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

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++ 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 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 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# 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.

  <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>

32

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