# Count of subarrays of size K having at least one pair with absolute difference divisible by K-1

Given an arr[] consisting of N elements, the task is to count all subarrays of size K having atleast one pair whose absolute difference is divisible by K – 1.

Examples:

Input: arr[] = {1, 5, 3, 2, 17, 18}, K = 4
Output: 3
Explanation:
The three subarrays of size 4 are:
{1, 5, 3, 2}: Pair {5, 2} have difference divisible by 3
{5, 3, 2, 17}: Pairs {5, 2}, {5, 17}, {2, 17} have difference divisible by 3
{3, 2, 17, 18}: Pairs {3, 18}, {2, 17} have difference divisible by 3

Input: arr[] = {1, 2, 3, 4, 5}, K = 5
Output: 1
Explanation:
{1, 2, 3, 4, 5}: Pair {1, 5} is divisble by 4

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

Naive Approach:
The simplest approach to solve the problem is to iterate over all subarrays of size K and check if there exists any pair whose difference is divisible by K – 1.
Time Complexity: O(N * K * K)

Efficient Approach: The above approach can be optimized using Pigeonhole Principle. Follow the steps below to solve the problem:

• Consider K-1 boxes labeled 0, 1, 2, …, K-2 respectively. They represent the remainders when any number x from the array is divided by K-1, which means the boxes store the modulo K-1 of array elements.
• Now, in a subarray of size K, according to the Pigeonhole Principle, there must be atleast one pair of boxes with same remainders. It means that there is atleast one pair whose difference or even the summation will be divisible by K.
• From this theorem we can conclude that every subarray of size K, will always have atleast one pair whose difference is divisible by K-1.
• So, the answer will be equal to the number of subarrays of size K possible from the given array, which is equal to N – K + 1.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the ` `// above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the required ` `// number of subarrays ` `int` `findSubarrays(``int` `arr[], ` `                  ``int` `N, ` `                  ``int` `K) ` `{ ` `    ``// Return number of possible ` `    ``// subarrays of length K ` `    ``return` `N - K + 1; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 5, 3, 2, 17, 18 }; ` `    ``int` `K = 4; ` `    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``cout << findSubarrays(arr, N, K); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the ` `// above approach ` `class` `GFG{ ` ` `  `// Function to return the required ` `// number of subarrays ` `static` `int` `findSubarrays(``int` `arr[], ``int` `N,  ` `                                    ``int` `K) ` `{ ` `     `  `    ``// Return number of possible ` `    ``// subarrays of length K ` `    ``return` `N - K + ``1``; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `arr[] = { ``1``, ``5``, ``3``, ``2``, ``17``, ``18` `}; ` `    ``int` `K = ``4``; ` `    ``int` `N = arr.length; ` ` `  `    ``System.out.print(findSubarrays(arr, N, K)); ` `} ` `} ` ` `  `// This code is contributed by shivanisinghss2110 `

Output:

```3
```

Time complexity: O(1)
Auxiliary Space: O(1)

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Improved By : shivanisinghss2110

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