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Maximum difference of prefix sum for all indices of given two Arrays

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  • Last Updated : 13 Jan, 2022

Given 2 arrays of integers a[] and s[] both of size N. The task is to find the maximum difference of prefix sum for all indices of the given arrays.

Examples:

Input: N = 5, a[] = {20, 20, 35, 20, 35}, s[] = {21, 31, 34, 41, 14}
Output: 32
Explanation: After prefix sum the arrays are a[] = {20, 40, 75, 95, 130} 
and b[] = {21, 52, 86, 127, 141} for S. The maximum difference is (127 – 95) = 32 for 4th position.

Input: N = 1, a[] = {32}, s[] = {15}
Output: A, 17
Explanation: The highest difference (since only one element) is 32 – 15 = 17.

 

Approach: This problem can be solved by calculating the prefix sum in both the arrays and then comparing the differences for every index. Follow the steps below to solve the problem:

  • Iterate over the range [0, n) using the variable i and calculate the prefix sum for both the arrays.
  • Calculate the difference of prefix sum for every index.
  • Return the maximum difference.

Below is the implementation of the above approach.

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to get the winner
int getWinner(int a[], int s[], int N)
{
    int maxDiff = abs(a[0] - s[0]);
 
    // Calculating prefix sum
    for (int i = 1; i < N; i++) {
        a[i] += a[i - 1];
        s[i] += s[i - 1];
        maxDiff = max(maxDiff,
                      abs(a[i] - s[i]));
    }
 
    // Return the result
    return maxDiff;
}
 
// Driver Code
int main()
{
    int N = 5;
 
    int a[] = { 20, 20, 35, 20, 35 },
        s[] = { 21, 31, 34, 41, 14 };
 
    cout << getWinner(a, s, N);
    return 0;
}


Java




// Java program for the above approach
class GFG{
 
// Function to get the winner
static int getWinner(int a[], int s[], int N)
{
    int maxDiff = Math.abs(a[0] - s[0]);
 
    // Calculating prefix sum
    for (int i = 1; i < N; i++) {
        a[i] += a[i - 1];
        s[i] += s[i - 1];
        maxDiff = Math.max(maxDiff,
                      Math.abs(a[i] - s[i]));
    }
 
    // Return the result
    return maxDiff;
}
 
// Driver Code
public static void main(String[] args)
{
    int N = 5;
 
    int a[] = { 20, 20, 35, 20, 35 },
        s[] = { 21, 31, 34, 41, 14 };
 
    System.out.print(getWinner(a, s, N));
}
}
 
// This code is contributed by Rajput-Ji


Python3




# python3 program for the above approach
 
# Function to get the winner
 
 
def getWinner(a, s, N):
 
    maxDiff = abs(a[0] - s[0])
 
    # Calculating prefix sum
    for i in range(1, N):
        a[i] += a[i - 1]
        s[i] += s[i - 1]
        maxDiff = max(maxDiff, abs(a[i] - s[i]))
 
    # Return the result
    return maxDiff
 
 
# Driver Code
if __name__ == "__main__":
 
    N = 5
 
    a = [20, 20, 35, 20, 35]
    s = [21, 31, 34, 41, 14]
 
    print(getWinner(a, s, N))
 
# This code is contributed by rakeshsahni


C#




// C# program for the above approach
using System;
class GFG {
 
  // Function to get the winner
  static int getWinner(int[] a, int[] s, int N)
  {
    int maxDiff = Math.Abs(a[0] - s[0]);
 
    // Calculating prefix sum
    for (int i = 1; i < N; i++) {
      a[i] += a[i - 1];
      s[i] += s[i - 1];
      maxDiff
        = Math.Max(maxDiff, Math.Abs(a[i] - s[i]));
    }
 
    // Return the result
    return maxDiff;
  }
 
  // Driver Code
  public static void Main(string[] args)
  {
    int N = 5;
 
    int[] a = { 20, 20, 35, 20, 35 };
    int[] s = { 21, 31, 34, 41, 14 };
 
    Console.WriteLine(getWinner(a, s, N));
  }
}
 
// This code is contributed by ukasp.


Javascript




<script>
      // JavaScript code for the above approach
      // Function to get the winner
      function getWinner(a, s, N) {
          let maxDiff = Math.abs(a[0] - s[0]);
 
          // Calculating prefix sum
          for (let i = 1; i < N; i++) {
              a[i] += a[i - 1];
              s[i] += s[i - 1];
              maxDiff = Math.max(maxDiff,
                  Math.abs(a[i] - s[i]));
          }
 
          // Return the result
          return maxDiff;
      }
 
      // Driver Code
 
      let N = 5;
 
      let a = [20, 20, 35, 20, 35],
          s = [21, 31, 34, 41, 14];
 
      document.write(getWinner(a, s, N));
 
// This code is contributed by Potta Lokesh
  </script>


 
 

Output

32

Time Complexity: O(N)
Auxiliary Space: O(1) 


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