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Maximum length of subarray with same sum at corresponding indices from two Arrays

  • Last Updated : 25 Nov, 2021

Given two arrays A[] and B[] both consisting of N integers, the task is to find the maximum length of subarray [i, j] such that the sum of A[i… j] is equal to B[i… j].

Examples:

Input: A[] = {1, 1, 0, 1}, B[] = {0, 1, 1, 0}
Output: 3
Explanation: For (i, j) = (0, 2), sum of A[0… 2] = sum of B[0… 2] (i.e, A[0]+A[1]+A[2] = B[0]+B[1]+B[2] => 1+1+0 = 0+1+1 => 2 = 2). Similarly, for (i, j) = (1, 3), sum of A[1… 3] = B[1… 3]. Therefore, the length of the subarray with equal sum is 3 which is the maximum possible.

Input: A[] = {1, 2, 3, 4}, B[] = {4, 3, 2, 1}
Output: 4

Approach: The given problem can be solved by using a Greedy Approach with the help of unordered maps. It can be observed that for a pair (i, j), if the sum of A[i… j] = sum of B[i… j], then \sum_{x=i}^{j} (A[x] - B[x]) = 0         must hold true. Therefore, a prefix sum array of the difference (A[x] – B[x]) can be created. It can be observed that the repeated values in the prefix sum array represent that the sum of a subarray between the two repeated values must be 0. Hence, keep a track of the maximum size of such subarrays in a variable maxSize which will be the required answer.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include "bits/stdc++.h"
using namespace std;
 
// Function to find maximum length of subarray
// of array A and B having equal sum
int maxLength(vector<int>& A, vector<int>& B)
{
    int n = A.size();
 
    // Stores the maximum size of valid subarray
    int maxSize = 0;
 
    // Stores the prefix sum of the difference
    // of the given arrays
    unordered_map<int, int> pre;
    int diff = 0;
    pre[0] = 0;
 
    // Traverse the given array
    for (int i = 0; i < n; i++) {
 
        // Add the difference of the
        // corresponding array element
        diff += (A[i] - B[i]);
 
        // If current difference is not present
        if (pre.find(diff) == pre.end()) {
            pre = i + 1;
        }
 
        // If current difference is present,
        // update the value of maxSize
        else {
            maxSize = max(maxSize, i - pre + 1);
        }
    }
 
    // Return the maximum length
    return maxSize;
}
 
// Driver Code
int main()
{
    vector<int> A = { 1, 2, 3, 4 };
    vector<int> B = { 4, 3, 2, 1 };
 
    cout << maxLength(A, B);
 
    return 0;
}

Java




// Java program for the above approach
import java.util.HashMap;
class GFG {
 
    // Function to find maximum length of subarray
    // of array A and B having equal sum
    public static int maxLength(int[] A, int[] B) {
        int n = A.length;
 
        // Stores the maximum size of valid subarray
        int maxSize = 0;
 
        // Stores the prefix sum of the difference
        // of the given arrays
        HashMap<Integer, Integer> pre = new HashMap<Integer, Integer>();
        int diff = 0;
        pre.put(0, 0);
 
        // Traverse the given array
        for (int i = 0; i < n; i++) {
 
            // Add the difference of the
            // corresponding array element
            diff += (A[i] - B[i]);
 
            // If current difference is not present
            if (!pre.containsKey(diff)) {
                pre.put(diff, i + 1);
            }
 
            // If current difference is present,
            // update the value of maxSize
            else {
                maxSize = Math.max(maxSize, i - pre.get(diff));
            }
        }
 
        // Return the maximum length
        return maxSize;
    }
 
    // Driver Code
    public static void main(String args[]) {
        int[] A = { 1, 2, 3, 4 };
        int[] B = { 4, 3, 2, 1 };
 
        System.out.println(maxLength(A, B));
 
    }
}
 
// This code is contributed by gfgking.

Python3




# python program for the above approach
 
# Function to find maximum length of subarray
# of array A and B having equal sum
 
 
def maxLength(A,  B):
 
    n = len(A)
 
    # Stores the maximum size of valid subarray
    maxSize = 0
 
    # Stores the prefix sum of the difference
    # of the given arrays
    pre = {}
    diff = 0
    pre[0] = 0
 
    # Traverse the given array
    for i in range(0, n):
 
        # Add the difference of the
        # corresponding array element
        diff += (A[i] - B[i])
 
        # If current difference is not present
        if (not (diff in pre)):
            pre = i + 1
 
        # If current difference is present,
        # update the value of maxSize
        else:
            maxSize = max(maxSize, i - pre + 1)
 
    # Return the maximum length
    return maxSize
 
 
# Driver Code
if __name__ == "__main__":
 
    A = [1, 2, 3, 4]
    B = [4, 3, 2, 1]
 
    print(maxLength(A, B))
 
# This code is contributed by rakeshsahni

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
    // Function to find maximum length of subarray
    // of array A and B having equal sum
    public static int maxLength(int[] A, int[] B) {
        int n = A.Length;
 
        // Stores the maximum size of valid subarray
        int maxSize = 0;
 
        // Stores the prefix sum of the difference
        // of the given arrays
        Dictionary<int, int> pre =
                new Dictionary<int, int>();
                 
        int diff = 0;
        pre.Add(0, 0);
 
        // Traverse the given array
        for (int i = 0; i < n; i++) {
 
            // Add the difference of the
            // corresponding array element
            diff += (A[i] - B[i]);
 
            // If current difference is not present
            if (!pre.ContainsKey(diff)) {
                pre.Add(diff, i + 1);
            }
 
            // If current difference is present,
            // update the value of maxSize
            else {
                maxSize = Math.Max(maxSize, i - pre[(diff)]);
            }
        }
 
        // Return the maximum length
        return maxSize;
    }
 
// Driver Code
public static void Main()
{
    int[] A = { 1, 2, 3, 4 };
    int[] B = { 4, 3, 2, 1 };
 
    Console.Write(maxLength(A, B));
}
}
 
// This code is contributed by sanjoy_62.

Javascript




<script>
    // JavaScript program for the above approach
 
 
    // Function to find maximum length of subarray
    // of array A and B having equal sum
    const maxLength = (A, B) => {
        let n = A.length;
 
        // Stores the maximum size of valid subarray
        let maxSize = 0;
 
        // Stores the prefix sum of the difference
        // of the given arrays
        let pre = {};
        let diff = 0;
        pre[0] = 0;
 
        // Traverse the given array
        for (let i = 0; i < n; i++) {
 
            // Add the difference of the
            // corresponding array element
            diff += (A[i] - B[i]);
 
            // If current difference is not present
 
            if (!(diff in pre)) {
                pre = i + 1;
            }
 
            // If current difference is present,
            // update the value of maxSize
            else {
                maxSize = Math.max(maxSize, i - pre + 1);
            }
        }
 
        // Return the maximum length
        return maxSize;
    }
 
    // Driver Code
 
    let A = [1, 2, 3, 4];
    let B = [4, 3, 2, 1];
 
    document.write(maxLength(A, B));
 
// This code is contributed by rakeshsahni.
</script>

 
 

Output
3

 

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

 


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