Longest Span with same Sum in two Binary arrays

Given two binary arrays, arr1[] and arr2[] of the same size n. Find the length of the longest common span (i, j) where j >= i such that arr1[i] + arr1[i+1] + …. + arr1[j] = arr2[i] + arr2[i+1] + …. + arr2[j].
The expected time complexity is Θ(n).

Examples:

Input: arr1[] = {0, 1, 0, 0, 0, 0};
       arr2[] = {1, 0, 1, 0, 0, 1};
Output: 4
The longest span with same sum is from index 1 to 4.

Input: arr1[] = {0, 1, 0, 1, 1, 1, 1};
       arr2[] = {1, 1, 1, 1, 1, 0, 1};
Output: 6
The longest span with same sum is from index 1 to 6.

Input: arr1[] = {0, 0, 0};
       arr2[] = {1, 1, 1};
Output: 0

Input: arr1[] = {0, 0, 1, 0};
       arr2[] = {1, 1, 1, 1};
Output: 1 

We strongly recommend that you click here and practice it, before moving on to the solution.

Method 1 (Simple Solution)
One by one by consider same subarrays of both arrays. For all subarrays, compute sums and if sums are same and current length is more than max length, then update max length. Below is C++ implementation of the simple approach.  

C++

// A Simple C++ program to find longest common
// subarray of two binary arrays with same sum
#include<bits/stdc++.h>
using namespace std;

// Returns length of the longest common subarray
// with same sum
int longestCommonSum(bool arr1[], bool arr2[], int n)
{
    // Initialize result
    int maxLen = 0;

    // One by one pick all possible starting points
    // of subarrays
    for (int i=0; i<n; i++)
    {
       // Initialize sums of current subarrays
       int sum1 = 0, sum2 = 0;

       // Consider all points for starting with arr[i]
       for (int j=i; j<n; j++)
       {
           // Update sums
           sum1 += arr1[j];
           sum2 += arr2[j];

           // If sums are same and current length is
           // more than maxLen, update maxLen
           if (sum1 == sum2)
           {
             int len = j-i+1;
             if (len > maxLen)
                maxLen = len;
           }
       }
    }
    return maxLen;
}

// Driver program to test above function
int main()
{
    bool  arr1[] = {0, 1, 0, 1, 1, 1, 1};
    bool  arr2[] = {1, 1, 1, 1, 1, 0, 1};
    int n = sizeof(arr1)/sizeof(arr1[0]);
    cout << "Length of the longest common span with same "
            "sum is "<< longestCommonSum(arr1, arr2, n);
    return 0;
}

Java

// A Simple Java program to find longest common
// subarray of two binary arrays with same sum

class Test
{
    static int arr1[] = new int[]{0, 1, 0, 1, 1, 1, 1};
    static int arr2[] = new int[]{1, 1, 1, 1, 1, 0, 1};

    // Returns length of the longest common sum in arr1[]
    // and arr2[]. Both are of same size n.
    static int longestCommonSum(int n)
    {
        // Initialize result
        int maxLen = 0;

        // One by one pick all possible starting points
        // of subarrays
        for (int i=0; i<n; i++)
        {
           // Initialize sums of current subarrays
           int sum1 = 0, sum2 = 0;

           // Consider all points for starting with arr[i]
           for (int j=i; j<n; j++)
           {
               // Update sums
               sum1 += arr1[j];
               sum2 += arr2[j];

               // If sums are same and current length is
               // more than maxLen, update maxLen
               if (sum1 == sum2)
               {
                 int len = j-i+1;
                 if (len > maxLen)
                    maxLen = len;
               }
           }
        }
        return maxLen;
    }

    // Driver method to test the above function
    public static void main(String[] args)
    {
        System.out.print("Length of the longest common span with same sum is ");
        System.out.println(longestCommonSum(arr1.length));
    }
}

Python3

# A Simple python program to find longest common
# subarray of two binary arrays with same sum

# Returns length of the longest common subarray
# with same sum
def longestCommonSum(arr1, arr2, n):

    # Initialize result
    maxLen = 0

    # One by one pick all possible starting points
    # of subarrays
    for i in range(0,n):

        # Initialize sums of current subarrays
        sum1 = 0
        sum2 = 0

        # Consider all points for starting with arr[i]
        for j in range(i,n):

            # Update sums
            sum1 += arr1[j]
            sum2 += arr2[j]

            # If sums are same and current length is
            # more than maxLen, update maxLen
            if (sum1 == sum2):
                len = j-i+1
                if (len > maxLen):
                    maxLen = len

    return maxLen


# Driver program to test above function
arr1 = [0, 1, 0, 1, 1, 1, 1]
arr2 = [1, 1, 1, 1, 1, 0, 1]
n = len(arr1)
print("Length of the longest common span with same "
            "sum is",longestCommonSum(arr1, arr2, n))

# This code is contributed by
# Smitha Dinesh Semwal

C#

// A Simple C# program to find
// longest common subarray of
// two binary arrays with same sum
using System;

class GFG
{
static int[] arr1 = new int[]{0, 1, 0, 1, 1, 1, 1};
static int[] arr2 = new int[]{1, 1, 1, 1, 1, 0, 1};

// Returns length of the longest
// common sum in arr1[] and arr2[].
// Both are of same size n.
static int longestCommonSum(int n)
{
    // Initialize result
    int maxLen = 0;

    // One by one pick all possible
    // starting points of subarrays
    for (int i = 0; i < n; i++)
    {
    // Initialize sums of current
    // subarrays
    int sum1 = 0, sum2 = 0;

    // Consider all points for
    // starting with arr[i]
    for (int j = i; j < n; j++)
    {
        // Update sums
        sum1 += arr1[j];
        sum2 += arr2[j];

        // If sums are same and current
        // length is more than maxLen,
        // update maxLen
        if (sum1 == sum2)
        {
            int len = j - i + 1;
            if (len > maxLen)
                maxLen = len;
        }
    }
    }
    return maxLen;
}

// Driver Code
public static void Main()
{
    Console.Write("Length of the longest " +
           "common span with same sum is ");
    Console.Write(longestCommonSum(arr1.Length));
}
}

// This code is contributed
// by ChitraNayal

PHP

<?php
// A Simple PHP program to find
// longest common subarray of
// two binary arrays with same sum

// Returns length of the longest
// common subarray with same sum
function longestCommonSum($arr1, $arr2, $n)
{
    // Initialize result
    $maxLen = 0;

    // One by one pick all possible
    // starting points of subarrays
    for ($i = 0; $i < $n; $i++)
    {
    // Initialize sums of
    // current subarrays
    $sum1 = 0; $sum2 = 0;

    // Consider all points
    // for starting with arr[i]
    for ($j = $i; $j < $n; $j++)
    {
        // Update sums
        $sum1 += $arr1[$j];
        $sum2 += $arr2[$j];

        // If sums are same and current
        // length is more than maxLen,
        // update maxLen
        if ($sum1 == $sum2)
        {
            $len = $j - $i + 1;
            if ($len > $maxLen)
                $maxLen = $len;
        }
    }
    }
    return $maxLen;
}

// Driver Code
$arr1 = array(0, 1, 0, 1, 1, 1, 1);
$arr2 = array (1, 1, 1, 1, 1, 0, 1);
$n = sizeof($arr1);
echo "Length of the longest common span ".
                  "with same ", "sum is ",
       longestCommonSum($arr1, $arr2, $n);

// This code is contributed by aj_36
?>

Javascript

<script>

    // A Simple Javascript program to find
    // longest common subarray of
    // two binary arrays with same sum

    let arr1 = [0, 1, 0, 1, 1, 1, 1];
    let arr2 = [1, 1, 1, 1, 1, 0, 1];

    // Returns length of the longest
    // common sum in arr1[] and arr2[].
    // Both are of same size n.
    function longestCommonSum(n)
    {
        // Initialize result
        let maxLen = 0;

        // One by one pick all possible
        // starting points of subarrays
        for (let i = 0; i < n; i++)
        {
          // Initialize sums of current
          // subarrays
          let sum1 = 0, sum2 = 0;

          // Consider all points for
          // starting with arr[i]
          for (let j = i; j < n; j++)
          {
              // Update sums
              sum1 += arr1[j];
              sum2 += arr2[j];

              // If sums are same and current
              // length is more than maxLen,
              // update maxLen
              if (sum1 == sum2)
              {
                  let len = j - i + 1;
                  if (len > maxLen)
                      maxLen = len;
              }
          }
        }
        return maxLen;
    }

    document.write("Length of the longest " + "common span with same sum is ");
    document.write(longestCommonSum(arr1.length));

</script>

Output :

Length of the longest common span with same sum is 6

Time Complexity : O(n²) 
Auxiliary Space : O(1)

Method 2 (Using Auxiliary Array)
The idea is based on the below observations. 

1. Since there are total n elements, maximum sum is n for both arrays.
2. The difference between two sums varies from -n to n. So there are total 2n + 1 possible values of difference.
3. If differences between prefix sums of two arrays become same at two points, then subarrays between these two points have same sum.

Below is the Complete Algorithm.  

1. Create an auxiliary array of size 2n+1 to store starting points of all possible values of differences (Note that possible values of differences vary from -n to n, i.e., there are total 2n+1 possible values)
2. Initialize starting points of all differences as -1.
3. Initialize maxLen as 0 and prefix sums of both arrays as 0, preSum1 = 0, preSum2 = 0
4. Traverse both arrays from i = 0 to n-1. 
   1. Update prefix sums: preSum1 += arr1[i], preSum2 += arr2[i]
   2. Compute difference of current prefix sums: curr_diff = preSum1 – preSum2
   3. Find index in diff array: diffIndex = n + curr_diff // curr_diff can be negative and can go till -n
   4. If curr_diff is 0, then i+1 is maxLen so far
   5. Else If curr_diff is seen first time, i.e., starting point of current diff is -1, then update starting point as i
   6. Else (curr_diff is NOT seen first time), then consider i as ending point and find length of current same sum span. If this length is more, then update maxLen
5. Return maxLen

Below is the implementation of above algorithm.  

C++

// A O(n) and O(n) extra space C++ program to find
// longest common subarray of two binary arrays with
// same sum
#include<bits/stdc++.h>
using namespace std;

// Returns length of the longest common sum in arr1[]
// and arr2[]. Both are of same size n.
int longestCommonSum(bool arr1[], bool arr2[], int n)
{
    // Initialize result
    int maxLen = 0;

    // Initialize prefix sums of two arrays
    int preSum1 = 0, preSum2 = 0;

    // Create an array to store starting and ending
    // indexes of all possible diff values. diff[i]
    // would store starting and ending points for
    // difference "i-n"
    int diff[2*n+1];

    // Initialize all starting and ending values as -1.
    memset(diff, -1, sizeof(diff));

    // Traverse both arrays
    for (int i=0; i<n; i++)
    {
        // Update prefix sums
        preSum1 += arr1[i];
        preSum2 += arr2[i];

        // Compute current diff and index to be used
        // in diff array. Note that diff can be negative
        // and can have minimum value as -1.
        int curr_diff = preSum1 - preSum2;
        int diffIndex = n + curr_diff;

        // If current diff is 0, then there are same number
        // of 1's and 0's so far in both arrays, i.e., (i+1) is
        // maximum length.
        if (curr_diff == 0)
            maxLen = i+1;

        // If current diff is seen first time, then update
        // starting index of diff.
        else if ( diff[diffIndex] == -1)
            diff[diffIndex] = i;

        // Current diff is already seen which means there lies curr_diff= 0 in between
        else
        {
            // Find length of this same sum by calculating the common span
            int len = i - diff[diffIndex];

            // Update max len if needed
                maxLen = max(maxLen,len);
        }
    }
    return maxLen;
}

// Driver code
int main()
{
    bool  arr1[] = {0, 1, 0, 1, 1, 1, 1};
    bool  arr2[] = {1, 1, 1, 1, 1, 0, 1};
    int n = sizeof(arr1)/sizeof(arr1[0]);
    cout << "Length of the longest common span with same "
            "sum is "<< longestCommonSum(arr1, arr2, n);
    return 0;
}

Java

// A O(n) and O(n) extra space Java program to find
// longest common subarray of two binary arrays with
// same sum

class Test
{
    static int arr1[] = new int[]{0, 1, 0, 1, 1, 1, 1};
    static int arr2[] = new int[]{1, 1, 1, 1, 1, 0, 1};

    // Returns length of the longest common sum in arr1[]
    // and arr2[]. Both are of same size n.
    static int longestCommonSum(int n)
    {
        // Initialize result
        int maxLen = 0;

        // Initialize prefix sums of two arrays
        int preSum1 = 0, preSum2 = 0;

        // Create an array to store starting and ending
        // indexes of all possible diff values. diff[i]
        // would store starting and ending points for
        // difference "i-n"
        int diff[] = new int[2*n+1];

        // Initialize all starting and ending values as -1.
        for (int i = 0; i < diff.length; i++) {
            diff[i] = -1;
        }

        // Traverse both arrays
        for (int i=0; i<n; i++)
        {
            // Update prefix sums
            preSum1 += arr1[i];
            preSum2 += arr2[i];

            // Compute current diff and index to be used
            // in diff array. Note that diff can be negative
            // and can have minimum value as -1.
            int curr_diff = preSum1 - preSum2;
            int diffIndex = n + curr_diff;

            // If current diff is 0, then there are same number
            // of 1's so far in both arrays, i.e., (i+1) is
            // maximum length.
            if (curr_diff == 0)
                maxLen = i+1;

            // If current diff is seen first time, then update
            // starting index of diff.
            else if ( diff[diffIndex] == -1)
                diff[diffIndex] = i;

            // Current diff is already seen
            else
            {
                // Find length of this same sum common span
                int len = i - diff[diffIndex];

                // Update max len if needed
                if (len > maxLen)
                    maxLen = len;
            }
        }
        return maxLen;
    }

    // Driver method to test the above function
    public static void main(String[] args)
    {
        System.out.print("Length of the longest common span with same sum is ");
        System.out.println(longestCommonSum(arr1.length));
    }
}

Python

# Python program to find longest common
# subarray of two binary arrays with
# same sum

def longestCommonSum(arr1, arr2, n):

    # Initialize result
    maxLen = 0

    # Initialize prefix sums of two arrays
    presum1 = presum2 = 0

    # Create a dictionary to store indices
    # of all possible sums
    diff = {}

    # Traverse both arrays
    for i in range(n):

        # Update prefix sums
        presum1 += arr1[i]
        presum2 += arr2[i]

        # Compute current diff which will be
        # used as index in diff dictionary
        curr_diff = presum1 - presum2

        # If current diff is 0, then there
        # are same number of 1's so far in
        # both arrays, i.e., (i+1) is
        # maximum length.
        if curr_diff == 0:
            maxLen = i+1
        elif curr_diff not in diff:
            # save the index for this diff
            diff[curr_diff] = i
        else:
            # calculate the span length
            length = i - diff[curr_diff]
            maxLen = max(maxLen, length)

    return maxLen

# Driver program
arr1 = [0, 1, 0, 1, 1, 1, 1]
arr2 = [1, 1, 1, 1, 1, 0, 1]
print("Length of the longest common",
    " span with same", end = " ")
print("sum is",longestCommonSum(arr1,
                   arr2, len(arr1)))

# This code is contributed by Abhijeet Nautiyal

C#

// A O(n) and O(n) extra space C# program
// to find longest common subarray of two
// binary arrays with same sum
using System;

class GFG
{
static int[] arr1 = new int[]{0, 1, 0, 1, 1, 1, 1};
static int[] arr2 = new int[]{1, 1, 1, 1, 1, 0, 1};

// Returns length of the longest
// common sum in arr1[] and arr2[].
// Both are of same size n.
static int longestCommonSum(int n)
{
    // Initialize result
    int maxLen = 0;

    // Initialize prefix sums of
    // two arrays
    int preSum1 = 0, preSum2 = 0;

    // Create an array to store starting
    // and ending indexes of all possible
    // diff values. diff[i] would store
    // starting and ending points for
    // difference "i-n"
    int[] diff = new int[2 * n + 1];

    // Initialize all starting and ending
    // values as -1.
    for (int i = 0; i < diff.Length; i++)
    {
        diff[i] = -1;
    }

    // Traverse both arrays
    for (int i = 0; i < n; i++)
    {
        // Update prefix sums
        preSum1 += arr1[i];
        preSum2 += arr2[i];

        // Compute current diff and index to
        // be used in diff array. Note that
        // diff can be negative and can have
        // minimum value as -1.
        int curr_diff = preSum1 - preSum2;
        int diffIndex = n + curr_diff;

        // If current diff is 0, then there
        // are same number of 1's so far in
        // both arrays, i.e., (i+1) is
        // maximum length.
        if (curr_diff == 0)
            maxLen = i + 1;

        // If current diff is seen first time,
        // then update starting index of diff.
        else if ( diff[diffIndex] == -1)
            diff[diffIndex] = i;

        // Current diff is already seen
        else
        {
            // Find length of this same
            // sum common span
            int len = i - diff[diffIndex];

            // Update max len if needed
            if (len > maxLen)
                maxLen = len;
        }
    }
    return maxLen;
}

// Driver Code
public static void Main()
{
    Console.Write("Length of the longest common " +
                         "span with same sum is ");
    Console.WriteLine(longestCommonSum(arr1.Length));
}
}

// This code is contributed
// by Akanksha Rai(Abby_akku)

Javascript

<script>

    // A O(n) and O(n) extra space
    // Javascript program to find longest
    // common subarray of two binary arrays
    // with same sum

    let arr1 = [0, 1, 0, 1, 1, 1, 1];
    let arr2 = [1, 1, 1, 1, 1, 0, 1];

    // Returns length of the longest
    // common sum in arr1[] and arr2[].
    // Both are of same size n.
    function longestCommonSum(n)
    {
        // Initialize result
        let maxLen = 0;

        // Initialize prefix sums of
        // two arrays
        let preSum1 = 0, preSum2 = 0;

        // Create an array to store starting
        // and ending indexes of all possible
        // diff values. diff[i] would store
        // starting and ending points for
        // difference "i-n"
        let diff = new Array(2 * n + 1);

        // Initialize all starting and ending
        // values as -1.
        for (let i = 0; i < diff.length; i++)
        {
            diff[i] = -1;
        }

        // Traverse both arrays
        for (let i = 0; i < n; i++)
        {
            // Update prefix sums
            preSum1 += arr1[i];
            preSum2 += arr2[i];

            // Compute current diff and index to
            // be used in diff array. Note that
            // diff can be negative and can have
            // minimum value as -1.
            let curr_diff = preSum1 - preSum2;
            let diffIndex = n + curr_diff;

            // If current diff is 0, then there
            // are same number of 1's so far in
            // both arrays, i.e., (i+1) is
            // maximum length.
            if (curr_diff == 0)
                maxLen = i + 1;

            // If current diff is seen first time,
            // then update starting index of diff.
            else if ( diff[diffIndex] == -1)
                diff[diffIndex] = i;

            // Current diff is already seen
            else
            {
                // Find length of this same
                // sum common span
                let len = i - diff[diffIndex];

                // Update max len if needed
                if (len > maxLen)
                    maxLen = len;
            }
        }
        return maxLen;
    }

    document.write("Length of the longest common "
    + "span with same sum is ");
    document.write(longestCommonSum(arr1.length));

</script>

Output:

Length of the longest common span with same sum is 6

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

Method 3 (Using Hashing)

1. Find difference array arr[] such that arr[i] = arr1[i] – arr2[i].
2. Largest subarray with equal number of 0s and 1s in the difference array.

C++

// C++ program to find largest subarray
// with equal number of 0's and 1's.
#include <bits/stdc++.h>
using namespace std;

// Returns largest common subarray with equal
// number of 0s and 1s in both of t
int longestCommonSum(bool arr1[], bool arr2[], int n)
{
    // Find difference between the two
    int arr[n];
    for (int i=0; i<n; i++)
      arr[i] = arr1[i] - arr2[i];

    // Creates an empty hashMap hM
    unordered_map<int, int> hM;

    int sum = 0;     // Initialize sum of elements
    int max_len = 0; // Initialize result

    // Traverse through the given array
    for (int i = 0; i < n; i++)
    {
        // Add current element to sum
        sum += arr[i];

        // To handle sum=0 at last index
        if (sum == 0)
            max_len = i + 1;

        // If this sum is seen before,
        // then update max_len if required
        if (hM.find(sum) != hM.end())
          max_len = max(max_len, i - hM[sum]);

        else // Else put this sum in hash table
            hM[sum] = i;
    }

    return max_len;
}

// Driver program to test above function
int main()
{
    bool  arr1[] = {0, 1, 0, 1, 1, 1, 1};
    bool  arr2[] = {1, 1, 1, 1, 1, 0, 1};
    int n = sizeof(arr1)/sizeof(arr1[0]);
    cout << longestCommonSum(arr1, arr2, n);
    return 0;
}

Java

// Java program to find largest subarray
// with equal number of 0's and 1's.
import java.io.*;
import java.util.*;

class GFG
{

    // Returns largest common subarray with equal
    // number of 0s and 1s
    static int longestCommonSum(int[] arr1, int[] arr2, int n)
    {
        // Find difference between the two
        int[] arr = new int[n];
        for (int i = 0; i < n; i++)
            arr[i] = arr1[i] - arr2[i];

        // Creates an empty hashMap hM
        HashMap<Integer, Integer> hM = new HashMap<>();

        int sum = 0;     // Initialize sum of elements
        int max_len = 0; // Initialize result

        // Traverse through the given array
        for (int i = 0; i < n; i++)
        {
            // Add current element to sum
            sum += arr[i];

            // To handle sum=0 at last index
            if (sum == 0)
                max_len = i + 1;

            // If this sum is seen before,
            // then update max_len if required
            if (hM.containsKey(sum))
                max_len = Math.max(max_len, i - hM.get(sum));

            else // Else put this sum in hash table
                hM.put(sum, i);
        }
        return max_len;
    }

    // Driver code
    public static void main(String args[])
    {
            int[] arr1 = {0, 1, 0, 1, 1, 1, 1};
            int[] arr2 = {1, 1, 1, 1, 1, 0, 1};
            int n = arr1.length;
            System.out.println(longestCommonSum(arr1, arr2, n));
    }
}

// This code is contributed by rachana soma

Python3

# Python program to find largest subarray
# with equal number of 0's and 1's.

# Returns largest common subarray with equal
# number of 0s and 1s
def longestCommonSum(arr1, arr2, n):

    # Find difference between the two
    arr = [0 for i in range(n)]

    for i in range(n):
        arr[i] = arr1[i] - arr2[i];

    # Creates an empty hashMap hM
    hm = {}
    sum = 0     # Initialize sum of elements
    max_len = 0     #Initialize result

    # Traverse through the given array
    for i in range(n):

        # Add current element to sum
        sum += arr[i]

        # To handle sum=0 at last index
        if (sum == 0):
            max_len = i + 1

        # If this sum is seen before,
        # then update max_len if required
        if sum in hm:
            max_len = max(max_len, i - hm[sum])
        else:   # Else put this sum in hash table
            hm[sum] = i
    return max_len

# Driver code
arr1 = [0, 1, 0, 1, 1, 1, 1]
arr2 = [1, 1, 1, 1, 1, 0, 1]
n = len(arr1)
print(longestCommonSum(arr1, arr2, n))

# This code is contributed by rag2127

C#

// C# program to find largest subarray
// with equal number of 0's and 1's.
using System;
using System.Collections.Generic;
public class GFG
{

  // Returns largest common subarray with equal
  // number of 0s and 1s
  static int longestCommonSum(int[] arr1, int[] arr2, int n)
  {

    // Find difference between the two
    int[] arr = new int[n];
    for (int i = 0; i < n; i++)
      arr[i] = arr1[i] - arr2[i];

    // Creates an empty hashMap hM
    Dictionary<int,int> hM = new Dictionary<int,int>();

    int sum = 0;     // Initialize sum of elements
    int max_len = 0; // Initialize result

    // Traverse through the given array
    for (int i = 0; i < n; i++)
    {

      // Add current element to sum
      sum += arr[i];

      // To handle sum=0 at last index
      if (sum == 0)
        max_len = i + 1;

      // If this sum is seen before,
      // then update max_len if required
      if (hM.ContainsKey(sum))
        max_len = Math.Max(max_len, i - hM[sum]);

      else // Else put this sum in hash table
        hM[sum] = i;
    }
    return max_len;
  }

  // Driver code
  static public void Main ()
  {
    int[] arr1 = {0, 1, 0, 1, 1, 1, 1};
    int[] arr2 = {1, 1, 1, 1, 1, 0, 1};
    int n = arr1.Length;
    Console.WriteLine(longestCommonSum(arr1, arr2, n));
  }
}

// This code is contributed by avanitrachhadiya2155

Javascript

<script>
// Javascript program to find largest subarray
// with equal number of 0's and 1's.

// Returns largest common subarray with equal
// number of 0s and 1s
function longestCommonSum(arr1,arr2,n)
{
    // Find difference between the two
        let arr = new Array(n);
        for (let i = 0; i < n; i++)
            arr[i] = arr1[i] - arr2[i];

        // Creates an empty hashMap hM
        let hM = new Map();

        let sum = 0;     // Initialize sum of elements
        let max_len = 0; // Initialize result

        // Traverse through the given array
        for (let i = 0; i < n; i++)
        {
            // Add current element to sum
            sum += arr[i];

            // To handle sum=0 at last index
            if (sum == 0)
                max_len = i + 1;

            // If this sum is seen before,
            // then update max_len if required
            if (hM.has(sum))
                max_len = Math.max(max_len, i - hM.get(sum));

            else // Else put this sum in hash table
                hM.set(sum, i);
        }
        return max_len;
}

// Driver code
let arr1=[0, 1, 0, 1, 1, 1, 1];
let arr2=[1, 1, 1, 1, 1, 0, 1];
let n = arr1.length;
document.write(longestCommonSum(arr1, arr2, n));


// This code is contributed by ab2127
</script>

Output: 

6

Time Complexity: O(n)  (As the array is traversed only once.)
Auxiliary Space: O(n) (As hashmap has been used which takes extra space.)

https://www.youtube.com/watch?v=xtfj4

-r_Ahs
This article is contributed by Sumit Gupta. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above