# 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 = a; ` `    ``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); ` `    ``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 = a; ` `        ``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
```

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