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++
// 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
// 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
# 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#
// 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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