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:

Below is the implementation of the above approach:

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// 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;
}
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// 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
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# Python3 implementation of the
# above approach
  
# Function to return the required
# number of subarrays
def findSubarrays(arr, N, K):
      
    # Return number of possible
    # subarrays of length K
    return N - K + 1;
  
# Driver Code
if __name__ == '__main__':
      
    arr = [ 1, 5, 3, 2, 17, 18 ];
    K = 4;
    N = len(arr);
  
    print(findSubarrays(arr, N, K));
  
# This code is contributed by Rohit_ranjan
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// C# implementation of the
// above approach
using System;
  
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;
  
    Console.Write(findSubarrays(arr, N, K));
}
}
  
// This code is contributed by Amit Katiyar
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Output:
3


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

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