Thank u .

exactly my point also is the same. no use of rotation further when orginal array and rotated array are same. Could you tell me whether logic will work ?

static int[] circularArrayRotation(int[] a, int k, int[] queries) { int[] rotationArray = new int[a.Length]; int[] result = new int[queries.Length]; int rotationStartIndex = (a.Length - k); int rotationStopIndex = (a.Length - k) - 1; int index = 0; while (rotationStartIndex <= a.Length -1) { rotationArray[index] = a[rotationStartIndex]; index++; rotationStartIndex++; } rotationStartIndex = index; index = 0; while (index <= rotationStopIndex) { rotationArray[rotationStartIndex] = a[index]; index++; rotationStartIndex++; }

        for (int i = 0; i < queries.Length; i++)
        {
            if (queries[i] < result.Length)
            {
                result[i] = rotationArray[queries[i]];
            }
        }
        return result;

    }

Hi, dk201966. Probably you are facing problems because the order you are doing the changes. If you copy the value from the 1st position to the 2nd one and, after this, you copy the 2nd to the 3rd, at the end you will have all the positions with the 1st value. And don't forget saving the last value before overwriting it! PS: you'll probably face other issues with this challenge...

Dude please fix your "include"s they seem ugly :)

The main problem in the code is the order of the for loop used for rotating the elenets in the array. You should store the last element value in some temporary variable..and then use a decreasing counter in the mentioned for loop ..then u can store the value in temporary variable to the first element The error in ur code of overwriting values is eleminated by this code!👍

if(j+1!=n-1)

in your code there is assignment of values....shifting of values is absent..

hi Change your condition as a[j-1]=a[j]; and j=n-1;j>=0;j-- in for loop

your code is wrong due to elae statement

I did the same and realized the values of the elements of the array that you assign later has already replaced the other element in the array. For eg: a[1]=a[0] then a[2]=a[1]=a[0] which is equal to each other

#include<bits/stdc++.h>

using namespace std;


int main() {
   long long int k,n,q;
    cin>>n>>k>>q;
    k=k%n;
   long long int i,ar[n],x,s=0;
    for(i=0;i<n;i++)
        cin>>ar[i];
    for(i=1;i<=q;i++)
    {
          cin>>x;
     s=n-k+x;
     if(s<n)
         cout<<ar[s]<<endl;
     else if(s>=n)
     { s=s-n;
         cout<<ar[s]<<endl;
    }}

    return 0;
}

it will work

refer to my code

#include<bits/stdc++.h>

using namespace std;


int main() {
   long long int k,n,q;
    cin>>n>>k>>q;
    k=k%n;
   long long int i,ar[n],x,s=0;
    for(i=0;i<n;i++)
        cin>>ar[i];
    for(i=1;i<=q;i++)
    {
          cin>>x;
     s=n-k+x;
     if(s<n)
         cout<<ar[s]<<endl;
     else if(s>=n)
     { s=s-n;
         cout<<ar[s]<<endl;
    }}

    return 0;
}

refer this

same to you. but if you dry run the code ,then can you understand what is wrong with the code !!

First, Intialize starting position k= k%2. then,from this first postion (k) use a for loop till (n-1) to assign the values in the new array and finally use another for loop from 0 to (k-1).

the O(n) of this algorithm is n and it works for all test cases.

        k=k%n;
        j=0;
        for(i=0+k;i<n;i++)
        {
            newarr[i]=arr[j];
            j++;
        }
        for(i=0;i<k;i++)
        {
            newarr[i]=arr[j];
            j++;
        }

i actually tried running the code in eclipse, and it is matching the expected output, expect in the last case where it doesnot return the result before enter is pressed, as when the return key is pressed the output becomes fully correct.

here is my code----------------------

import java.io.; import java.util.;

public class Solution {

public static void main(String[] args) {
    Scanner s1=new Scanner(System.in);
    int n=s1.nextInt();
    int k=s1.nextInt();
    int q=s1.nextInt();
    int a[]=new int[n];
    for(int i=0;i<n;i++)
        {
        a[i]=s1.nextInt();
    }
    for(int i=0;i<q;i++)
        {
        int m=s1.nextInt();
        System.out.println(a[(n-k+m)%n]);
    }
}

}

I posted my code, if u could look at it in the comment section above that would be great!

include

using namespace std; int main() { int n,k,q,temp,i,p; cin>>n>>k>>q; int a[n]; for(i=0;i>a[i]; } while(k>0) { temp=a[n-1]; for(i=n-1;i>=0;i--) { a[i]=a[i-1]; } a[0]=temp; k--; for(i=0;i>p; cout<

return 0;

}

problem : Terminated due to time out??

Helpful!!

you figured out the time problem bro!!!! great

Thanks!!

I encountered the same problem. The test case 4 has k bigger than n, which causes segmentation error due to index out of bounds when you do n - k. Solved the problem by setting k = k % n when it is larger than n.

@hasan2020mainul can you explain the operation a(m-k)%len(a) you did?

or we can simply do like this by using python collections

import collections
def circularArrayRotation(a, k, queries):
    d = collections.deque(a)
    d.rotate(k)
    return (d[i] for i in queries)

hi @vasilij_kolomie1 your code doesn't work for test case 4

If K=2n then the loop will run twice and k will be set to 0. If k is a multiple of n then the array after k rotations will be same as initial array. k=k%n could also do the trick.

thank you sir.

how it worked please can u explain me??

hi @poon12, the solution is working fine, you must have done something wrong. Can you post your code here so that i can check.

can you give that link it is not opening.

found out what the problem was, new JS code:

function circularArrayRotation(a, k, queries) {

    var rarr = [];
    var aind = [];
    var res = [];

    for (var i = 0; i < a.length; i++){
        if ((i + k) < a.length) {
            aind[i] = i + k;
        } else {
            aind[i] = (i + k) % a.length;
        }
    }

    for (var j = 0; j < a.length; j++) {
        rarr[aind[j]] = a[j];
    }


    for (var k = 0; k < queries.length; k++){
        res[k] = rarr[queries[k]];
    }

    return (res);
}

function circularArrayRotation(a, k, queries) { // Write your code here return queries.map(value => a.reduce((target, item, index) => { let focus = (index + k) % a.length; target[focus] = item; return target; }, [])[value]); }

This is O(n) where n is the length of the queries array. It is independent of the length of the input array, since it grabs the input array elements by index and only has to loop through the queries to grab each one.

Again you misunderstood what i am saying. I am saying that the algorithmn part(i.e. the if and else statment) is O(1) not the whole code, since in this problem we cannot run away from array. And i am claming that my solution is the most optimized one.

And BTW the example that you gave has O(n)time complexity not O(1) since for ever new element you are doing a computational opperation(i.e. adding the sum and the new input)