# 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, s2) or (s1, s2)
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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 `// C++ implementation of the approach ` `#include ` `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 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 mp = ``new` `HashMap();  ` ` `  `        ``// 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 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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