Count of pairs between two arrays such that the sums are distinct

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Given two arrays a[] and b[], the task is to find the count of all pairs (a[i], b[j]) such that a[i] + b[j] is unique among all the pairs i.e. if two pairs have equal sum then only one will be counted in the result.
Examples:

Input: a[] = {3, 3}, b[] = {3}
Output: 1
The two possible pairs are (a[0], b[0]) and (a[1], b[0]).
Pair 1: 3 + 3 = 6
Pair 2: 3 + 3 = 6
Input: a[] = {12, 2, 7}, b[] = {4, 3, 8}
Output: 7

Approach: Initialise count = 0 and run two loops to consider all possible pairs and store the sum of every pair in an unordered_set to check whether the sum has been obtained before. If it has then ignore the current pair else increment the count.
Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the count
// of pairs with distinct sum
int countPairs(int a[], int b[], int n, int m)
{
 
    // To store the required count
    int cnt = 0;
 
    // Set to store the sum
    // obtained for each pair
    unordered_set<int> s;
 
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
 
            // Sum of the current pair
            int sum = a[i] + b[j];
 
            // If the sum obtained is distinct
            if (s.count(sum) == 0) {
 
                // Increment the count
                cnt++;
 
                // Insert sum in the set
                s.insert(sum);
            }
        }
    }
 
    return cnt;
}
 
// Driver code
int main()
{
    int a[] = { 12, 2, 7 };
    int n = sizeof(a) / sizeof(a[0]);
    int b[] = { 4, 3, 8 };
    int m = sizeof(b) / sizeof(b[0]);
 
    cout << countPairs(a, b, n, m);
 
    return 0;
}

Java

// Java implementation of the approach 
import java.util.*;
 
class GFG
{
     
    // Function to return the count 
    // of pairs with distinct sum 
    static int countPairs(int a[], int b[], int n, int m) 
    { 
     
        // To store the required count 
        int cnt = 0; 
     
        // Set to store the sum 
        // obtained for each pair 
        HashSet<Integer> s = new HashSet<Integer>();
 
        for (int i = 0; i < n; i++) 
        { 
            for (int j = 0; j < m; j++) 
            { 
     
                // Sum of the current pair 
                int sum = a[i] + b[j]; 
     
                // If the sum obtained is distinct 
                if (s.contains(sum) == false) 
                { 
     
                    // Increment the count 
                    cnt++; 
     
                    // Insert sum in the set 
                    s.add(sum); 
                } 
            } 
        } 
     
        return cnt; 
    } 
     
    // Driver code 
    static public void main (String args[])
    {
        int a[] = { 12, 2, 7 }; 
        int n = a.length; 
        int b[] = { 4, 3, 8 }; 
        int m = b.length; 
     
        System.out.println(countPairs(a, b, n, m)); 
    }
}
 
// This code is contributed by AnkitRai01

Python3

# Python3 implementation of the approach
 
# Function to return the count
# of pairs with distinct sum
def countPairs(a, b, n, m):
 
    # To store the required count
    cnt = 0
 
    # Set to store the sum
    # obtained for each pair
    s=dict()
 
    for i in range(n):
        for j in range(m):
 
            # Sum of the current pair
            sum = a[i] + b[j]
 
            # If the sum obtained is distinct
            if (sum not in s.keys()):
                # Increment the count
                cnt+=1
 
                # Insert sum in the set
                s[sum]=1
 
    return cnt
 
 
# Driver code
 
a =[ 12, 2, 7]
n = len(a)
b =[ 4, 3, 8 ]
m = len(b)
 
print(countPairs(a, b, n, m))
 
# This code is contributed by mohit kumar 29

C#

// C# implementation of the approach 
using System;
using System.Collections.Generic; 
 
class GFG
{
     
    // Function to return the count 
    // of pairs with distinct sum 
    static int countPairs(int []a, int []b,
                          int n, int m) 
    { 
     
        // To store the required count 
        int cnt = 0; 
     
        // Set to store the sum 
        // obtained for each pair 
        HashSet<int> s = new HashSet<int>();
 
        for (int i = 0; i < n; i++) 
        { 
            for (int j = 0; j < m; j++) 
            { 
     
                // Sum of the current pair 
                int sum = a[i] + b[j]; 
     
                // If the sum obtained is distinct 
                if (s.Contains(sum) == false) 
                { 
     
                    // Increment the count 
                    cnt++; 
     
                    // Insert sum in the set 
                    s.Add(sum); 
                } 
            } 
        } 
     
        return cnt; 
    } 
     
    // Driver code 
    static public void Main (String []args)
    {
        int []a = { 12, 2, 7 }; 
        int n = a.Length; 
        int []b = { 4, 3, 8 }; 
        int m = b.Length; 
     
        Console.WriteLine(countPairs(a, b, n, m)); 
    }
}
 
// This code is contributed by PrinciRaj1992 

Javascript

<script>
 
    // Javascript implementation of the approach
 
    // Function to return the count 
    // of pairs with distinct sum 
    function countPairs(a, b, n, m) 
    { 
       
        // To store the required count 
        let cnt = 0; 
       
        // Set to store the sum 
        // obtained for each pair 
        let s = new Set();
   
        for (let i = 0; i < n; i++) 
        { 
            for (let j = 0; j < m; j++) 
            { 
       
                // Sum of the current pair 
                let sum = a[i] + b[j]; 
       
                // If the sum obtained is distinct 
                if (s.has(sum) == false) 
                { 
       
                    // Increment the count 
                    cnt++; 
       
                    // Insert sum in the set 
                    s.add(sum); 
                } 
            } 
        } 
       
        return cnt; 
    }
     
    // Driver code
     
        let a = [ 12, 2, 7 ]; 
        let n = a.length; 
        let b = [ 4, 3, 8 ]; 
        let m = b.length; 
       
        document.write(countPairs(a, b, n, m)); 
         
    // This code is contributed by susmitakundugoaldanga.      
</script>

Output:

Time complexity: O(N * M).
Auxiliary Space: O(1).

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