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 divisible 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-1
- 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 <bits/stdc++.h>using namespace std;// Function to return the required// number of subarraysint findSubarrays(int arr[], int N, int K){ // Return number of possible // subarrays of length K return N - K + 1;}// Driver Codeint 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;} |
Java
// Java implementation of the// above approachclass GFG{// Function to return the required// number of subarraysstatic int findSubarrays(int arr[], int N, int K){ // Return number of possible // subarrays of length K return N - K + 1;}// Driver Codepublic 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 |
Python3
# Python3 implementation of the# above approach# Function to return the required# number of subarraysdef findSubarrays(arr, N, K): # Return number of possible # subarrays of length K return N - K + 1;# Driver Codeif __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 |
C#
// C# implementation of the// above approachusing System;class GFG{// Function to return the required// number of subarraysstatic int findSubarrays(int []arr, int N, int K){ // Return number of possible // subarrays of length K return N - K + 1;}// Driver Codepublic 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 |
Javascript
<script>// Javascript implementation for the above approach// Function to return the required// number of subarraysfunction findSubarrays(arr, N, K){ // Return number of possible // subarrays of length K return N - K + 1;} // Driver Code let arr = [ 1, 5, 3, 2, 17, 18 ]; let K = 4; let N = arr.length; document.write(findSubarrays(arr, N, K));</script> |
3
Time complexity: O(1)
Auxiliary Space: O(1)
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