Count pairs from two arrays having sum equal to K - GeeksforGeeks

25 Mar, 2022
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++
Java
Python3
C#
Javascript
// C++ implementation of above approach. #include <bits/stdc++.h>using namespace std; // Function to return the count of pairs// having sum equal to Kint 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 programint 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;}
// Java implementation of above approach.import java.util.*;class GfG { // Function to return the count of pairs// having sum equal to Kstatic 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 programpublic 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));}}
# Python3 implementation of above approach # Function to return the count of# pairs having sum equal to Kdef 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 CodeA1 = [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 answerprint(countPairs(A1, A2, n1, n2, K))         # This code is contributed# by Shashank_Sharma
// C# implementation of above approach.using System;using System.Collections.Generic; class GfG{ // Function to return the count of pairs// having sum equal to Kstatic 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 programpublic 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 */
<script> // JavaScript implementation of above approach.  // Function to return the count of pairs// having sum equal to Kfunction countPairs(A1, A2, n1, n2, K){    // Initialize pairs to 0    let res = 0;     // create map of elements of array A1    let m = new Map();    for (let i = 0; i < n1; ++i){        if(m.has(A1[i])){            m.set(A1[i],m.get(A1[i])+1);        }        else m.set(A1[i],1);    }     // count total pairs    for (let i = 0; i < n2; ++i) {        let temp = K - A2[i];         if (m.has(temp)) {            res++;             // Every element can be part            // of at most one pair.            m.set(temp,m.get(temp)-1);        }    }     // return total pairs    return res;} // Driver program let A1 = [ 1, 1, 3, 4, 5, 6, 6 ];let A2 = [ 1, 4, 4, 5, 7 ], K = 10; let n1 = A1.length;let n2 = A2.length; // function call to print required answerdocument.write(countPairs(A1, A2, n1, n2, K)); //  This code is contributed by shinjanpatra </script>
4
 
Time Complexity: O(N+M)
Auxiliary Space: O(N+M)
Shashank_Sharma
prerna saini
princiraj1992
rohitsingh07052
shinjanpatra
cpp-unordered_map
frequency-counting
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