Longest Span with same Sum in two Binary arrays

22 Jun, 2022
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 
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++
Java
Python3
C#
PHP
Javascript
// 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 sumint 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 functionint 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;}
// 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));    }}
# 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 sumdef 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 functionarr1 = [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
// A Simple C# program to find// longest common subarray of// two binary arrays with same sumusing 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 Codepublic 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// 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 sumfunction 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?>
<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)   
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Method 2 (Using Auxiliary Array) The idea is based on the below observations. 
Since there are total n elements, maximum sum is n for both arrays.The difference between two sums varies from -n to n. So there are total 2n + 1 possible values of difference.If differences between prefix sums of two arrays become same at two points, then subarrays between these two points have same sum.
Since there are total n elements, maximum sum is n for both arrays.
The difference between two sums varies from -n to n. So there are total 2n + 1 possible values of difference.
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.  
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)Initialize starting points of all differences as -1.Initialize maxLen as 0 and prefix sums of both arrays as 0, preSum1 = 0, preSum2 = 0Traverse both arrays from i = 0 to n-1. Update prefix sums: preSum1 += arr1[i], preSum2 += arr2[i]Compute difference of current prefix sums: curr_diff = preSum1 – preSum2Find index in diff array: diffIndex = n + curr_diff // curr_diff can be negative and can go till -nIf curr_diff is 0, then i+1 is maxLen so farElse If curr_diff is seen first time, i.e., starting point of current diff is -1, then update starting point as iElse (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 maxLenReturn maxLen
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)
Initialize starting points of all differences as -1.
Initialize maxLen as 0 and prefix sums of both arrays as 0, preSum1 = 0, preSum2 = 0
Traverse both arrays from i = 0 to n-1. Update prefix sums: preSum1 += arr1[i], preSum2 += arr2[i]Compute difference of current prefix sums: curr_diff = preSum1 – preSum2Find index in diff array: diffIndex = n + curr_diff // curr_diff can be negative and can go till -nIf curr_diff is 0, then i+1 is maxLen so farElse If curr_diff is seen first time, i.e., starting point of current diff is -1, then update starting point as iElse (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
Update prefix sums: preSum1 += arr1[i], preSum2 += arr2[i]Compute difference of current prefix sums: curr_diff = preSum1 – preSum2Find index in diff array: diffIndex = n + curr_diff // curr_diff can be negative and can go till -nIf curr_diff is 0, then i+1 is maxLen so farElse If curr_diff is seen first time, i.e., starting point of current diff is -1, then update starting point as iElse (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
Update prefix sums: preSum1 += arr1[i], preSum2 += arr2[i]
Compute difference of current prefix sums: curr_diff = preSum1 – preSum2
Find index in diff array: diffIndex = n + curr_diff // curr_diff can be negative and can go till -n
If curr_diff is 0, then i+1 is maxLen so far
Else If curr_diff is seen first time, i.e., starting point of current diff is -1, then update starting point as i
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
Return maxLen
Below is the implementation of above algorithm.  
C++
Java
Python
C#
Javascript
// 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 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 codeint 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;}
// 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 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
// A O(n) and O(n) extra space C# program// to find longest common subarray of two// binary arrays with same sumusing 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 Codepublic 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)
<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)
Find difference array arr[] such that arr[i] = arr1[i] – arr2[i].Largest subarray with equal number of 0s and 1s in the difference array.
Find difference array arr[] such that arr[i] = arr1[i] – arr2[i].
Largest subarray with equal number of 0s and 1s in the difference array.
C++
Java
Python3
C#
Javascript
// 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 tint 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 functionint 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 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
# 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 1sdef 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 codearr1 = [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# 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
<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 1sfunction 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 codelet 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.)
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-r_AhsThis 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 
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