Count equal pairs from given string arrays

Last Updated : 03 Feb, 2023

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 pairs
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
int 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 pairs
def 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 code
s1 = [ "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 idea
using 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 pairs
function 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

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 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;
}

Java

// 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

Python3

# 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

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 pairs
function 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 code
var 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

Time Complexity: O(n1+n2), Where n1 and n2 are the lengths of given string array s1 and s2 respectively.
Auxiliary Space: O(n1)

Count unequal element pairs from the given Array

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Count equal pairs from given string arrays

C++

Java

Python3

C#

Javascript

C++

Java

Python3

C#

Javascript

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