for a 1 indexed list and the a,b values read in are read in starting from 1 not 0. If you do not use (n+1)if b happens to be the last number of your list it will go out of bounds.
The 0 index value will always be 0 so it doesn't affect your answer when you sum for max difference.
Just improved your code a bit.
- split() is slightly faster than split("")
- an if was not needed as always true
- variables names fit those used in the exercise
I took the ideas explained here to put also into python in a bit more spelled-out way to help make sense of this more abstract work-around.
def arrayManipulation(n, queries):
# Big O (N)
res = [0]*(n+1) # we only really need one terminal row, since we're applying each pass to all rows below
# loop through all the queries and apply the increments/decrements for each
# Big O (M) (size of queires)
for row in range(len(queries)):
a = queries[row][0]
b = queries[row][1]
k = queries[row][2]
res[a-1] += k # increment the starting position
# this is where a loop would increment everything else between a & b by k
# but instead of taking b-a steps, we take a constant 2 steps, saving huge on time
res[b] -= k # decrement the position AFTER the ending position
# now loop through res one time - Big O (N) (size of res)
sm = 0 # running sum
mx = 0 # maximum value found so far
for i in range(len(res)):
sm += res[i]
if sm > mx:
mx = sm
# total run time is Big O (2*N + M) >> Big O(N)
return mx
The key concepts in my opinion here are:
1) we don't need to build the full aray, since it's impossible for any row but the last to have the max value. This is impossible because we apply the adjustments to every subsequent row of the resulting 2-D array.
2) we don't need to actually increment each value for indices 'a' through 'b'. While that's the most straight-forward way, that also requires x many (a minus b) steps for each pass of the loop. By noting instead where to start incrementing and where to stop incrementing (noted by decrementing after what would be the final increment), we can note the adjustments to apply without having to take every step each time. We can then run a separate single loop to go over each of the increments and keep a running sum of all the overlapping increments. The decrement allows us to know when that range where the increment applied is over by reducing the running sum by that number - in other words when that range is ended, we would have to look at overlapping increments excluding that one that just ended going forward to see if anything is bigger.
Someone else in here gave an image of a stair/hill which I found extremely helpful in visualizing and understanding this concept. The basic idea here is that instead of actually applying each increment, we can just note what the increment and range is and one by one go through each place and apply all the compounded increments at once. Loop 1 saves the increments in a different format, Loop 2 checks the overlap. And by using two separate loops we have basically Big O (N) rather than Big O (N^2) - or more specifically Big O (2N + M) instead of Big O (NM + N), where N is the size of the resulting array and M is the size of the queries array.
def arrayManipulation(n, queries):
arr=[0]*n
for _ in queries:
start=[0]-1
end=[1]-1
val=_[2]
topVal=end+1
for i in range(start,topVal):
arr[i]+=val
return max(arr)
m = input().split()
n = int(nm[0])
m = int(nm[1])
queries = []
for _ in range(m):
queries.append(list(map(int, input().rstrip().split())))
Most of the peope solved this problem but time complexity of solution is O(n*m) (due to two nested for loops)which can not be used to solve this problem for given time constraint, so you need better approach which beats O(n*m)
I have created a video tutorial for you and uploaded the same on youtube with complete explanation along with code complexity analysis.
Here is the video tutorial for my solution O(n+m) complexity passed all test cases.
Most of the peope solved this problem but time complexity of solution is O(n*m) (due to two nested for loops)which can not be used to solve this problem for given time constraint, so you need better approach which beats O(n*m)
I have created a video tutorial for you and uploaded the same on youtube with complete explanation along with code complexity analysis.
Here is the video tutorial for my solution O(n+m) complexity passed all test cases.
can the same be written like this:
def arrayManipulation(n, queries):
arr_n=[0]*n
for i in range(len(queries)):
n1=queries[i][0]
n2=queries[i][1]
for j in range(n1-1,n2):
arr_n[j]=arr_n[j]+queries[i][2]
return max(arr_n)
This is giving me a tiemout error while submitting.
can u assist here?
yeah,at the beginning, l got the same problem as well. This algorithm's complexity could be o(n). Try to learn something about prefix sum array.I hope this can help.
This breaks when y == n. The if when decrementing list[y] should be if y < len(list): list[y] -= incr. Overall great solution! Thanks for doing it in python!
if((y)<=len(list))
will always return true since len(list)==n+1 to start with and hence redundant. This condition makes more sense if len(list)==n.
This code will work perfectly fine though, just pointing out in case someone is confused regarding why this condition check is there.
max=list1[0]
for j in range(1,n):
list1[j]=list1[j]+list1[j-1]
if max
here list1 is prefix sum arraywhy did my code failed few test cases?here's my entire code :
def arrayManipulation(n, queries):
list1=[0]*(n+1)
for i in range(len(queries)):
list1[queries[i][0]-1]+=queries[i][2]
list1[queries[i][1]]+=-queries[i][2]
"""prefix sum"""
max=list1[0]
for j in range(1,n):
list1[j]=list1[j]+list1[j-1]
if max<list1[j]:
max=list1[j]
return(max)
if max<list1[j]:
max=list1[j]
I still have some doubt if anyone could explain it. I understand that it is making use of difference array to keep track of the difference between items rather than updating every item.
However as I understand, size of difference array is one less than total items in the actual array.
However the logic used seems to be a variant of difference array that I am not able to understand. It would be great if someone could help me connect the dots from this difference array to the actual working solution of this problem. Thanks.
so finally, 1st col is always the real value, and others are accumlated of previous values and it self.
so actual values are 100 200 200 200 100, max is 200
As i have understood
(5 , 3) 0 0 0 0 0
(1 , 2) 100 100 0 0 0 -> first and second places i.e a=1,b=2 , k=100, now a+k = 0+100 = 100, b+K = 0+100 = 100.
(2 , 5) now from 2 to 5 i.e 2,3,4,5 are added with K i.e 100.
already 2=100 now 2 becomes 100+100 =200, simialrly 3 = 0+100=100,4=0+100=100,5=0+100=100.
(3 , 4) now 3 and 4 are added with 100.
ie 3 value is 100 it becomes 100+100=200, 4 becomes 100+100=200.
you aren't allocating space for a that way. He uses new to create an array on the heap. You are trying to declare it on the stack, but you can't do that with a dynamic value like 'n'
If you have a problem understanding dynamic memory then make use of vector . It allocates memory dynamically wherein the user doesn't have to deal with pointers .
vector arr(n); declares an array of long int of length n.
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Array Manipulation
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Important points to note in this solution.
1)the first element of array a will always remain zero since 1<= a <=b <=n; 2
2)the second element of array a is the second element of the array after m operations.
Same solution translated in python -
And the same approach in ruby -- I claim no credit for working this out -- I wrote this after reading the comments and code posted here.
Just thought I'd add a ruby solution for anyone looking for one.
Same solution in C
Edit 1: I had used pointers so much because I was learning about pointers at that time
Edit 2: Many were asking about the logic, I will try to give you the intuition The basic idea is doing lazy update,i.e., not doing actual update of range every time and instead store the net update somwhere so that it can be easily done in one go
Suppose we have an array 3,2,4,5,7,9,4 and we want to make update, from 0 to 2 increase value by 1 We create update arr of equal size filled with 0 . For first update,
update[0]+=1
andupdate[3]-=1
. We add where the interval begins, we subtract just after the interval ends which just helps in excluding all indices from 3 and after that from being incremented. How so? When we have to do final update, we actually sum values from left. After suming values from left(update[i]+=update[i-1]
) and filling update array you would get,update = [1,1,1,0,0,0,0]
. Notice that we succesfully excluded all indices from 3 and after. If we hadn't subtracted at position 3, the whole update array would have been filled with 1. Similarly we can make more range updates using same logic. Sum at each index gives the net value to be updated. Hope this helps!Same solution in C#
Same solution in Java
It is the same logic as mentioned above.
Hi i dont understand how the difference array works. What is the logic behind adding at one index and subtracting at the other and taking its sum?
You can try to visualize the array as steps / stairs
We are just noting down the bump ups and bump downs
I still havent understood this logic.Even though i implemented this logic in java with ease,i dont understand how this logic helps us arrive at the solution.
me netheir, I am looking for the maths here, I am pretty sure the solution has a math method. Somebody here wrote "Prefix sum".
I tried an answer in the spirit of digital signal processing here.
After thinking like that i also understood the logic the solution.
Let's think our summing part input like that {A B S} = {1 3 100} {2 5 150} {3 4 110} {2 4 160}
Instead of writing all elements of array we can write maximum value at just starting and ending indexes to have less writing operation. So, after first input row, array can be something like that.
0 100 0 100 0 0 0 0 0
But the problem is here that even we didn't write anything, value of index 2 is also 100. When we wanted to continue with second step we have to check whether index 2 is between indexes of first row operation or not.
Instead of doing like that we can write S value to index A and -S value to B+1, so it is still similar logic. Starting from A to B all indexes have S value and rest of them have less than these indexes as S as. Now the array is like that:
0 100 0 0 -100 0 0 0 0
While calculating second row, we are writing 150 to index 2 and -150 to index 6. It will be like that: 0 100 150 0 -100 0 -150 0 0
If we write array with old method, which means that all numbers calculated one, it will be: 0 100 250 250 150 150 0 0 0
It shows that value of index 2 is : 100+150 = 250. Value of index 5: 100 + 150 + (-100) = 150. So by calculating with the solution written above, instead of writing all numbers, we are writing changes at edge indexes.
check it out here, you will get all your doubts solved https://www.geeksforgeeks.org/difference-array-range-update-query-o1/
Below link will also help to understand theory behind it. https://www.geeksforgeeks.org/constant-time-range-add-operation-array/
Same solution in Javascript
Hey, I did the code in Java8 and my code is getting failed for input type - where only single value is present in a row of array. meaning only left index value is provided and right and k value is missing from array. So can you help me how to solve this issue?
you could post your code,and we can check it out
simpler in es6:
function arrayManipulation(n, queries) { let arr = new Array(2*n).fill(0); let max = 0;
}
I did something pretty similar, just with a little bit more readable forEach:
This produces wrong answer in some of the tests.
Hi,
try this. Here is the video tutorial for my solution O(n+m) complexity.
https://www.youtube.com/watch?v=hDhf04AJIRs&list=PLSIpQf0NbcCltzNFrOJkQ4J4AAjW3TSmA
Would really appreciate your feedback like and comment etc. on my video.
It is a good video. I understood the algorithm clearly.
thanks @chandraprabha90.
but it would be great if you can please provide your comments, like, dislike on my video how i did it..It motivates me to create better content for my audience.
Could you add subtitles? I tried watching it but couldn't quite understand your accent through the audio
Hi Grozny,
I appologize for my bad sound quality but i am trying to improve it. but it will be very difficult to add subtitle for this video because its around half an hour which explains all the concepts in deep.
Making this long video took lot of effort and time now adding a subtitle will be very tedious without any support.
Will suggest you to try automatic transcribe feature from youtube to translate it.
Anyway thanks for watching.
Hi Grozny,
I have added subtitle for this tutorial and I hope it will you to understand logic with more clarity.
Nice approach you got there!
I had the same issue. my mistake was decrementing lower and upper. you don't decrement upper, the difference array needs to show it went down AFTER then last index, not within.
Thought to myself ....no js solutions here? then I find 3 of them, Js, ES6, readable wow.
Your last for loop isn't needed. You can move Math.max to the previous for loop.
Added some ES6 syntax suger...
Got stuck because of this
Thanks man!
You could simplify your code a smidge, and save a little processing power, by removing your final for loop, and putting in an if check in the second to last loop, like this:
Perfect! Could you please explain me the thought process behind the solution?
My js solution ran out of time for a few test cases
function arrayManipulation(n, queries) { let a = new Array(n).fill(0); for(let j = 0; j < queries.length; j++) { let temp = queries[j] let k = temp[0]; while(k <= temp[1]) { a[k-1] += temp[2]; k++; }
} return Math.max.apply(Math, a); }
this one is the one that finally made it click for me...crazy how many different explanations are "whoosh" until the right person finds the "right" explanations.
But let's keep doing standardized testing in schools /s
Thanks for the post, it really made me understand the logic clearly
Thank you @Kemal_caymaz for the explanation, I have found this very useful.
Thank you!
thnx
super awesome X 1000!!!
can you expalin this: But the problem is here that even we didn't write anything, value of index 2 is also 100. When we wanted to continue with second step we have to check whether index 2 is between indexes of first row operation or not.
Hi,
I have created a video tutorial for you and uploaded the same on youtube. Here is the video tutorial for my solution O(n+m) complexity.
https://youtu.be/hDhf04AJIRs
Would really appreciate your feedback like, dislike , comment etc. on my video.
fantastic video, thank you!
most welcome.
but if you dont mind can you please leave the same comment along with like, dislike on my video. it motivates me and help others too find the solution over internet.
impressive ...
thanks bro..
Hi,
Most of the peope solved this problem but time complexity of solution is O(n*m) (due to two nested for loops)which can not be used to solve this problem for given time constraint, so you need better approach which beats O(n*m)
I have created a video tutorial for you and uploaded the same on youtube with complete explanation along with code complexity analysis.
Here is the video tutorial for my solution O(n+m) complexity passed all test cases.
https://youtu.be/hDhf04AJIRs
Would really appreciate your feedback like, dislike , comment etc. on my video.
Thanks a lot budy for your fantastic explanation ! Your thinking is really amazing like you . Good job.
most welcome. It would be great, if you can provide your feedback like, dislike , comment etc. on my video. It motivate me to do more for you all
hey used this logic its failing for 10 test cases with large inputs
Hi,
Most of the peope solved this problem but time complexity of solution is O(n*m) (due to two nested for loops)which can not be used to solve this problem for given time constraint, so you need better approach which beats O(n*m)
I have created a video tutorial for you and uploaded the same on youtube with complete explanation along with code complexity analysis.
Here is the video tutorial for my solution O(n+m) complexity passed all test cases.
https://youtu.be/hDhf04AJIRs
Would really appreciate your feedback like, dislike , comment etc. on my video.
Thanks, your explanation is very helpful
Thank you so much for clearly explaining this.
did you got it?
getting it correct for few cases but when both indexes are same then its giving a error message
How can you implement the logic with ease without understanding the logic? Copy and pasting is not the same as "implementing"
To explain further for people confused:
We are creating a "difference array" Which shows how many steps up or down have occurred (the difference between 0 and result of each operation) and where in the array they have occurred. This way, you can see how high the max ends up and return that for the solution.
I found this explanation helpful to have this fact dawn on me after much noodling: https://www.geeksforgeeks.org/difference-array-range-update-query-o1/
Like piling up blocks? Adding a number -> going up one step and subtracting -> down . Finally, we count how high we can go.
I'm still trying to figure it out myself. But if you graph result after doing the operations, you would see some rise and fall in the graph.
It looks like his solution tracks the differences between each data point. It went up by x, down by y, remained the same...etc. And his solutions finds the highest increase.
Example: 5 3
1 2 100
2 5 100
3 4 100
After doing the operations you get [100, 200, 200, 200, 100] His solutions final array is [0, 100, 100, 0, 0, -100] Meaning starting at 0 the graph went up by 100, went up by 100 again, remained the same, then went back down by 100.
So the highest point is 200, the solution.
you add up all the numbers > 0 in the final list, which is 100 + 100 = 200
Hi,
I have created a video tutorial for you and uploaded the same on youtube. Here is the video tutorial for my solution O(n+m) complexity.
https://youtu.be/hDhf04AJIRs
Would really appreciate your feedback like, dislike , comment etc. on my video.
One insight that might help is that we're keeping track of the change in values rather than the values themselves at each index. Therefore, by adding the k at a and subtracting it after b, we are saying that at a the total value increases by k compared to the previous element, and after b the total value decreases by k compared to b. Hope that helps!
Here's the same solution in swift in case anyone needs it :).
func arrayManipulation(n: Int, queries: [[Int]]) -> Int { var sums = Int: Int for query in queries { sums[query[0]] = (sums[query[0]] ?? 0) + query[2] sums[query[1] + 1] = (sums[query[1] + 1] ?? 0) - query[2] }
}
refer to prefix sum algorithm.....you'll be clear!
Hi , I have a doubt.
Here the indices are starting from 1. So, we should be subtracting 1 from both lower index and upper index. Here you have done so for lower index, but haven't done for upper index.
Can you please explain the reason behind this ?
It's because it doesn't go back down until the element after the section ends.
eg: n = 4, a = 1, b = 2 k = 3. So we have 3 3 0 0 after reading in that line. In his array he represents this as 3 0 -3 0 ie the subtraction is the element after the last element in the section.
The reason the lower value has a "-1" is because java uses 0-indexed arrays, ie they start at 0. But the input for this question only starts at 1. So he puts the values one index lower in the array. The upper value has no "-1" for the reason in the above paragraph about subtracting after the last element in the section
You don't have to do this:
because long by default is 0
could you please explain me the working of the code?
i think test code bigger than long int,so we need a larger data structure
Hi,
I have uploaded a video tutorial which can help you to understand the problem as well as logic.
Here is the video tutorial for my solution O(n+m) complexity.
https://youtu.be/hDhf04AJIRs
Would really appreciate your feedback like, dislike , comment etc. on my video.
tnks @kanahaiya
no need to init array elements to 0 in Java
Hi, your solution works but I am not convinced!
I don't see how do you increment all elements between lower and upper!
Can you please explain? Thanks.
It's a difference array. He is storing the difference/the changes that should be made in each index and then runs a loop to add up all these changes. You increment the lower bound because everything inbetween the lower and upper bound should be incremented but you dont want this change to continue for the rest of the array so you decrement at (upper+1).
but how can we know about the upper index of a particular increment,when we are adding all at a once in a loop?
Using Java 8 syntaxis:
large amount of lines will crash your solution. And map fucntion need to make Boxing methinks and this takes much time
WTF!!! It's true! I had a time out problem with test case 7 up to 12 I think and my code is good enough and didn't know what to do...until I read your comment, I removed all the empty spaces and the debugging prints (System.out.println for Java 8) and it worked :O LOL thanks.
This is true, I spent too much time trying figure out why my solution was timing out. Deleted some whitespace and it ran fine.
Hi,
I have uploaded a video tutorial which can help you to understand the problem as well as logic.
Here is the video tutorial for my solution O(n+m) complexity.
https://youtu.be/hDhf04AJIRs
Would really appreciate your feedback like, dislike , comment etc. on my video.
why temp += arr[i];
Array is already 0, you dont have to assign 0s to each array elements
scanner on large lines of input is not suitable solution, it reads too long
You don't need to loop the arr to put 0. Bydefault, it has the value zero when you initialize.
In java, we don't need to loop explicitly to assign zero to all the arr locations. When we create it, it has the zero as default values. I like your solution, thanks.
using an if statement inside the for loop will just contribute in complexity. you can replace it with Math.max function.
Is there a reason all the solutions posted above are written inside main() and not the provided function arrayManipulation() ? Or did hackerrank just change this over the past few years for readability?
Probably just for readability.
I was wondering the same thing. The instructions say to complete the manipulation method, not to rewrite the main method. I assumed that it should work without timing out if I just get the manipulation method to the point where it is efficient enough.
Same solution in Golang
Clever solution!
Can you help me see what is wrong with my code here?
Wrong name in first line:
func arrayManipulation(n int32, queries [][]int32) int64 {
}
why it is a[q + 1] -= sum, it should be a[q + 1] += sum, right????
import java.io.; import java.math.; import java.security.; import java.text.; import java.util.; import java.util.concurrent.; import java.util.regex.*;
public class Solution {
}
What is wrong with this code. Please help.
it may be showing Terminated due to timeout error because in test case 7 there is a huge number of input 10000000 100000 (like that)
you should try prefix array sum it is so simple
Hi,
Your solution is good but time complexity of your solution is O(n*m) which can not be used to solve this problem for given time constraint, so you need better approach which beats O(n*m)
Here is the video tutorial for my solution O(n+m) complexity.
https://www.youtube.com/watch?v=hDhf04AJIRs&list=PLSIpQf0NbcCltzNFrOJkQ4J4AAjW3TSmA
Would really appreciate your feedback like and comment etc. on my video.
your video is helpful
thanks. But if would be great ,if you can provide your comments, like, dislike on my video how i did it..It motivates me to create better content for my audience.
Your video helped me understand the algorithm. I liked the video. Thank you so much
thanks. But if would be great ,if you can provide your comments, like, dislike on my video how i did it..It motivates me to create better content for my audience.
Thanks @Kanahaiya for sharing. Your video is very helpful. I really liked the way you explained by dry running the solution. One of the best programming tutroial ever seen. Surely, will look into other videos as well.. Thanks.
you will get TLE. Try different algorithm which takes less time.
Thanks for the clean code and the comment! It helped more than the entire discussion!
This line is unnecessary due to the fact that the default value of long in Java is 0L and there is no need for assigning them again to 0.
for(int i=0;i<n;i++) arr[i]=0;
Thanks for the enlightenment.
Here is the same solution in Swift:
I think
is not needed.
Same solution in javascript
very smart,here is follow output to help me understand the idea
Guys, if you look for a clear understanding of the solution, I read a pretty clear comment down the road that clarified my mind.
Basically, when you add value from a to b you just need to know that it goes up from k and goes down of k after.
What this algo does is to register the slopes only, so we just need 2 entry, with O(1) complexity.
We just need to know that we are upping from k at the beginning and decreasing at the end.
Finally, the maximum would be...
The addition of all the slopes, that is why it's max(sum) of the tables, because the tables itself registers the slopes
Thanks a lot!!That really helped!
Can you explain the concept of just adding add subtracting at a particular index? I mean how have we arrived to this logic?
Hi,
I have uploaded a video tutorial which can help you to understand the problem as well as logic.
Here is the video tutorial for my solution O(n+m) complexity.
https://youtu.be/hDhf04AJIRs
Would really appreciate your feedback like, dislike , comment etc. on my video.
Hi,
Here is the video tutorial for my solution O(n+m) complexity.
https://youtu.be/hDhf04AJIRs
Would really appreciate your feedback like and comment etc. on my video.
I've add it to my playlist
haha..thank you.
but if you dont mind, Would really appreciate your feedback like, dislike , comment etc. on my video.
I didnt well understand that what will happen if b+1 is out of array bounds?
it can't be out of bounds, it saids that b>n in the problem statement.
if b+1 > n then the the addition of k from position a will continue till the last element of the array.
Hi,
I have uploaded a video tutorial which can help you to understand the problem as well as logic.
Here is the video tutorial for my solution O(n+m) complexity.
https://youtu.be/hDhf04AJIRs
Would really appreciate your feedback like, dislike , comment etc. on my video.
Jesus christ, it all makes sense now after that graph lol, I kept wondering what drug these people were taking to arrive at this conclusion.
can you explain ? :)
Very well explained (Y)
i have got correct output of your test cases but 3 test cases of hackerrank are getting wrong. iam not understanding whats wrong.please help me
brilliant idea!
what about the custom input:
5 1
1 5 -100
wouldn't the output be 0? But the max would be -100 since all elements would be -100.
There's a constraint: 0 <= k <= 10^9
In your case of negative k, the minimum value can be obtained by the similar approach. Imagine a valley rather than a mountain.
why are you subtracting???? "( b < n ) res [ b ] -= k;"
How is your method faster than the following?
The "Apply operation part" is O(k) here. In the diff array version, apply operation is O(1)
hey did u passed all with this...i used same logic in C#..but passes till 6
your database is int? some tests data is too big
your code is only passing 3 test cases out of 10 . Don't post the code unless it pass all the test cases dude !!!!
bro, this is a discussion forum. Why are you demotivating a begineers ? Anyway I have submitted that code successfully. Thanks for the advice.
I used the same logic, it passes most of the tests but I get timeout error.
Hi,
I have uploaded a video tutorial which can help you to understand the problem as well as logic.
Here is the video tutorial for my solution O(n+m) complexity.
https://youtu.be/hDhf04AJIRs
Would really appreciate your feedback like, dislike , comment etc. on my video.
if(b+1 <= n) numList[b+1] -= k;
Why we are substracting?
It is failing test case 4 to 13. Not sure why
It is failing test case 4 to 13. Can you please check
Used the idea o modifiers, but without the n-sized array. Also in C#. Complexity is O(m log m), that can be less than O(n).
In the last for, i should be initialized to 0, not 1
*(a+c)+=g; if(d+1<=n){ *(a+d+1)-=g; }
This is a very elegant and beautiful solution.
could you please explain me the logic?
Hi,
Most of the peope solved this problem but time complexity of your solution is O(n*m) (due to two nested for loops)which can not be used to solve this problem for given time constraint, so you need better approach which beats O(n*m)
I have created a video tutorial for you and uploaded the same on youtube. Here is the video tutorial for my solution O(n+m) complexity passed all test cases.
https://youtu.be/hDhf04AJIRs
Would really appreciate your feedback like, dislike , comment etc. on my video.
Thanks, that helped understanding the optimization.
pls explain logic
Hi,
Most of the peope solved this problem but time complexity of solution is O(n*m) (due to two nested for loops)which can not be used to solve this problem for given time constraint, so you need better approach which beats O(n*m)
I have created a video tutorial for you and uploaded the same on youtube with complete explanation along with code complexity analysis.
Here is the video tutorial for my solution O(n+m) complexity passed all test cases.
https://youtu.be/hDhf04AJIRs
Would really appreciate your feedback like, dislike , comment etc. on my video.
Hi,
I have uploaded a video tutorial which can help you to understand the problem as well as logic.
Here is the video tutorial for my solution O(n+m) complexity passed all test cases.
https://youtu.be/hDhf04AJIRs
Would really appreciate your feedback like, dislike , comment etc. on my video.
Will fail if max result is negative
i cant pass some test cases, those with very long inputs, i dont know what to do! here is my code: -
long arrayManipulation(long long int n, vector> queries) {
}
hi,
I am sure this comment will definately help you.
https://www.hackerrank.com/challenges/crush/forum/comments/570284
No need to explicitly assign 0 to every item in the array. malloc() does that automatically.
calloc would be faster i guess instead of malloc + zeroing the mem
Could you explain more, please?
in the code there are two steps to acquire a new array filled with zeroes:
In my opinion, we can do the same with single call of "calloc". It supposed to be faster in theory, cause calloc is hardware dependent and probably we don't have to loop over N items again just for get them zero valued, since calloc could know is it actually required or not, so on some harware it could be already zeroes or hardware knows better how to zero during allocation faster
okay thanks for the information!
Can you please explain this code to me
This does not work for the current version of this problem. The prompt is asking for the solution of a method with two args, one n for the size ofthe array and a queries array of arrays with start index, end index, and value. There should be no need for gets.chomp().
And you can easily make solution "sparse" to conserve memory on big arrays, just replace
arr = Array.new(N + 1, 0)
with
arr = Hash.new(0)
and
arr.each do |value|
with
1.upto(N) do |n| value = arr[n]
Here, an easiest way to understand in ruby, at least for me.
This worked, but I'm still trying to understand why this was the correct answer.
Because it takes less time to execute. Otherwise solution execution got time outed
You are not answering the question.
I believe what he's trying to say is this: There are two approaches here - 1. The difference array approach (the one every one is praising as being more efficient) 2. The intuitive approach - i.e., just going through each group of a,b,k values and incrementing elements located between a and b in the array, then finally searching through for a max)
The reason approach 1 is more efficient is because the operation for the difference array only requires a maximum 2 adjustments per transformation (you increment the value at the start index by k, and decrement the value after the end index by k). Contrast this with approach 2, where actually going through the array and incrementing every value within the specified 'a-b' range could result in N operations.
So approach 2 could take a max of O(N * M) time- where 'M' is the number of operations, and N is the size of the array And approach 1 could take a max of O(2 * M) time, which is considered equivalent to O(M)
Does that make sense? Someone correct me if I'm wrong! Cheers :)
Yeah, if we went with the brute force solution with two nested for loops, we will get a graph as below for the array if we used the data below
but if we used the second method which is adding at the first index and subtracting at the index afer the last we get this graph
and then we can get the maximum out of summing the results as below
but I can't understand the logic of least step ( summing step )
the best way to understand it is form a simple example.
say there are 4 of us in a line: 1. me 2. you 3. bob 4. sue and we all start out with zero points.
Thcan be represented as (where my points are in the first index, your in the second, bob's in the third, sue's in fourth):
0 0 0 0
Furthermore, we go through rounds, where in each round a contiguous block of us can receive some set amount of points.
So in round one, say 2 points are awarded to anything in the range of start index = 1, and end index = 2. This means that you and bob, who are in the range, get 2 points.
But rather than write the current score as: 0 2 2 0
We instead want to write the score as:
0 2 0 -2
Because we want each value to represent how much greater/less than it is from the previous-index value. Doing this allows us to only ever need to change two elements in the list for a given round.
Now say we play one more round, and 1 point is awarded to all people in range of index = 0 to index = 1. This gives you
1 2 -1 -2
All I did was add the point value to the start index, and subtract it from the "end index + 1".
Then calculating the max is simply a matter of going through that list, adding each element to an accumulator variable (as this summing process reveals the actual value of an element at any given point - e.g., you would have 3 points at the end because your score is the result of 1 + 2), and having a variable which keeps track of the running max.
Thank you for your detailed explanation. It helped!
This is the best explanation of the difference array I have read in these discussions -- thank you for making it click in my head.
Hi,
Most of the peope solved this problem but time complexity of solution is O(n*m) (due to two nested for loops)which can not be used to solve this problem for given time constraint, so you need better approach which beats O(n*m)
I have created a video tutorial for you and uploaded the same on youtube with complete explanation along with code complexity analysis.
Here is the video tutorial for my solution O(n+m) complexity passed all test cases.
https://youtu.be/hDhf04AJIRs
Would really appreciate your feedback like, dislike , comment etc. on my video.
I would not suggest eclipsing
list
Can some one please help as to why my solution gives a segmentation fault?
include
include
include
include
int main() {
}
Maximum array size has to be 10^7 which is not possible in case of C. I tried Dynamic memory allocation (malloc) which worked but got TLE for bigger test cases
yeah it is getting a tle. We need to use a different algorithm. I wanted to know what the problem in my code was so i posted my solution.
use long instead of int.
Hi Vineet could u explain what logic have you followed?
Hi,
Most of the peope solved this problem but time complexity of solution is O(n*m) (due to two nested for loops)which can not be used to solve this problem for given time constraint, so you need better approach which beats O(n*m)
I have created a video tutorial for you and uploaded the same on youtube with complete explanation along with code complexity analysis.
Here is the video tutorial for my solution O(n+m) complexity passed all test cases.
https://youtu.be/hDhf04AJIRs
Would really appreciate your feedback like, dislike , comment etc. on my video.
Same solution but optimized:
That's so good! You are awesome!
can you explain why it's (n+1) and not (n)?
The question asks for a 1 indexed list and the a,b values read in are read in starting from 1 not 0. If you do not use (n+1)if b happens to be the last number of your list it will go out of bounds.
The 0 index value will always be 0 so it doesn't affect your answer when you sum for max difference.
why are you subtracting dx[b] -= c ?????
Walk through the array it may help.
Query 1 -> [1, 2, 100] Array 1 -> [0, 100, 0, -100, 0, 0, 0] Query 2 -> [2, 5, 100] Array 2 -> [0, 100, 100, -100, 0, 0, -100] Query 3 -> [3, 4, 100] Array 3 -> [0, 100, 100, 0, 0, -100, -100] Array Accumulated -> [0, 100, 200, 200, 200, 100, 0] Result -> 200
The use of accumulate is so brilliant!
Nice
How does that bottom for list return the maximum value in the list? Could you explain the logic to me please?
Why are you checking if "y" is <= len(list) ? This is given by the problem, right?
Really liked this solution! One question, shouldn't it be max=-float('inf')?
P.S - Talking about the python version
Just improved your code a bit. - split() is slightly faster than split("") - an if was not needed as always true - variables names fit those used in the exercise
I took the ideas explained here to put also into python in a bit more spelled-out way to help make sense of this more abstract work-around.
The key concepts in my opinion here are:
1) we don't need to build the full aray, since it's impossible for any row but the last to have the max value. This is impossible because we apply the adjustments to every subsequent row of the resulting 2-D array.
2) we don't need to actually increment each value for indices 'a' through 'b'. While that's the most straight-forward way, that also requires x many (a minus b) steps for each pass of the loop. By noting instead where to start incrementing and where to stop incrementing (noted by decrementing after what would be the final increment), we can note the adjustments to apply without having to take every step each time. We can then run a separate single loop to go over each of the increments and keep a running sum of all the overlapping increments. The decrement allows us to know when that range where the increment applied is over by reducing the running sum by that number - in other words when that range is ended, we would have to look at overlapping increments excluding that one that just ended going forward to see if anything is bigger.
Someone else in here gave an image of a stair/hill which I found extremely helpful in visualizing and understanding this concept. The basic idea here is that instead of actually applying each increment, we can just note what the increment and range is and one by one go through each place and apply all the compounded increments at once. Loop 1 saves the increments in a different format, Loop 2 checks the overlap. And by using two separate loops we have basically Big O (N) rather than Big O (N^2) - or more specifically Big O (2N + M) instead of Big O (NM + N), where N is the size of the resulting array and M is the size of the queries array.
def arrayManipulation(n, queries): arr=[0]*n for _ in queries: start=[0]-1 end=[1]-1 val=_[2] topVal=end+1 for i in range(start,topVal): arr[i]+=val return max(arr)
m = input().split()
n = int(nm[0])
m = int(nm[1])
queries = []
for _ in range(m): queries.append(list(map(int, input().rstrip().split())))
result=arrayManipulation(n, queries)
not sure why, but when I run your code as it is I get an error: if((y)<=len(list)): list[y] -= incr; IndexError: list index out of range
but when I do run that line like this:
if((y)<=len(list)): list[y-1] -= incr;
that works
A bit less efficient, but more readable
Python 3 more clear code
can you explain your logic ??
Hi,
Most of the peope solved this problem but time complexity of solution is O(n*m) (due to two nested for loops)which can not be used to solve this problem for given time constraint, so you need better approach which beats O(n*m)
I have created a video tutorial for you and uploaded the same on youtube with complete explanation along with code complexity analysis.
Here is the video tutorial for my solution O(n+m) complexity passed all test cases.
https://youtu.be/hDhf04AJIRs
Would really appreciate your feedback like, dislike , comment etc. on my video.
Thank you. This is clear.
You could also find the max using:
max(accumulate(my_array))
See below:
However, I have not understood the theory behind the difference array algorithm, why it works the way it works and how to derive it.
can anyone debug this program
Brilliant man, mine was working for initial test cases but getting timeout for rest.
Hi,
Most of the peope solved this problem but time complexity of solution is O(n*m) (due to two nested for loops)which can not be used to solve this problem for given time constraint, so you need better approach which beats O(n*m)
I have created a video tutorial for you and uploaded the same on youtube with complete explanation along with code complexity analysis.
Here is the video tutorial for my solution O(n+m) complexity passed all test cases.
https://youtu.be/hDhf04AJIRs
Would really appreciate your feedback like, dislike , comment etc. on my video.
Thanks , max(list) wil be simpler
can the same be written like this: def arrayManipulation(n, queries): arr_n=[0]*n for i in range(len(queries)): n1=queries[i][0] n2=queries[i][1] for j in range(n1-1,n2): arr_n[j]=arr_n[j]+queries[i][2] return max(arr_n) This is giving me a tiemout error while submitting. can u assist here?
yeah,at the beginning, l got the same problem as well. This algorithm's complexity could be o(n). Try to learn something about prefix sum array.I hope this can help.
But when y=len(list), list(y) will throw an error right? out of index error?
Another python translation:
Brilliant!
can u explain why we are subracting i dont understand the if statement ......
Hi,
I have uploaded a video tutorial which can help you to understand the problem as well as logic.
Here is the video tutorial for my solution O(n+m) complexity passed all test cases.
https://youtu.be/hDhf04AJIRs
Would really appreciate your feedback like, dislike , comment etc. on my video.
This breaks when
y == n
. The if when decrementinglist[y]
should beif y < len(list): list[y] -= incr
. Overall great solution! Thanks for doing it in python!Excellent, but why you are using this step.. list[y] -= incr; and that "if" condition is always correct, right..! Can you explain please...
for some test cases its showing teminated due to timeout what can i do
these solutions give runtime error
yup for 7 testcase its showing time out what should I do??
hi,
I am sure this comment will definately help you to solve your problem.
https://www.hackerrank.com/challenges/crush/forum/comments/570284
okay its great but why you using if((y)<=len(list)):
Hi, I have an even simpler solution. But, I am still getting a run time error. Can anyone please help me with this. Here is my code in Python 3:
def arrayManipulation(arr, query): a, b, k = query[0], query[1], query[2] arr[a - 1 : b] = [arr[i] + k for i in range(a - 1, b)] return arr
tq
The condition check
I replaced this part of your code :
max = x = 0 for i in list: x=x+i; if(max
by this:
max=list1[0] for j in range(1,n): list1[j]=list1[j]+list1[j-1] if max
here list1 is prefix sum array why did my code failed few test cases? here's my entire code : def arrayManipulation(n, queries): list1=[0]*(n+1)
I still have some doubt if anyone could explain it. I understand that it is making use of difference array to keep track of the difference between items rather than updating every item.
However as I understand, size of difference array is one less than total items in the actual array.
So as per the demo program given,
However the logic used seems to be a variant of difference array that I am not able to understand. It would be great if someone could help me connect the dots from this difference array to the actual working solution of this problem. Thanks.
Hello, I don't really get your DIFF representation, but here is the deal:
you need to keep in mind that at the end of queries to get the real value of i we need to sum the values of A[0]->A[i].
for a query (l, r, x) if we add x to A[l] then we're sure later for every k in [l, r] sums(0->k) will include x.
the problem is that every k > r will also include that x, therefore we'll decrement l[r+1] by x.
Now the sum through 0 -> k will include x only if k >= l and will exclude it only if k > r which is equivalent to:
sum(0->k) will consider x only if k is in [l, r]
I think the DIFF matrix should be:
then x is 100+100+100-100-100 = 200
still don't quite get it though
I think the Diff should like this
so finally, 1st col is always the real value, and others are accumlated of previous values and it self. so actual values are 100 200 200 200 100, max is 200
Thanks this helps me to understand how it works.
@julie_larson
100+100+100-100-100 = 100 still not 200 I've been banging my head still not able to understand this solution.
The final diff array should be like this:
100 100 0 0 -100 (straight foward from the code)
So when you iterate over it to get the accumalated sum you'll get this:
The maximum value of the accumulated sum is 200.
They're always checking the sum against the count, so when the sum is less than the count (which is the greatest sum so far) the count doesn't update.
As i have understood (5 , 3) 0 0 0 0 0 (1 , 2) 100 100 0 0 0 -> first and second places i.e a=1,b=2 , k=100, now a+k = 0+100 = 100, b+K = 0+100 = 100. (2 , 5) now from 2 to 5 i.e 2,3,4,5 are added with K i.e 100. already 2=100 now 2 becomes 100+100 =200, simialrly 3 = 0+100=100,4=0+100=100,5=0+100=100. (3 , 4) now 3 and 4 are added with 100. ie 3 value is 100 it becomes 100+100=200, 4 becomes 100+100=200.
among these 200 is max value.
instead of writing long int *a=long int[n+1] if I write long int a[n+1]; It shows segmentation fault after test case 7. why? Explain please.
yes ,the same with you, i am confused
you aren't allocating space for a that way. He uses new to create an array on the heap. You are trying to declare it on the stack, but you can't do that with a dynamic value like 'n'
As per my knowledge, u can create an array to size upto 10^6 in stack and for size above it we need to create in heap so we do it dynamically. https://discuss.codechef.com/questions/28604/maximum-size-of-an-array refer this
Thanks @ami3443
for me the best explanation for the dynamic allocation was:-here!
Nice explanation.
If you have a problem understanding dynamic memory then make use of vector . It allocates memory dynamically wherein the user doesn't have to deal with pointers .
vector arr(n); declares an array of long int of length n.