Related Articles

# Count equal pairs from given string arrays

• Difficulty Level : Easy
• Last Updated : 10 May, 2021

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:
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:

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`

## Javascript

 ``
Output:
`1`

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes arrow_drop_up