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

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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the
// above approach
#include <bits/stdc++.h>
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[0]);
  
    cout << findSubarrays(arr, N, K);
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


Output:

3


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

Don’t stop now and take your learning to the next level. Learn all the important concepts of Data Structures and Algorithms with the help of the most trusted course: DSA Self Paced. Become industry ready at a student-friendly price.




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.



Improved By : shivanisinghss2110