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Count pairs from two arrays whose modulo operation yields K
  • Difficulty Level : Basic
  • Last Updated : 10 Sep, 2019

Given an integer k and two arrays arr1 and arr2, the task is to count the total pairs (formed after choosing an element from arr1 and another from arr2) from these arrays whose modulo operation yields k.
Note: If in a pair (a, b), a > b then the modulo must be performed as a % b. Also, pairs occurring more than once will be counted only once.

Examples:

Input: arr1[] = {1, 3, 7}, arr2[] = {5, 3, 1}, K = 2
Output: 2
(3, 5) and (7, 5) are the only possible pairs.
Since, 5 % 3 = 2 and 7 % 5 = 2

Input: arr1[] = {2, 5, 99}, arr2[] = {2, 8, 1, 4}, K = 0
Output: 6
All possible pairs are (2, 2), (2, 8), (2, 4), (2, 1), (5, 1) and (99, 1).

Approach:



  • Take one element from arr1 at a time and perform it’s modulo operation with all the other elements of arr2 one by one.
  • If the result from the previous step is equal to k then store the pair (a, b) in a set in order to avoid duplicates where a is the smaller element and b is the larger one.
  • Total required pairs will be the size of the set in the end.

Below is the implementation of the above approach:

C++

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// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the total pairs
// of elements whose modulo yield K
int totalPairs(int arr1[], int arr2[], int K, int n, int m)
{
  
    // set is used to avoid duplicate pairs
    set<pair<int, int> > s;
  
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
  
            // check which element is greater and
            // proceed according to it
            if (arr1[i] > arr2[j]) {
  
                // check if modulo is equal to K
                if (arr1[i] % arr2[j] == K)
                    s.insert(make_pair(arr1[i], arr2[j]));
            }
            else {
                if (arr2[j] % arr1[i] == K)
                    s.insert(make_pair(arr2[j], arr1[i]));
            }
        }
    }
  
    // return size of the set
    return s.size();
}
  
// Driver code
int main()
{
    int arr1[] = { 8, 3, 7, 50 };
    int arr2[] = { 5, 1, 10, 4 };
    int K = 3;
    int n = sizeof(arr1) / sizeof(arr1[0]);
    int m = sizeof(arr2) / sizeof(arr2[0]);
  
    cout << totalPairs(arr1, arr2, K, n, m);
    return 0;
}

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Java

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// Java implementation of above approach 
import java.util.*;
  
class GFG
{
    static class pair
    {
        int first, second;
        public pair(int first, int second)
        {
            this.first = first;
            this.second = second;
        }
    }
      
    // Function to return the total pairs 
    // of elements whose modulo yield K 
    static int totalPairs(int []arr1, int []arr2,
                          int K, int n, int m) 
    
      
        // set is used to avoid duplicate pairs 
        HashSet<pair> s = new HashSet<pair>(); 
      
        for (int i = 0; i < n; i++) 
        
            for (int j = 0; j < m; j++) 
            
      
                // check which element is greater and 
                // proceed according to it 
                if (arr1[i] > arr2[j]) 
                
      
                    // check if modulo is equal to K 
                    if (arr1[i] % arr2[j] == K) 
                        s.add(new pair(arr1[i], arr2[j])); 
                
                else
                
                    if (arr2[j] % arr1[i] == K) 
                        s.add(new pair(arr2[j], arr1[i])); 
                
            
        
      
        // return size of the set 
        return s.size(); 
    
      
    // Driver code 
    public static void main(String []args) 
    
        int []arr1 = { 8, 3, 7, 50 }; 
        int []arr2 = { 5, 1, 10, 4 }; 
        int K = 3
        int n = arr1.length; 
        int m = arr2.length;
      
        System.out.println(totalPairs(arr1, arr2, K, n, m)); 
    
}
  
// This code is contributed by Princi Singh

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Python3

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# Python3 implementation of above approach
  
# Function to return the total pairs
# of elements whose modulo yield K
def totalPairs(arr1, arr2, K, n, m):
  
    # set is used to avoid duplicate pairs
    s={}
  
    for i in range(n):
        for j in range(m):
  
            # check which element is greater and
            # proceed according to it
            if (arr1[i] > arr2[j]):
  
                # check if modulo is equal to K
                if (arr1[i] % arr2[j] == K):
                    s[(arr1[i], arr2[j])]=1
            else:
                if (arr2[j] % arr1[i] == K):
                    s[(arr2[j], arr1[i])]=1
  
  
  
    # return size of the set
    return len(s)
  
# Driver code
  
arr1 = [ 8, 3, 7, 50 ]
arr2 = [5, 1, 10, 4 ]
K = 3
n = len(arr1)
m = len(arr2)
  
print(totalPairs(arr1, arr2, K, n, m))
  
# This code is contributed by mohit kumar 29

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C#

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// C# implementation of the approach
using System;
using System.Collections.Generic;
      
class GFG
{
    public class pair
    {
        public int first, second;
        public pair(int first, int second)
        {
            this.first = first;
            this.second = second;
        }
    }
      
    // Function to return the total pairs 
    // of elements whose modulo yield K 
    static int totalPairs(int []arr1, int []arr2,
                          int K, int n, int m) 
    
      
        // set is used to avoid duplicate pairs 
        HashSet<pair> s = new HashSet<pair>(); 
      
        for (int i = 0; i < n; i++) 
        
            for (int j = 0; j < m; j++) 
            
      
                // check which element is greater and 
                // proceed according to it 
                if (arr1[i] > arr2[j]) 
                
      
                    // check if modulo is equal to K 
                    if (arr1[i] % arr2[j] == K) 
                        s.Add(new pair(arr1[i], arr2[j])); 
                
                else
                
                    if (arr2[j] % arr1[i] == K) 
                        s.Add(new pair(arr2[j], arr1[i])); 
                
            
        
      
        // return size of the set 
        return s.Count; 
    
      
    // Driver code 
    public static void Main(String []args) 
    
        int []arr1 = { 8, 3, 7, 50 }; 
        int []arr2 = { 5, 1, 10, 4 }; 
        int K = 3; 
        int n = arr1.Length; 
        int m = arr2.Length;
      
        Console.WriteLine(totalPairs(arr1, arr2, K, n, m)); 
    
}
  
// This code is contributed by PrinciRaj1992

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Output:

3

Note: To print all the pairs just print the elements of set.

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