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# 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``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 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 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

 ` ``\$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

 ``

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``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

## 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 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

 ``

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 ``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 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 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

 ``

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.)