Count equal pairs from given string arrays

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
Only valid pair is (s1[0], s2[0]) or (s1[0], s2[1])
Note that even though “abc” apperas 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

Approach:

  • Create an unordered_map to store the frequencies of all the string 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++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the count of required pairs
int 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 code
int 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;
}

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Java

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// Java implementation of the approach
import 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

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Python3

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# python 3 implementation of the approach
  
# Function to return the count of required pairs
def 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 code
if __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

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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 mp = new Dictionary();

// 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

Output:

1


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