Count pairs from two arrays whose modulo operation yields K

Last Updated : 08 Sep, 2022

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++

// 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;
}

Java

// 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

Python3

# 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

// 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

JavaScript

<script>

// Javascript implementation of above approach

// Function to return the total pairs
// of elements whose modulo yield K
function totalPairs(arr1, arr2, K, n, m)
{

    // set is used to avoid duplicate pairs
    var s = new Set();

    for (var i = 0; i < n; i++) {
        for (var 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([arr1[i], arr2[j]]);
            }
            else {
                if (arr2[j] % arr1[i] == K)
                    s.add([arr2[j], arr1[i]]);
            }
        }
    }

    // return size of the set
    return s.size;
}

// Driver code
var arr1 = [ 8, 3, 7, 50 ];
var arr2 = [ 5, 1, 10, 4 ];
var K = 3;
var n = arr1.length;
var m = arr2.length;
document.write( totalPairs(arr1, arr2, K, n, m));

</script>

Output

Complexity Analysis:

Time Complexity: O(n * m)
Auxiliary Space: O(n * m)

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

Count pairs whose products exist in array

Shashank_Sharma

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Count pairs from two arrays whose modulo operation yields K

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