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# Count common characters in two strings

• Difficulty Level : Basic
• Last Updated : 02 Aug, 2021

Given two strings s1 and s2 consisting of lowercase English alphabets, the task is to count all the pairs of indices (i, j) from the given strings such that s1[i] = s2[j] and all the indices are distinct i.e. if s1[i] pairs with some s2[j] then these two characters will not be paired with any other character.
Example

Input: s1 = “abcd”, s2 = “aad”
Output:
(s1, s2) and (s1, s2) are the only valid pairs.
(s1, s2) is not includes because s1 has already been paired with s2
Input: s1 = “geeksforgeeks”, s2 = “platformforgeeks”
Output:

Approach: Count the frequencies of all the characters from both strings. Now, for every character if the frequency of this character in string s1 is freq1 and in string s2 is freq2 then total valid pairs with this character will be min(freq1, freq2). The sum of this value for all the characters is the required answer.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to return the count of``// valid indices pairs``int` `countPairs(string s1, ``int` `n1, string s2, ``int` `n2)``{` `    ``// To store the frequencies of characters``    ``// of string s1 and s2``    ``int` `freq1 = { 0 };``    ``int` `freq2 = { 0 };` `    ``// To store the count of valid pairs``    ``int` `i, count = 0;` `    ``// Update the frequencies of``    ``// the characters of string s1``    ``for` `(i = 0; i < n1; i++)``        ``freq1[s1[i] - ``'a'``]++;` `    ``// Update the frequencies of``    ``// the characters of string s2``    ``for` `(i = 0; i < n2; i++)``        ``freq2[s2[i] - ``'a'``]++;` `    ``// Find the count of valid pairs``    ``for` `(i = 0; i < 26; i++)``        ``count += (min(freq1[i], freq2[i]));` `    ``return` `count;``}` `// Driver code``int` `main()``{``    ``string s1 = ``"geeksforgeeks"``, s2 = ``"platformforgeeks"``;``    ``int` `n1 = s1.length(), n2 = s2.length();``    ``cout << countPairs(s1, n1, s2, n2);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``import` `java.util.*;` `class` `GFG``{` `// Function to return the count of``// valid indices pairs``static` `int` `countPairs(String s1, ``int` `n1,``                        ``String s2, ``int` `n2)``{` `    ``// To store the frequencies of characters``    ``// of string s1 and s2``    ``int` `[]freq1 = ``new` `int``[``26``];``    ``int` `[]freq2 = ``new` `int``[``26``];``    ``Arrays.fill(freq1, ``0``);``    ``Arrays.fill(freq2, ``0``);` `    ``// To store the count of valid pairs``    ``int` `i, count = ``0``;` `    ``// Update the frequencies of``    ``// the characters of string s1``    ``for` `(i = ``0``; i < n1; i++)``        ``freq1[s1.charAt(i) - ``'a'``]++;` `    ``// Update the frequencies of``    ``// the characters of string s2``    ``for` `(i = ``0``; i < n2; i++)``        ``freq2[s2.charAt(i) - ``'a'``]++;` `    ``// Find the count of valid pairs``    ``for` `(i = ``0``; i < ``26``; i++)``        ``count += (Math.min(freq1[i], freq2[i]));` `    ``return` `count;``}` `// Driver code``public` `static` `void` `main(String args[])``{``    ``String s1 = ``"geeksforgeeks"``, s2 = ``"platformforgeeks"``;``    ``int` `n1 = s1.length(), n2 = s2.length();``    ``System.out.println(countPairs(s1, n1, s2, n2));``}``}` `// This code is contributed by``// Surendra_Gangwar`

## Python3

 `# Python3 implementation of the approach` `# Function to return the count of``# valid indices pairs``def` `countPairs(s1, n1, s2, n2) :` `    ``# To store the frequencies of characters``    ``# of string s1 and s2``    ``freq1 ``=` `[``0``] ``*` `26``;``    ``freq2 ``=` `[``0``] ``*` `26``;` `    ``# To store the count of valid pairs``    ``count ``=` `0``;` `    ``# Update the frequencies of``    ``# the characters of string s1``    ``for` `i ``in` `range``(n1) :``        ``freq1[``ord``(s1[i]) ``-` `ord``(``'a'``)] ``+``=` `1``;` `    ``# Update the frequencies of``    ``# the characters of string s2``    ``for` `i ``in` `range``(n2) :``        ``freq2[``ord``(s2[i]) ``-` `ord``(``'a'``)] ``+``=` `1``;` `    ``# Find the count of valid pairs``    ``for` `i ``in` `range``(``26``) :``        ``count ``+``=` `min``(freq1[i], freq2[i]);` `    ``return` `count;` `# Driver code``if` `__name__ ``=``=` `"__main__"` `:` `    ``s1 ``=` `"geeksforgeeks"``;``    ``s2 ``=` `"platformforgeeks"``;``    ` `    ``n1 ``=` `len``(s1) ;``    ``n2 ``=` `len``(s2);``    ` `    ``print``(countPairs(s1, n1, s2, n2));` `# This code is contributed by Ryuga`

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG``{` `// Function to return the count of``// valid indices pairs``static` `int` `countPairs(``string` `s1, ``int` `n1,``                      ``string` `s2, ``int` `n2)``{` `    ``// To store the frequencies of``    ``// characters of string s1 and s2``    ``int` `[]freq1 = ``new` `int``;``    ``int` `[]freq2 = ``new` `int``;``    ``Array.Fill(freq1, 0);``    ``Array.Fill(freq2, 0);` `    ``// To store the count of valid pairs``    ``int` `i, count = 0;` `    ``// Update the frequencies of``    ``// the characters of string s1``    ``for` `(i = 0; i < n1; i++)``        ``freq1[s1[i] - ``'a'``]++;` `    ``// Update the frequencies of``    ``// the characters of string s2``    ``for` `(i = 0; i < n2; i++)``        ``freq2[s2[i] - ``'a'``]++;` `    ``// Find the count of valid pairs``    ``for` `(i = 0; i < 26; i++)``        ``count += (Math.Min(freq1[i],``                           ``freq2[i]));` `    ``return` `count;``}` `// Driver code``public` `static` `void` `Main()``{``    ``string` `s1 = ``"geeksforgeeks"``,``           ``s2 = ``"platformforgeeks"``;``    ``int` `n1 = s1.Length, n2 = s2.Length;``    ``Console.WriteLine(countPairs(s1, n1, s2, n2));``}``}` `// This code is contributed by``// Akanksha Rai`

## Javascript

 ``
Output:

`8`

Time Complexity: O(n1 + n2)
Auxiliary Space: O(1)

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