You are given an array of positive and/or negative integers and a value K . The task is to find count of all sub-arrays whose sum is divisible by K?

**Examples :**

Input : arr[] = {4, 5, 0, -2, -3, 1}, K = 5 Output : 7 // there are 7 sub-arrays whose is divisible by K // {4, 5, 0, -2, -3, 1} // {5} // {5, 0} // {5, 0, -2, -3} // {0} // {0, -2, -3} // {-2, -3}

A **simple solution** for this problem is to one by one calculate sum of all sub-arrays possible and check divisible by K. The time complexity for this approach will be O(n^2).

An **efficient solution** is based on below observation.

Let there be a subarray (i, j) whose sum is divisible by k sum(i, j) = sum(0, j) - sum(0, i-1) Sum for any subarray can be written as q*k + rem where q is a quotient and rem is remainder Thus, sum(i, j) = (q1 * k + rem1) - (q2 * k + rem2) sum(i, j) = (q1 - q2)k + rem1-rem2 We see, for sum(i, j) i.e. for sum of any subarray to be divisible by k, the RHS should also be divisible by k. (q1 - q2)k is obviously divisible by k, for (rem1-rem2) to follow the same, rem1 = rem2 where rem1 = Sum of subarray (0, j) % k rem2 = Sum of subarray (0, i-1) % k

So if any sub-array sum from index i’th to j’th is divisible by k then we can say**a[0]+…a[i-1] (mod k) = a[0]+…+a[j] (mod k)**

The above explanation is provided by Ekta Goel.

So we need to find such a pair of indices (i, j) that they satisfy the above condition. Here is the algorithm :

- Make an auxiliary array of size k as
**Mod[k]**. This array holds the count of each remainder we are getting after dividing cumulative sum till any index in arr[]. - Now start calculating cumulative sum and simultaneously take it’s mod with K, whichever remainder we get increment count by 1 for remainder as index in Mod[] auxiliary array. Sub-array by each pair of positions with same value of ( cumSum % k) constitute a continuous range whose sum is divisible by K.
- Now traverse Mod[] auxiliary array, for any Mod[i] > 1 we can choose any two pair of indices for sub-array by
**(Mod[i]*(Mod[i] – 1))/2**number of ways . Do the same for all remainders < k and sum up the result that will be the number all possible sub-arrays divisible by K.

## C++

`// C++ program to find count of subarrays with ` `// sum divisible by k. ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Handles all the cases ` `// function to find all sub-arrays divisible by k ` `// modified to handle negative numbers as well ` `int` `subCount(` `int` `arr[], ` `int` `n, ` `int` `k) ` `{ ` ` ` `// create auxiliary hash array to count frequency ` ` ` `// of remainders ` ` ` `int` `mod[k]; ` ` ` `memset` `(mod, 0, ` `sizeof` `(mod)); ` ` ` ` ` `// Traverse original array and compute cumulative ` ` ` `// sum take remainder of this current cumulative ` ` ` `// sum and increase count by 1 for this remainder ` ` ` `// in mod[] array ` ` ` `int` `cumSum = 0; ` ` ` `for` `(` `int` `i = 0; i < n; i++) { ` ` ` `cumSum += arr[i]; ` ` ` ` ` `// as the sum can be negative, taking modulo twice ` ` ` `mod[((cumSum % k) + k) % k]++; ` ` ` `} ` ` ` ` ` `int` `result = 0; ` `// Initialize result ` ` ` ` ` `// Traverse mod[] ` ` ` `for` `(` `int` `i = 0; i < k; i++) ` ` ` ` ` `// If there are more than one prefix subarrays ` ` ` `// with a particular mod value. ` ` ` `if` `(mod[i] > 1) ` ` ` `result += (mod[i] * (mod[i] - 1)) / 2; ` ` ` ` ` `// add the elements which are divisible by k itself ` ` ` `// i.e., the elements whose sum = 0 ` ` ` `result += mod[0]; ` ` ` ` ` `return` `result; ` `} ` ` ` `// Driver program to run the case ` `int` `main() ` `{ ` ` ` `int` `arr[] = { 4, 5, 0, -2, -3, 1 }; ` ` ` `int` `k = 5; ` ` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]); ` ` ` `cout << subCount(arr, n, k) << endl; ` ` ` `int` `arr1[] = { 4, 5, 0, -12, -23, 1 }; ` ` ` `int` `k1 = 5; ` ` ` `int` `n1 = ` `sizeof` `(arr1) / ` `sizeof` `(arr1[0]); ` ` ` `cout << subCount(arr1, n1, k1) << endl; ` ` ` `return` `0; ` `} ` ` ` `// This code is corrected by Ashutosh Kumar` |

## Java

`// Java program to find count of ` `// subarrays with sum divisible by k. ` `import` `java.util.*; ` ` ` `class` `GFG { ` ` ` ` ` `// Handles all the cases ` ` ` `// function to find all sub-arrays divisible by k ` ` ` `// modified to handle negative numbers as well ` ` ` `static` `int` `subCount(` `int` `arr[], ` `int` `n, ` `int` `k) ` ` ` `{ ` ` ` ` ` `// create auxiliary hash array to ` ` ` `// count frequency of remainders ` ` ` `int` `mod[] = ` `new` `int` `[k]; ` ` ` `Arrays.fill(mod, ` `0` `); ` ` ` ` ` `// Traverse original array and compute cumulative ` ` ` `// sum take remainder of this current cumulative ` ` ` `// sum and increase count by 1 for this remainder ` ` ` `// in mod[] array ` ` ` `int` `cumSum = ` `0` `; ` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++) { ` ` ` `cumSum += arr[i]; ` ` ` ` ` `// as the sum can be negative, taking modulo twice ` ` ` `mod[((cumSum % k) + k) % k]++; ` ` ` `} ` ` ` ` ` `// Initialize result ` ` ` `int` `result = ` `0` `; ` ` ` ` ` `// Traverse mod[] ` ` ` `for` `(` `int` `i = ` `0` `; i < k; i++) ` ` ` ` ` `// If there are more than one prefix subarrays ` ` ` `// with a particular mod value. ` ` ` `if` `(mod[i] > ` `1` `) ` ` ` `result += (mod[i] * (mod[i] - ` `1` `)) / ` `2` `; ` ` ` ` ` `// add the elements which are divisible by k itself ` ` ` `// i.e., the elements whose sum = 0 ` ` ` `result += mod[` `0` `]; ` ` ` ` ` `return` `result; ` ` ` `} ` ` ` ` ` `// Driver code ` ` ` `public` `static` `void` `main(String[] args) ` ` ` `{ ` ` ` ` ` `int` `arr[] = { ` `4` `, ` `5` `, ` `0` `, -` `2` `, -` `3` `, ` `1` `}; ` ` ` `int` `k = ` `5` `; ` ` ` `int` `n = arr.length; ` ` ` `System.out.println(subCount(arr, n, k)); ` ` ` `int` `arr1[] = { ` `4` `, ` `5` `, ` `0` `, -` `12` `, -` `23` `, ` `1` `}; ` ` ` `int` `k1 = ` `5` `; ` ` ` `int` `n1 = arr1.length; ` ` ` `System.out.println(subCount(arr1, n1, k1)); ` ` ` `} ` `} ` ` ` `// This code is contributed by Anant Agarwal. ` |

## Python3

`# Python program to find ` `# count of subarrays with ` `# sum divisible by k. ` ` ` `# Handles all the cases ` `# function to find all ` `# sub-arrays divisible by k ` `# modified to handle ` `# negative numbers as well ` `def` `subCount(arr, n, k): ` ` ` ` ` `# create auxiliary hash ` ` ` `# array to count frequency ` ` ` `# of remainders ` ` ` `mod ` `=` `[] ` ` ` `for` `i ` `in` `range` `(k ` `+` `1` `): ` ` ` `mod.append(` `0` `) ` ` ` ` ` `# Traverse original array ` ` ` `# and compute cumulative ` ` ` `# sum take remainder of this ` ` ` `# current cumulative ` ` ` `# sum and increase count by ` ` ` `# 1 for this remainder ` ` ` `# in mod[] array ` ` ` `cumSum ` `=` `0` ` ` `for` `i ` `in` `range` `(n): ` ` ` `cumSum ` `=` `cumSum ` `+` `arr[i] ` ` ` ` ` `# as the sum can be negative, ` ` ` `# taking modulo twice ` ` ` `mod[((cumSum ` `%` `k)` `+` `k)` `%` `k]` `=` `mod[((cumSum ` `%` `k)` `+` `k)` `%` `k] ` `+` `1` ` ` ` ` ` ` `result ` `=` `0` `# Initialize result ` ` ` ` ` `# Traverse mod[] ` ` ` `for` `i ` `in` `range` `(k): ` ` ` ` ` `# If there are more than ` ` ` `# one prefix subarrays ` ` ` `# with a particular mod value. ` ` ` `if` `(mod[i] > ` `1` `): ` ` ` `result ` `=` `result ` `+` `(mod[i]` `*` `(mod[i]` `-` `1` `))` `/` `/` `2` ` ` ` ` `# add the elements which ` ` ` `# are divisible by k itself ` ` ` `# i.e., the elements whose sum = 0 ` ` ` `result ` `=` `result ` `+` `mod[` `0` `] ` ` ` ` ` `return` `result ` ` ` `# driver code ` ` ` `arr ` `=` `[` `4` `, ` `5` `, ` `0` `, ` `-` `2` `, ` `-` `3` `, ` `1` `] ` `k ` `=` `5` `n ` `=` `len` `(arr) ` ` ` `print` `(subCount(arr, n, k)) ` `arr1 ` `=` `[` `4` `, ` `5` `, ` `0` `, ` `-` `12` `, ` `-` `23` `, ` `1` `] ` ` ` `k1 ` `=` `5` `n1 ` `=` `len` `(arr1) ` `print` `(subCount(arr1, n1, k1)) ` ` ` `# This code is contributed ` `# by Anant Agarwal. ` |

## C#

`// C# program to find count of ` `// subarrays with sum divisible by k. ` `using` `System; ` ` ` `class` `GFG { ` ` ` ` ` `// Handles all the cases ` ` ` `// function to find all sub-arrays divisible by k ` ` ` `// modified to handle negative numbers as well ` ` ` `static` `int` `subCount(` `int` `[] arr, ` `int` `n, ` `int` `k) ` ` ` `{ ` ` ` `// create auxiliary hash array to ` ` ` `// count frequency of remainders ` ` ` `int` `[] mod = ` `new` `int` `[k]; ` ` ` ` ` `// Traverse original array and compute cumulative ` ` ` `// sum take remainder of this current cumulative ` ` ` `// sum and increase count by 1 for this remainder ` ` ` `// in mod[] array ` ` ` `int` `cumSum = 0; ` ` ` `for` `(` `int` `i = 0; i < n; i++) { ` ` ` `cumSum += arr[i]; ` ` ` ` ` `// as the sum can be negative, taking modulo twice ` ` ` `mod[((cumSum % k) + k) % k]++; ` ` ` `} ` ` ` ` ` `// Initialize result ` ` ` `int` `result = 0; ` ` ` ` ` `// Traverse mod[] ` ` ` `for` `(` `int` `i = 0; i < k; i++) ` ` ` ` ` `// If there are more than one prefix subarrays ` ` ` `// with a particular mod value. ` ` ` `if` `(mod[i] > 1) ` ` ` `result += (mod[i] * (mod[i] - 1)) / 2; ` ` ` ` ` `// add the elements which are divisible by k itself ` ` ` `// i.e., the elements whose sum = 0 ` ` ` `result += mod[0]; ` ` ` ` ` `return` `result; ` ` ` `} ` ` ` ` ` `// Driver code ` ` ` `public` `static` `void` `Main() ` ` ` `{ ` ` ` `int` `[] arr = { 4, 5, 0, -2, -3, 1 }; ` ` ` `int` `k = 5; ` ` ` `int` `n = arr.Length; ` ` ` `Console.WriteLine(subCount(arr, n, k)); ` ` ` `int` `[] arr1 = { 4, 5, 0, -12, -23, 1 }; ` ` ` `int` `k1 = 5; ` ` ` `int` `n1 = arr1.Length; ` ` ` `Console.WriteLine(subCount(arr1, n1, k1)); ` ` ` `} ` `} ` ` ` `// This code is contributed by vt_m. ` |

Output:

7 7

Time complexity : O(n + k)

Auxiliary Space : O(k)

**References **:

http://stackoverflow.com/questions/16605991/number-of-subarrays-divisible-by-k

This article is contributed by **Shashank Mishra ( Gullu )**. 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.

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