Given two string arrays s1[] and s2[]. The task is to find the count of pairs (s1[i], s2[j]) such that s1[i] = s2[j]. Note that an element s1[i] can only participate in a single pair.
Examples:
Input: s1[] = {“abc”, “def”}, s2[] = {“abc”, “abc”}
Output: 1
Explanation: Only valid pair is (s1[0], s2[0]) or (s1[0], s2[1])
Note that even though “abc” appears twice in the
array s2[] but it can only make a single pair
as “abc” only appears once in the array s1[]Input: s1[] = {“aaa”, “aaa”}, s2[] = {“aaa”, “aaa”}
Output: 2
Naive approach:
Generate all the pairs and check for the given condition, if it satisfy then increment the count by 1.
- Initialise a variable count = 0
- Generate all the pair
- Check if this pair is already considered, then continue
- Check if the pair satisfies the given condition
- Replace the string in s2 with “-1”, just to mark that we have considered this string with some pair
- Increment the count by 1
- Return 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 required pairsint count_pairs(string s1[], string s2[], int n1, int n2){ // Initialise a variable count = 0 int count = 0; // Generate all the pair for (int i = 0; i < n1; i++) { for (int j = 0; j < n2; j++) { // Check if this pair is already considered // if true, then continue if (s2[j] == "-1") continue; // Check if pair satisfy the given condition if (s1[i] == s2[j]) { // Replace the string in s2 with -1, just to // mark that we have consider this string // with some pair s2[j] = "-1"; // Increment the count by 1 count++; break; } } } // Return the count return count;}// Driver codeint main(){ string s1[] = { "abc", "def", "abc" }; string s2[] = { "abc", "abc" }; int n1 = sizeof(s1) / sizeof(string); int n2 = sizeof(s2) / sizeof(string); cout << count_pairs(s1, s2, n1, n2); return 0;} |
Java
import java.util.Arrays;public class Gfg { // Function to return the count of required pairs public static int count_pairs(String s1[], String s2[], int n1, int n2) { // Initialise a variable count = 0 int count = 0; // Generate all the pair for (int i = 0; i < n1; i++) { for (int j = 0; j < n2; j++) { // Check if this pair is already considered // if true, then continue if (s2[j] == "-1") continue; // Check if pair satisfy the given condition if (s1[i].equals(s2[j])) { // Replace the string in s2 with -1, // just to mark that we have consider // this string with some pair s2[j] = "-1"; // Increment the count by 1 count++; break; } } } // Return the count return count; } // Driver code public static void main(String[] args) { String s1[] = { "abc", "def", "abc" }; String s2[] = { "abc", "abc" }; int n1 = s1.length; int n2 = s2.length; System.out.println(count_pairs(s1, s2, n1, n2)); }} |
Python3
# Python implementation of the approach# Function to return the count of required pairsdef count_pairs( s1, s2, n1, n2): # Initialise a variable count = 0 count = 0; # Generate all the pair for i in range(0,n1): for j in range(0,n2): # Check if this pair is already considered # if true, then continue if (s2[j] == "-1"): continue; # Check if pair satisfy the given condition if (s1[i] == s2[j]) : # Replace the string in s2 with -1, just to # mark that we have consider this string # with some pair s2[j] = "-1"; # Increment the count by 1 count+=1; break; # Return the count return count;# Driver codes1 = [ "abc", "def", "abc" ];s2 = [ "abc", "abc" ];n1 = len(s1);n2 = len(s2);print(count_pairs(s1, s2, n1, n2)); # This code is contributed by ratiagrawal. |
C#
// C# code to implement the above ideausing System;using System.Collections.Generic;class GFG { static int count_pairs(string[] s1, string[] s2, int n1, int n2) { // Initialise a variable count = 0 int count = 0; // Generate all the pair for (int i = 0; i < n1; i++) { for (int j = 0; j < n2; j++) { // Check if this pair is already considered // if true, then continue if (s2[j] == "-1") continue; // Check if pair satisfy the given condition if (s1[i] == s2[j]) { // Replace the string in s2 with -1, just to // mark that we have consider this string // with some pair s2[j] = "-1"; // Increment the count by 1 count++; break; } } } // Return the count return count; } // Driver code public static void Main() { string[] s1 = { "abc", "def", "abc" }; string[] s2 = { "abc", "abc" }; int n1 = s1.Length; int n2 = s2.Length; Console.Write(count_pairs(s1, s2, n1, n2)); }}// This code is contributed by poojaagarwal2. |
Javascript
// Javascript implementation of the approach// Function to return the count of required pairsfunction count_pairs(s1, s2, n1, n2){ // Initialise a variable count = 0 let count = 0; // Generate all the pair for (let i = 0; i < n1; i++) { for (let j = 0; j < n2; j++) { // Check if this pair is already considered // if true, then continue if (s2[j] == "-1") continue; // Check if pair satisfy the given condition if (s1[i] == s2[j]) { // Replace the string in s2 with -1, just to // mark that we have consider this string // with some pair s2[j] = "-1"; // Increment the count by 1 count++; break; } } } // Return the count return count;}// Driver code let s1 = [ "abc", "def", "abc" ]; let s2 = [ "abc", "abc" ]; let n1 = s1.length; let n2 = s2.length; document.write(count_pairs(s1, s2, n1, n2)); |
Output
2
Time Complexity: O(n1*n2), Where n1 and n2 are the lengths of given string array s1 and s2 respectively.
Auxiliary Space: O(1)
Approach:
- Create an unordered_map to store the frequencies of all the strings of the array s1[].
- Now for every string of the array, check whether a string equal to the current string is present in the map or not.
- If yes then increment the count and decrement the frequency of the string from the map. This is because a string can only make a pair once.
- Print the count in the end.
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 required pairsint count_pairs(string s1[], string s2[], int n1, int n2){ // Map to store the frequencies of // all the strings of array s1[] unordered_map<string, int> mp; // Update the frequencies for (int i = 0; i < n1; i++) mp[s1[i]]++; // To store the count of pairs int cnt = 0; // For every string of array s2[] for (int i = 0; i < n2; i++) { // If current string can make a pair if (mp[s2[i]] > 0) { // Increment the count of pairs cnt++; // Decrement the frequency of the // string as once occurrence has been // used in the current pair mp[s2[i]]--; } } // Return the count return cnt;}// Driver codeint main(){ string s1[] = { "abc", "def" }; string s2[] = { "abc", "abc" }; int n1 = sizeof(s1) / sizeof(string); int n2 = sizeof(s2) / sizeof(string); cout << count_pairs(s1, s2, n1, n2); return 0;} |
Java
// Java implementation of the approachimport java.util.*; class GFG { // Function to return // the count of required pairs static int count_pairs(String s1[], String s2[], int n1, int n2) { // Map to store the frequencies of // all the strings of array s1[] HashMap<String, Integer> mp = new HashMap<String, Integer>(); // Update the frequencies for (int i = 0; i < n1; i++) mp.put(s1[i], 0); // Update the frequencies for (int i = 0; i < n1; i++) mp.put(s1[i], mp.get(s1[i]) + 1); // To store the count of pairs int cnt = 0; // For every string of array s2[] for (int i = 0; i < n2; i++) { // If current string can make a pair if (mp.get(s2[i]) > 0) { // Increment the count of pairs cnt++; // Decrement the frequency of the // string as once occurrence has been // used in the current pair mp.put(s2[i], mp.get(s2[i]) - 1); } } // Return the count return cnt; } // Driver code public static void main (String[] args) { String s1[] = { "abc", "def" }; String s2[] = { "abc", "abc" }; int n1 = s1.length; int n2 = s2.length; System.out.println(count_pairs(s1, s2, n1, n2)); }}// This code is contributed by AnkitRai01 |
Python3
# python 3 implementation of the approach# Function to return the count of required pairsdef count_pairs(s1, s2,n1,n2): # Map to store the frequencies of # all the strings of array s1[] mp = {s1[i]:0 for i in range(len(s1))} # Update the frequencies for i in range(n1): mp[s1[i]] += 1 # To store the count of pairs cnt = 0 # For every string of array s2[] for i in range(n2): # If current string can make a pair if (mp[s2[i]] > 0): # Increment the count of pairs cnt += 1 # Decrement the frequency of the # string as once occurrence has been # used in the current pair mp[s2[i]] -= 1 # Return the count return cnt# Driver codeif __name__ == '__main__': s1 = ["abc", "def"] s2 = ["abc", "abc"] n1 = len(s1) n2 = len(s2) print(count_pairs(s1, s2, n1, n2))# This code is contributed by# Surendra_Gangwar |
C#
// C# implementation of the approach using System;using System.Collections.Generic; class GFG { // Function to return // the count of required pairs static int count_pairs(String []s1, String []s2, int n1, int n2) { // Map to store the frequencies of // all the strings of array s1[] Dictionary<String, int> mp = new Dictionary<String, int>(); // Update the frequencies for (int i = 0; i < n1; i++) mp.Add(s1[i], 0); // Update the frequencies for (int i = 0; i < n1; i++) { var v = mp[s1[i]] + 1; mp.Remove(s1[i]); mp.Add(s1[i], v); } // To store the count of pairs int cnt = 0; // For every string of array s2[] for (int i = 0; i < n2; i++) { // If current string can make a pair if (mp[s2[i]] > 0) { // Increment the count of pairs cnt++; // Decrement the frequency of the // string as once occurrence has been // used in the current pair if(mp.ContainsKey(s2[i])) { var v = mp[s2[i]] - 1; mp.Remove(s2[i]); mp.Add(s2[i], v); } else mp.Add(s2[i], mp[s2[i]] - 1); } } // Return the count return cnt; } // Driver code public static void Main (String[] args) { String []s1 = { "abc", "def" }; String []s2 = { "abc", "abc" }; int n1 = s1.Length; int n2 = s2.Length; Console.WriteLine(count_pairs(s1, s2, n1, n2)); }}// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript implementation of the approach// Function to return the count of required pairsfunction count_pairs(s1, s2, n1, n2){ // Map to store the frequencies of // all the strings of array s1[] var mp = new Map(); // Update the frequencies for (var i = 0; i < n1; i++) { if(mp.has(s1[i])) { mp.set(s1[i], mp.get(s1[i])+1) } else { mp.set(s1[i], 1) } } // To store the count of pairs var cnt = 0; // For every string of array s2[] for (var i = 0; i < n2; i++) { // If current string can make a pair if (mp.get(s2[i]) > 0) { // Increment the count of pairs cnt++; // Decrement the frequency of the // string as once occurrence has been // used in the current pair mp.set(s2[i], mp.get(s2[i])-1) } } // Return the count return cnt;}// Driver codevar s1 = ["abc", "def" ];var s2 = ["abc", "abc" ];var n1 = s1.length;var n2 = s2.length;document.write( count_pairs(s1, s2, n1, n2));</script> |
Output
1
Time Complexity: O(n1+n2), Where n1 and n2 are the lengths of given string array s1 and s2 respectively.
Auxiliary Space: O(n1)
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