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 a 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++
// 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
// 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)); } } |
Python3
# 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 = 10n1 = 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#
// 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 */ |
Javascript
<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 programlet 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> |
Output:
4
Time Complexity: O(N+M), since two loops are running. One for N times and the other for M times.
Auxiliary Space: O(N+M)
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