Count pairs from two arrays having sum equal to K

Given an integer K and two arrays A1 and A2, the task is to return the total number of pairs (one element from A1 and one element from A2) with sum equal to K.
Note: Arrays can have duplicate elements. We consider every pair as different, the only constraint is, an element (of any array) can participate only in one pair. For example, A1[] = {3, 3}, A2[] = {4, 4} and K = 7, we consider only two pairs (3, 4) and (3, 4)

Examples:

Input: A1[] = {1, 1, 3, 4, 5, 6, 6}, A2[] = {1, 4, 4, 5, 7}, K = 10
Output: 4
All possible pairs are {3, 7}, {4, 6}, {5, 5} and {4, 6}

Input: A1[] = {1, 10, 13, 15}, A2[] = {3, 3, 12, 4}, K = 13
Output: 2



Approach:

  • Create a map of the elements of array A1.
  • For each element in array A2, check if temp = K – A2[i] exists in map created in previous step.
  • If map[temp] > 0 then increment result by 1 and decrement map[temp] by 1.
  • Print the total count 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 count of pairs 
// having sum equal to K
int countPairs(int A1[], int A2[]
                  , int n1, int n2, int K)
{
    // Initialize pairs to 0
    int res = 0;
  
    // create map of elements of array A1
    unordered_map<int, int> m;
    for (int i = 0; i < n1; ++i)
        m[A1[i]]++;
  
    // count total pairs
    for (int i = 0; i < n2; ++i) {
        int temp = K - A2[i];
  
        if (m[temp] != 0) {
            res++;
  
            // Every element can be part 
            // of at most one pair. 
            m[temp]--;
        }
    }
  
    // return total pairs
    return res;
}
  
// Driver program
int main()
{
    int A1[] = { 1, 1, 3, 4, 5, 6, 6 };
    int A2[] = { 1, 4, 4, 5, 7 }, K = 10;
  
    int n1 = sizeof(A1) / sizeof(A1[0]);
    int n2 = sizeof(A2) / sizeof(A2[0]);
  
    // function call to print required answer
    cout << countPairs(A1, A2, n1, n2, K);
  
    return 0;
}

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Java

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// Java implementation of above approach. 
import java.util.*;
class GfG {
  
// Function to return the count of pairs 
// having sum equal to K 
static int countPairs(int A1[], int A2[] , int n1, int n2, int K) 
    // Initialize pairs to 0 
    int res = 0
  
    // create map of elements of array A1 
    Map<Integer, Integer> m = new HashMap<Integer, Integer> (); 
    for (int i = 0; i < n1; ++i) 
    {
        if(m.containsKey(A1[i]))
        m.put(A1[i], m.get(A1[i]) + 1);
        else
        m.put(A1[i], 1);
    
  
    // count total pairs 
    for (int i = 0; i < n2; ++i) { 
        int temp = K - A2[i]; 
  
        if (m.containsKey(temp) && m.get(temp) != 0) { 
            res++; 
  
            // Every element can be part 
            // of at most one pair. 
            m.put(temp, m.get(A1[i]) - 1);
        
    
  
    // return total pairs 
    return res; 
  
// Driver program 
public static void main(String[] args) 
    int A1[] = { 1, 1, 3, 4, 5, 6, 6 }; 
    int A2[] = { 1, 4, 4, 5, 7 }, K = 10
  
    int n1 = A1.length;
    int n2 = A2.length; 
  
    // function call to print required answer 
    System.out.println(countPairs(A1, A2, n1, n2, K)); 
}

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Python3

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# Python3 implementation of above approach
  
# Function to return the count of 
# pairs having sum equal to K
def countPairs(A1, A2, n1, n2, K):
      
    # Initialize pairs to 0
    res = 0
      
    # Create dictionary of elements
    # of array A1
    m = dict()
    for i in range(0, n1):
        if A1[i] not in m.keys():
            m[A1[i]] = 1
        else:
            m[A1[i]] = m[A1[i]] + 1
          
    # count total pairs
    for i in range(0, n2):
        temp = K - A2[i]
        if temp in m.keys():
            res = res + 1
              
            # Every element can be part
            # of at most one pair
            m[temp] = m[temp] - 1
      
    # return total pairs
    return res
  
# Driver Code
A1 = [1, 1, 3, 4, 5, 6 ,6]
A2 = [1, 4, 4, 5, 7]
K = 10
  
n1 = len(A1)
n2 = len(A2)
  
# function call to print required answer
print(countPairs(A1, A2, n1, n2, K))
          
# This code is contributed 
# by Shashank_Sharma

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C#

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// C# implementation of above approach. 
using System;
using System.Collections.Generic;
  
class GfG
{
  
// Function to return the count of pairs 
// having sum equal to K 
static int countPairs(int []A1, int []A2 ,
                        int n1, int n2, int K) 
    // Initialize pairs to 0 
    int res = 0; 
  
    // create map of elements of array A1 
    Dictionary<int,int> m = new Dictionary<int,int> (); 
    for (int i = 0; i < n1; ++i) 
    {
        int a;
        if(m.ContainsKey(A1[i]))
        {
            a = m[A1[i]] + 1;
            m.Remove(A1[i]);
            m.Add(A1[i], a);
        }
        else
        m.Add(A1[i], 1);
    
  
    // count total pairs 
    for (int i = 0; i < n2; ++i)
    
        int temp = K - A2[i]; 
  
        if (m.ContainsKey(temp) && m[temp] != 0)
        
            res++; 
  
            // Every element can be part 
            // of at most one pair. 
            m.Remove(temp);
            m.Add(temp, m[A1[i]] - 1);
        
    
  
    // return total pairs 
    return res; 
  
// Driver program 
public static void Main() 
    int []A1 = { 1, 1, 3, 4, 5, 6, 6 }; 
    int []A2 = { 1, 4, 4, 5, 7 };
    int K = 10; 
  
    int n1 = A1.Length;
    int n2 = A2.Length; 
  
    // function call to print required answer 
    Console.WriteLine(countPairs(A1, A2, n1, n2, K)); 
  
/* This code contributed by PrinciRaj1992 */

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Output:

4


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