# Count of elements such that difference between sum of left and right sub arrays is equal to a multiple of k

Given an array arr[] of length n and an integer k, the task is to find the number of indices from 2 to n-1 in the array having a difference of the sum of left and right sub array equal to the multiple of given number k.

Examples:

Input: arr[] = {1, 2, 3, 3, 1, 1}, k = 4
Output: 2
Explanation: The only possible indices are 4 and 5

Input: arr[] = {1, 2, 3, 4, 5}, k = 1
Output: 3

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

• Create a prefix array which contains the sum of the elements in the left and the suffix array which contains the sum of elements in the right.
• Check for every index the difference of sum in the left and right and increase the counter if it is divisible by k

Below is the implementation of the above approach:

## CPP

 // C++ code to count of elements such that  // difference between the sum of left and right  // sub-arrays are equal to a multiple of k    #include using namespace std;    // Functions to find the no of elements int noOfElement(int a[], int n, int k) {     // Creating a prefix array     int prefix[n];        // Starting element of prefix array     // will be the first element     // of given array     prefix[0] = a[0];     for (int i = 1; i < n; i++) {         prefix[i] = prefix[i - 1] + a[i];     }        // Creating a suffix array;     int suffix[n];     // Last element of suffix array will     // be the last element of given array     suffix[n - 1] = a[n - 1];     for (int i = n - 2; i >= 0; i--) {         suffix[i] = suffix[i + 1] + a[i];     }        // Checking difference of left and right half     // is divisible by k or not.     int cnt = 0;     for (int i = 1; i < n - 1; i++) {         if ((prefix[i] - suffix[i]) % k == 0             || (suffix[i] - prefix[i]) % k == 0) {             cnt = cnt + 1;         }     }        return cnt; }    // Driver code int main() {     int a[] = { 1, 2, 3, 3, 1, 1 };     int k = 4;     int n = sizeof(a) / sizeof(a[0]);     cout << noOfElement(a, n, k);     return 0; }

## Java

 // Java code to count of elements such that  // difference between the sum of left and right  // sub-arrays are equal to a multiple of k class GFG {    // Functions to find the no of elements static int noOfElement(int a[], int n, int k) {     // Creating a prefix array     int []prefix = new int[n];        // Starting element of prefix array     // will be the first element     // of given array     prefix[0] = a[0];     for (int i = 1; i < n; i++)     {         prefix[i] = prefix[i - 1] + a[i];     }        // Creating a suffix array;     int []suffix = new int[n];            // Last element of suffix array will     // be the last element of given array     suffix[n - 1] = a[n - 1];     for (int i = n - 2; i >= 0; i--)     {         suffix[i] = suffix[i + 1] + a[i];     }        // Checking difference of left and right half     // is divisible by k or not.     int cnt = 0;     for (int i = 1; i < n - 1; i++)      {         if ((prefix[i] - suffix[i]) % k == 0             || (suffix[i] - prefix[i]) % k == 0)          {             cnt = cnt + 1;         }     }     return cnt; }    // Driver code public static void main(String[] args) {     int a[] = { 1, 2, 3, 3, 1, 1 };     int k = 4;     int n = a.length;     System.out.print(noOfElement(a, n, k)); } }    // This code is contributed by Rajput-Ji

## Python

 # Python3 code to count of elements such that # difference between the sum of left and right # sub-arrays are equal to a multiple of k    # Functions to find the no of elements def noOfElement(a, n, k):            # Creating a prefix array     prefix = [0] * n        # Starting element of prefix array     # will be the first element     # of given array     prefix[0] = a[0]     for i in range(1, n):         prefix[i] = prefix[i - 1] + a[i]        # Creating a suffix array     suffix = [0] * n            # Last element of suffix array will     # be the last element of given array     suffix[n - 1] = a[n - 1]     for i in range(n - 2, -1, -1):         suffix[i] = suffix[i + 1] + a[i]        # Checking difference of left and right half     # is divisible by k or not.     cnt = 0     for i in range(1, n - 1):         if ((prefix[i] - suffix[i]) % k == 0 or (suffix[i] - prefix[i]) % k == 0):             cnt = cnt + 1        return cnt    # Driver code    a = [ 1, 2, 3, 3, 1, 1 ] k = 4 n = len(a) print(noOfElement(a, n, k))    # This code is contributed by mohit kumar 29

## C#

 // C# code to count of elements such that  // difference between the sum of left and right  // sub-arrays are equal to a multiple of k using System;    class GFG {        // Functions to find the no of elements     static int noOfElement(int []a, int n, int k)     {         // Creating a prefix array         int []prefix = new int[n];                // Starting element of prefix array         // will be the first element         // of given array         prefix[0] = a[0];         for (int i = 1; i < n; i++)         {             prefix[i] = prefix[i - 1] + a[i];         }                // Creating a suffix array;         int []suffix = new int[n];                    // Last element of suffix array will         // be the last element of given array         suffix[n - 1] = a[n - 1];         for (int i = n - 2; i >= 0; i--)         {             suffix[i] = suffix[i + 1] + a[i];         }                // Checking difference of left and right half         // is divisible by k or not.         int cnt = 0;         for (int i = 1; i < n - 1; i++)          {             if ((prefix[i] - suffix[i]) % k == 0                 || (suffix[i] - prefix[i]) % k == 0)              {                 cnt = cnt + 1;             }         }         return cnt;     }            // Driver code     public static void Main()     {         int []a = { 1, 2, 3, 3, 1, 1 };         int k = 4;         int n = a.Length;         Console.Write(noOfElement(a, n, k));     } }    // This code is contributed by AnkitRai01

Output:

2

My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Article Tags :
Practice Tags :

Be the First to upvote.

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.