Longest Span with same Sum in two Binary arrays

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

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 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)); 
    } 
} 

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
  
        # Conider 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; 
  
    // 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 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; 
  
    // Conider 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 
?> 

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.

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
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
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 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; 
} 

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 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 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]; 
  
        // 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) 

Output:

Length of the longest common span with same sum is 6

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

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 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 << 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 

Output:

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes

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

C++

Java

Python3

C#

PHP

C++

Java

Python

C#

C++

Java

Recommended Posts: