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  1. Prepare
  2. Data Structures
  3. Arrays
  4. Array Manipulation
  5. Discussions

Array Manipulation

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2463 Discussions

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  • jayantnishad34
     + 0 comments

    public static long arrayManipulation(int n, List> queries) { // Write your code here // Use an array to store changes at boundaries long[] arr = new long[n + 2];

        // Process each query using the difference array technique
    for (List<Integer> query : queries) {
        int start = query.get(0);
        int end = query.get(1);
        int value = query.get(2);

        arr[start] += value;  // Add value at start index
        arr[end + 1] -= value; // Subtract value after the end index
    }

    // Find the maximum value by applying prefix sum
    long maxValue = 0;
    long currentSum = 0;

    for (int i = 1; i <= n; i++) {
        currentSum += arr[i];
        maxValue = Math.max(maxValue, currentSum);
    }

    return maxValue;
}

  • coneryj
     + 2 comments

    You can look at the queries as histograms so you've got a bunch of histograms and you want to combine them. Basically you don't want overlapping queries.

    A way to do this is to consider each index in the queries as a level transition. Doesn't matter if it's at the query start index or the end index. You can just make tuples of...( transition_index, level_change) A starting query index has positive level_change. An ending query index has a neg. level_change. Just track the sum level as you go

  • adadasaivenkat01
     + 0 comments
    It passes 9 test cases out of 16 test cases and remaining 7 test cases it says time limit exceeded......What to do??

def arrayManipulation(n, q):
    # Write your code here
    temp=[]
    for i in range(0,n):
        temp.append(0)
    for j in range(len(q)):
        a,b,k=q[j][0],q[j][1],q[j][2]
        for z in range(a-1,b):
            temp[z]=temp[z]+k
    return max(temp)
		
	
		UPDATED USED PREFIX SUM AND DIFFERENCE ARRAY TO OPTIMIZE THE CODE
		
		def arrayManipulation(n, q):
    # Write your code here
    temp=[]
    for i in range(0,n):
        temp.append(0)
    for j in range(len(q)):
        a,b,k=q[j][0],q[j][1],q[j][2]
        temp[a-1]=temp[a-1]+k
        if b<n:   # if b<len(temp):
            temp[b]=temp[b]-k
    sum=0
    maxi=0
    for z in range(len(temp)):
        sum=sum+temp[z]
        maxi=max(sum,maxi)
    return maxi

  • g_hamu
     + 0 comments

    Think of keeping an array that save the delta of operation. Then add up the array to see which index is the peak (max). By this way, you can find the optimal solution

  • cristopheralexi1
     + 0 comments

    ?