# Longest Span with same Sum in two Binary arrays

Hello Everyone,

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

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.

`// 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;`

`}`

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 below observations.

1. Since there are total n elements, maximum sum is n for both arrays.
2. 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 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.
5. Update prefix sums: preSum1 += arr1[i], preSum2 += arr2[i]
6. Compute difference of current prefix sums: curr_diff = preSum1 – preSum2
7. Find index in diff array: diffIndex = n + curr_diff // curr_diff can be negative and can go till -n
8. If curr_diff is 0, then i+1 is maxLen so far
9. Else If curr_diff is seen first time, i.e., starting point of current diff is -1, then update starting point as i
10. 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
11. Return maxLen

Below is the implementation of above algorithm.

`// 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 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`

`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;`

`}`

Output:

Length of the longest common span with same sum is 6

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