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

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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 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);     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            # 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

 \$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.

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. Travers 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 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 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);     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 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)

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 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);     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

Output:

6

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