Given two binary arrays arr1[] and arr2[] of same size n. Find 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].

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

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**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 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; // Conider 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; // Conider 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)); } }

Output:

Length of the longest common span with same sum is 6

Time Complexity: O(n^{2})

Auxiliary Space: O(1)

**Method 2 (Using Auxiliary Array)**

The idea is based on below observations.

- Since there are total n elements, maximum sum is n for both arrays.
- 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 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**= 0 - Travers 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 – 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++

// 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 staring 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]; // Comput 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 lenght of this same sum common span int len = i - diff[diffIndex]; // Update max len if needed if (len > maxLen) maxLen = len; } } return maxLen; } // Driver progra+m 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 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 staring 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]; // Comput 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 lenght 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

Output:

Length of the longest common span with same sum is 6

Time Complexity: Θ(n)

Auxiliary Space: Θ(n)

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