Count number of pairs in array having sum divisible by K | SET 2 - GeeksforGeeks

23 Apr, 2021
Given an array A[] and positive integer K, the task is to count the total number of pairs in the array whose sum is divisible by K.Examples: 
Input : A[] = {2, 2, 1, 7, 5, 3}, K = 4 Output : 5 There are five pairs possible whose sum Is divisible by ‘4’ i.e., (2, 2), (1, 7), (7, 5), (1, 3) and (5, 3)Input : A[] = {5, 9, 36, 74, 52, 31, 42}, K = 3 Output : 7  
Approach: In the previous post, an approach using hashing is discussed. In this article, another approach using hashing is discussed.
Analyzing the statement we can say we need to make pairs of (a, b) such that:
     (a + b) % K = 0
 =>    a%K + b%K = 0
 =>    a%K + b%K = K%K
 =>    b%K = K%K - a%K
 =>     b%K = (K - a%K) % K.     {Range of a%K => [0,K-1]}
The idea is a can be paired with (K — a%K) % K. Now we have to find the same for each a present in the given array.
The algorithm would create a hash-map:Keys: possible remainders for value%K i.e 0 to K-1 Values: count of values with value%K = key
The stepwise algorithm is: 
Find x = arr[i]%k.This array element can be paired with array elements having mod value k-x. This frequency count of array elements is stored in hash. So add that count to answer.Increment count for x in hash.In case the value of x is zero, then it can be paired only with elements having 0 mod value.
Find x = arr[i]%k.
This array element can be paired with array elements having mod value k-x. This frequency count of array elements is stored in hash. So add that count to answer.
Increment count for x in hash.
In case the value of x is zero, then it can be paired only with elements having 0 mod value.
Below is the implementation of the above approach:  
C++
Java
Python3
C#
Javascript
// C++ Program to count pairs// whose sum divisible by 'K'#include <bits/stdc++.h>using namespace std; // Program to count pairs whose sum divisible// by 'K'int countKdivPairs(int A[], int n, int K){    // Create a frequency array to count    // occurrences of all remainders when    // divided by K    int freq[K] = { 0 };     // To store count of pairs.    int ans = 0;     // Traverse the array, compute the remainder    // and add k-remainder value hash count to ans    for (int i = 0; i < n; i++) {        int rem = A[i] % K;               // Count number of ( A[i], (K - rem)%K ) pairs          ans += freq[(K - rem) % K];         // Increment count of remainder in hash map        freq[rem]++;    }     return ans;} // Driver codeint main(){     int A[] = { 2, 2, 1, 7, 5, 3 };    int n = sizeof(A) / sizeof(A[0]);    int K = 4;    cout << countKdivPairs(A, n, K);     return 0;}
// JAVA Program to count pairs whose sum divisible// by 'K'class GFG{   static int countKdivPairs(int A[], int n, int K)  {      // Create a frequency array to count      // occurrences of all remainders when      // divided by K      int []freq = new int[K];       // To store count of pairs.      int ans = 0;       // Traverse the array, compute the remainder      // and add k-remainder value hash count to ans      for (int i = 0; i < n; i++)      {          int rem = A[i] % K;           // Count number of ( A[i], (K - rem)%K ) pairs          ans += freq[(K - rem) % K];           // Increment count of remainder in hash map          freq[rem]++;      }       return ans;  } // Driver code  public static void main(String[] args)  {      int A[] = { 2, 2, 1, 7, 5, 3 };      int n = A.length;      int K = 4;      System.out.println(countKdivPairs(A, n, K));  }} // This code is contributed by Princi Singh, Yadvendra Naveen
# Python Program to count pairs whose sum divisible# by 'K'def countKdivPairs(A, n, K):         # Create a frequency array to count    # occurrences of all remainders when    # divided by K    freq = [0 for i in range(K)]     # To store count of pairs.    ans = 0     # Traverse the array, compute the remainder    # and add k-remainder value hash count to ans    for i in range(n):        rem = A[i] % K                 # Count number of ( A[i], (K - rem)%K ) pairs        ans += freq[(K - rem) % K]                 # Increment count of remainder in hash map        freq[rem] += 1     return ans # Driver codeif __name__ == '__main__':    A = [2, 2, 1, 7, 5, 3]    n = len(A)    K = 4    print(countKdivPairs(A, n, K)) # This code is contributed by# Surendra_Gangwar, Yadvendra Naveen
// C# Program to count pairs// whose sum divisible by 'K'using System; class GFG{ // Program to count pairs whose sum divisible// by 'K'static int countKdivPairs(int []A, int n, int K){    // Create a frequency array to count    // occurrences of all remainders when    // divided by K    int []freq = new int[K];     // To store count of pairs.    int ans = 0;     // Traverse the array, compute the remainder    // and add k-remainder value hash count to ans    for (int i = 0; i < n; i++)    {        int rem = A[i] % K;         // Count number of ( A[i], (K - rem)%K ) pairs          ans += freq[(K - rem) % K];         // Increment count of remainder in hash map        freq[rem]++;    }     return ans;} // Driver codepublic static void Main(String[] args){    int []A = { 2, 2, 1, 7, 5, 3 };    int n = A.Length;    int K = 4;    Console.WriteLine(countKdivPairs(A, n, K));}} // This code contributed by Rajput-Ji, Yadvendra Naveen
<script> // Javascript Program to count pairs// whose sum divisible by 'K' // Program to count pairs whose sum divisible// by 'K'function countKdivPairs( A, n, K){    // Create a frequency array to count    // occurrences of all remainders when    // divided by K    var freq = Array(K).fill(0);     // To store count of pairs.    var ans = 0;     // Traverse the array, compute the remainder    // and add k-remainder value hash count to ans    for (var i = 0; i < n; i++) {        var rem = A[i] % K;               // Count number of ( A[i], (K - rem)%K ) pairs          ans += freq[(K - rem) % K];         // Increment count of remainder in hash map        freq[rem]++;    }     return ans;} // Driver codevar A = [ 2, 2, 1, 7, 5, 3 ];var n = A.length;var K = 4;document.write( countKdivPairs(A, n, K)); // This code is contributed by itsok.</script>
5
 
Time Complexity: O(N) Auxiliary Space: O(K) 
princi singh
Rajput-Ji
SURENDRA_GANGWAR
Yadvendra Naveen
itsok
Hash
Arrays
Hash
Mathematical
Arrays
Hash
Mathematical
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