# 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: 4
{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: 7

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

Approach: Let’s define a sequence Sn such that Si = A1 + A2 + ··· + 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). The total time complexity is about O(NlogK).

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); ` ` `  `    ``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 `

Output:

```4
```

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.