# Count subarrays such that remainder after dividing sum of elements by K gives count of elements

Given an array arr[] of size N and an element K. The task is to find the number of sub-arrays of the given array such that the remainder when dividing the sum of its elements by K is equal to the number of elements in the subarray.

Examples:

Input: arr[] = {1, 4, 2, 3, 5}, K = 4
Output:
{1}, {1, 4, 2}, {4, 2} and {5}
are the only valid subarrays.
Input: arr[] = {4, 2, 4, 2, 4, 2, 4, 2}, K = 4
Output:

Approach: Letâ€™s define a sequence Sn such that Si = A1 + A2 + Â·Â·Â· + Ai and S0 = 0. Then, the condition that a contiguous subsequence Ai+1, …, Aj is valid can be represented as (Sj – Si) % K = j – i
This equation can then be transformed into the following equivalent conditions:
(Sj – j) % K = (Si – i) % K and j – i < K
Therefore, for each j(1 ? j ? N), count the number of j – K < i < j such that (Sj – j) % K = (Si – i) % K. For j the segment needed to be searched is (j – K, j), and for j + 1, it is (j – K + 1, j + 1), and these differ only by one element at the leftmost and rightmost, so in order to search for (j + 1)th after searching for jth element, only discard the leftmost element and add the rightmost element. Operations of discarding or adding can be performed quickly by managing the number of Si – i‘s by using associative arrays (such as map in C++ or dict in Python).

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach` `#include ` `using` `namespace` `std;`   `// Function to return the number of subarrays` `// of the given array such that the remainder` `// when dividing the sum of its elements` `// by K is equal to the number of its elements` `int` `sub_arrays(``int` `a[], ``int` `n, ``int` `k)` `{`   `    ``// To store prefix sum` `    ``int` `sum[n + 2] = { 0 };`   `    ``for` `(``int` `i = 0; i < n; i++) {`   `        ``// We are dealing with zero` `        ``// indexed array` `        ``a[i]--;`   `        ``// Taking modulus value` `        ``a[i] %= k;`   `        ``// Prefix sum` `        ``sum[i + 1] += sum[i] + a[i];` `        ``sum[i + 1] %= k;` `    ``}`   `    ``// To store the required answer, the left` `    ``// index and the right index` `    ``int` `ans = 0, l = 0, r = 0;`   `    ``// To store si - i value` `    ``map<``int``, ``int``> mp;`   `    ``for` `(``int` `i = 0; i < n + 1; i++) {`   `        ``// Include sum` `        ``ans += mp[sum[i]];` `        ``mp[sum[i]]++;`   `        ``// Increment the right index` `        ``r++;`   `        ``// If subarray has at least` `        ``// k elements` `        ``if` `(r - l >= k) {` `            ``mp[sum[l]]--;` `            ``l++;` `        ``}` `    ``}`   `    ``// Return the required answer` `    ``return` `ans;` `}`   `// Driver code` `int` `main()` `{` `    ``int` `a[] = { 1, 4, 2, 3, 5 };` `    ``int` `n = ``sizeof``(a) / ``sizeof``(a[0]);`   `    ``int` `k = 4;`   `    ``// Function call` `    ``cout << sub_arrays(a, n, k);`   `    ``return` `0;` `}`

## Java

 `// Java implementation of the approach ` `import` `java.util.*; `   `class` `gfg` `{` `    `  `    ``// Function to return the number of subarrays ` `    ``// of the given array such that the remainder ` `    ``// when dividing the sum of its elements ` `    ``// by K is equal to the number of its elements ` `    ``static` `int` `sub_arrays(``int` `[]a, ``int` `n, ``int` `k) ` `    ``{ ` `    `  `        ``// To store prefix sum ` `        ``int` `sum[] = ``new` `int``[n + ``2``] ; ` `        `  `        ``for` `(``int` `i = ``0``; i < n+``2``; i++)` `        ``{ ` `            ``sum[i] = ``0``;` `        ``}` `        `  `        ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``{ ` `    `  `            ``// We are dealing with zero ` `            ``// indexed array ` `            ``a[i]--; ` `    `  `            ``// Taking modulus value ` `            ``a[i] %= k; ` `    `  `            ``// Prefix sum ` `            ``sum[i + ``1``] += sum[i] + a[i]; ` `            ``sum[i + ``1``] %= k; ` `        ``} ` `    `  `        ``// To store the required answer, the left ` `        ``// index and the right index ` `        ``int` `ans = ``0``, l = ``0``, r = ``0``; ` `    `  `        ``// To store si - i value ` `        ``HashMap mp = ``new` `HashMap(); ` `    `  `        ``for` `(``int` `i = ``0``; i < n + ``1``; i++)` `        ``{ ` `            ``mp.put(sum[i], ``0``);` `        ``}` `        ``int` `temp;` `        `  `        ``for` `(``int` `i = ``0``; i < n + ``1``; i++) ` `        ``{ ` `    `  `            ``// Include sum ` `            ``ans += (``int``)mp.get(sum[i]); ` `            ``temp =(``int``)mp.get(sum[i]) + ``1``;` `            ``mp.put(sum[i], temp); ` `    `  `            ``// Increment the right index ` `            ``r++; ` `    `  `            ``// If subarray has at least ` `            ``// k elements ` `            ``if` `(r - l >= k)` `            ``{ ` `                ``//mp[sum[l]]--; ` `                ``temp = (``int``)mp.get(sum[l]) - ``1``;` `                ``mp.put(sum[l], temp);` `                ``l++; ` `            ``} ` `        ``} ` `    `  `        ``// Return the required answer ` `        ``return` `ans; ` `    ``} ` `    `  `    ``// Driver code ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``int` `[]a = { ``1``, ``4``, ``2``, ``3``, ``5` `}; ` `        `  `        ``int` `n = a.length; ` `    `  `        ``int` `k = ``4``; ` `    `  `        ``// Function call ` `        ``System.out.print(sub_arrays(a, n, k)); ` `    `  `    ``} ` `}`   `// This code is contributed by AnkitRai01`

## Python3

 `# Python3 implementation of the approach`   `# Function to return the number of ` `# subarrays of the given array ` `# such that the remainder when dividing ` `# the sum of its elements by K is ` `# equal to the number of its elements` `def` `sub_arrays(a, n, k):`   `    ``# To store prefix sum` `    ``sum` `=` `[``0` `for` `i ``in` `range``(n ``+` `2``)]`   `    ``for` `i ``in` `range``(n):`   `        ``# We are dealing with zero` `        ``# indexed array` `        ``a[i] ``-``=` `1`   `        ``# Taking modulus value` `        ``a[i] ``%``=` `k`   `        ``# Prefix sum` `        ``sum``[i ``+` `1``] ``+``=` `sum``[i] ``+` `a[i]` `        ``sum``[i ``+` `1``] ``%``=` `k`   `    ``# To store the required answer, ` `    ``# the left index and the right index` `    ``ans ``=` `0` `    ``l ``=` `0` `    ``r ``=` `0`   `    ``# To store si - i value` `    ``mp ``=` `dict``()`   `    ``for` `i ``in` `range``(n ``+` `1``):`   `        ``# Include sum` `        ``if` `sum``[i] ``in` `mp:` `            ``ans ``+``=` `mp[``sum``[i]]` `        ``mp[``sum``[i]] ``=` `mp.get(``sum``[i], ``0``) ``+` `1`   `        ``# Increment the right index` `        ``r ``+``=` `1`   `        ``# If subarray has at least` `        ``# k elements` `        ``if` `(r ``-` `l >``=` `k):` `            ``mp[``sum``[l]] ``-``=` `1` `            ``l ``+``=` `1`   `    ``# Return the required answer` `    ``return` `ans`   `# Driver code` `a ``=` `[``1``, ``4``, ``2``, ``3``, ``5``]` `n ``=` `len``(a)`   `k ``=` `4`   `# Function call` `print``(sub_arrays(a, n, k))`   `# This code is contributed by Mohit Kumar`

## C#

 `// C# implementation of the approach ` `using` `System;` `using` `System.Collections.Generic;`   `class` `gfg` `{` `    ``// Function to return the number of subarrays ` `    ``// of the given array such that the remainder ` `    ``// when dividing the sum of its elements ` `    ``// by K is equal to the number of its elements ` `    ``static` `int` `sub_arrays(``int` `[]a, ``int` `n, ``int` `k) ` `    ``{ ` `    `  `        ``// To store prefix sum ` `        ``int` `[]sum = ``new` `int``[n + 2] ; ` `        `  `        ``for` `(``int` `i = 0; i < n + 2; i++)` `        ``{ ` `            ``sum[i] = 0;` `        ``}` `        `  `        ``for` `(``int` `i = 0; i < n; i++) ` `        ``{ ` `    `  `            ``// We are dealing with zero ` `            ``// indexed array ` `            ``a[i]--; ` `    `  `            ``// Taking modulus value ` `            ``a[i] %= k; ` `    `  `            ``// Prefix sum ` `            ``sum[i + 1] += sum[i] + a[i]; ` `            ``sum[i + 1] %= k; ` `        ``} ` `    `  `        ``// To store the required answer, the left ` `        ``// index and the right index ` `        ``int` `ans = 0, l = 0, r = 0; ` `    `  `        ``// To store si - i value ` `        ``Dictionary<``int``, ``int``> mp = ``new` `Dictionary<``int``, ``int``>(); ` `    `  `        ``for` `(``int` `i = 0; i < n + 1; i++)` `        ``{ ` `            ``if``(!mp.ContainsKey(sum[i]))` `                ``mp.Add(sum[i], 0);` `        ``}` `        ``int` `temp;` `        `  `        ``for` `(``int` `i = 0; i < n + 1; i++) ` `        ``{ ` `    `  `            ``// Include sum ` `            ``ans += (``int``)mp[sum[i]]; ` `            ``temp =(``int``)mp[sum[i]] + 1;` `            ``mp[sum[i]] = temp; ` `    `  `            ``// Increment the right index ` `            ``r++; ` `    `  `            ``// If subarray has at least ` `            ``// k elements ` `            ``if` `(r - l >= k)` `            ``{ ` `                ``//mp[sum[l]]--; ` `                ``temp = (``int``)mp[sum[l]] - 1;` `                ``mp[sum[i]] = temp; ` `                ``l++; ` `            ``} ` `        ``} ` `    `  `        ``// Return the required answer ` `        ``return` `ans; ` `    ``} ` `    `  `    ``// Driver code ` `    ``public` `static` `void` `Main(String []args) ` `    ``{ ` `        ``int` `[]a = { 1, 4, 2, 3, 5 }; ` `        `  `        ``int` `n = a.Length; ` `    `  `        ``int` `k = 4; ` `    `  `        ``// Function call ` `        ``Console.Write(sub_arrays(a, n, k)); ` `    ``} ` `}`   `// This code is contributed by 29AjayKumar`

## Javascript

 ``

Output:

`4`

Time Complexity: O(N* log(N))
Auxiliary Space: O(N)

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