Count pairs in an array whose absolute difference is divisible by K
Given an array arr[] and a positive integer K. The task is to count the total number of pairs in the array whose absolute difference is divisible by K.
Examples:
Input: arr[] = {1, 2, 3, 4}, K = 2
Output: 2
Explanation:
Total 2 pairs exists in the array with absolute difference divisible by 2.
The pairs are: (1, 3), (2, 4).Input: arr[] = {3, 3, 3}, K = 3
Output: 3
Explanation:
Total 3 pairs exists in this array with absolute difference divisible by 3.
The pairs are: (3, 3), (3, 3), (3, 3).
Naive Approach: The idea is to check for each pair of the array one by one and count the total number pairs whose absolute difference is divisible by K.
C++
#include <bits/stdc++.h> using namespace std; // function to count pairs in an array // whose absolute difference is // divisible by k void countPairs(int arr[], int n, int k) { // initilize count as zero. int i, j, cnt = 0; // loop to count the valid pair for (i = 0; i < n - 1; i++) { for (j = i + 1; j < n; j++) { if ((arr[i] - arr[j] + k) % k == 0) cnt += 1; } } cout << cnt << endl; } // Driver code int main() { // input array int arr[] = {3, 3, 3}; int k = 3; // calculate the size of array int n = sizeof(arr) / sizeof(arr[0]); // function to count the valid pair countPairs(arr, n, k); return 0; } |
chevron_right
filter_none
Java
import java.util.*; class GFG { // function to count pairs in an array // whose absolute difference is // divisible by k static void countPairs(int arr[], int n, int k) { // initilize count as zero. int i, j, cnt = 0; // loop to count the valid pair for (i = 0; i < n - 1; i++) { for (j = i + 1; j < n; j++) { if ((arr[i] - arr[j] + k) % k == 0) cnt += 1; } } System.out.print(cnt +"\n"); } // Driver code public static void main(String[] args) { // input array int arr[] = {3, 3, 3}; int k = 3; // calculate the size of array int n = arr.length; // function to count the valid pair countPairs(arr, n, k); } } // This code is contributed by 29AjayKumar |
chevron_right
filter_none
Python3
# Python3 Code implementaton of the above approach # function to count pairs in an array # whose absolute difference is # divisible by k def countPairs(arr, n, k) : # initilize count as zero. cnt = 0; # loop to count the valid pair for i in range(n - 1) : for j in range(i + 1, n) : if ((arr[i] - arr[j] + k) % k == 0) : cnt += 1; print(cnt) ; # Driver code if __name__ == "__main__" : # input array arr = [3, 3, 3]; k = 3; # calculate the size of array n = len(arr); # function to count the valid pair countPairs(arr, n, k); # This code is contributed by AnkitRai01 |
chevron_right
filter_none
C#
using System; class GFG { // function to count pairs in an array // whose absolute difference is // divisible by k static void countPairs(int []arr, int n, int k) { // initilize count as zero. int i, j, cnt = 0; // loop to count the valid pair for (i = 0; i < n - 1; i++) { for (j = i + 1; j < n; j++) { if ((arr[i] - arr[j] + k) % k == 0) cnt += 1; } } Console.Write(cnt +"\n"); } // Driver code public static void Main(String[] args) { // input array int []arr = {3, 3, 3}; int k = 3; // calculate the size of array int n = arr.Length; // function to count the valid pair countPairs(arr, n, k); } } // This code is contributed by 29AjayKumar |
chevron_right
filter_none
Output:
3
Time Complexity: O(N2)
Space Complexity:O(1)
Efficient Approach:
Algorithm:
- Convert each elements
(A[i])of the array to((A[i]+K)%K)- Use hashing teching technique to store the number of times
(A[i]%K)occurs in the array- Now, if an element A[i] occurs x times in the array then add
x*(x-1)/2(choosing any2elements out ofxelements ) in the count pair where 1<=i<=n.This is because value of each elements of the array lies between0toK-1so the absolute difference is divisible only if value of both the elements of a pair are equal
CPP
// Write CPP code here #include <bits/stdc++.h> using namespace std; // function to Count pairs in an array whose // absolute difference is divisible by k void countPair(int arr[], int n, int k) { // intialize the count int cnt = 0; // making every element of arr in // range 0 to k - 1 for (int i = 0; i < n; i++) { arr[i] = (arr[i] + k) % k; } // create an array hash[] int hash[k] = { 0 }; // store to count of element of arr // in hash[] for (int i = 0; i < n; i++) { hash[arr[i]]++; } // count the pair whose absolute // difference is divisible by k for (int i = 0; i < k; i++) { cnt += (hash[i] * (hash[i] - 1)) / 2; } // print the value of count cout << cnt << endl; } // Driver Code int main() { // input array int arr[] = {1, 2, 3, 4}; int k = 2; // calculate the size of array int n = sizeof(arr) / sizeof(arr[0]); countPair(arr, n, k); return 0; } |
chevron_right
filter_none
Java
// JAVA Implementation of above approach import java.util.*; class GFG { // function to Count pairs in an array whose // absolute difference is divisible by k static void countPair(int arr[], int n, int k) { // intialize the count int cnt = 0; // making every element of arr in // range 0 to k - 1 for (int i = 0; i < n; i++) { arr[i] = (arr[i] + k) % k; } // create an array hash[] int hash[] = new int[k]; // store to count of element of arr // in hash[] for (int i = 0; i < n; i++) { hash[arr[i]]++; } // count the pair whose absolute // difference is divisible by k for (int i = 0; i < k; i++) { cnt += (hash[i] * (hash[i] - 1)) / 2; } // print the value of count System.out.print(cnt +"\n"); } // Driver Code public static void main(String[] args) { // input array int arr[] = {1, 2, 3, 4}; int k = 2; // calculate the size of array int n = arr.length; countPair(arr, n, k); } } // This code is contributed by PrinciRaj1992 |
chevron_right
filter_none
Python3
# Python3 Implementation of above approach # function to Count pairs in an array whose # absolute difference is divisible by k def countPair(arr, n, k): # intialize the count cnt = 0; # making every element of arr in # range 0 to k - 1 for i in range(n): arr[i] = (arr[i] + k) % k; # create an array hash hash = [0]*k; # store to count of element of arr # in hash for i in range(n): hash[arr[i]] += 1; # count the pair whose absolute # difference is divisible by k for i in range(k): cnt += (hash[i] * (hash[i] - 1)) / 2; # prthe value of count print(int(cnt)); # Driver Code if __name__ == '__main__': # input array arr = [1, 2, 3, 4]; k = 2; # calculate the size of array n = len(arr); countPair(arr, n, k); # This code is contributed by 29AjayKumar |
chevron_right
filter_none
C#
// C# Implementation of above approach using System; class GFG { // function to Count pairs in an array whose // absolute difference is divisible by k static void countPair(int []arr, int n, int k) { // intialize the count int cnt = 0; // making every element of arr in // range 0 to k - 1 for (int i = 0; i < n; i++) { arr[i] = (arr[i] + k) % k; } // create an array hash[] int []hash = new int[k]; // store to count of element of arr // in hash[] for (int i = 0; i < n; i++) { hash[arr[i]]++; } // count the pair whose absolute // difference is divisible by k for (int i = 0; i < k; i++) { cnt += (hash[i] * (hash[i] - 1)) / 2; } // print the value of count Console.Write(cnt +"\n"); } // Driver Code public static void Main(String[] args) { // input array int []arr = {1, 2, 3, 4}; int k = 2; // calculate the size of array int n = arr.Length; countPair(arr, n, k); } } // This code is contributed by 29AjayKumar |
chevron_right
filter_none
Output:
2
Time Complexity: O(n+k)
Auxiliary Space: O(k)
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
