Count common characters in two strings

25 Nov, 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: 2 (s1[0], s2[0]) and (s1[3], s2[2]) are the only valid pairs. (s1[0], s2[1]) is not includes because s1[0] has already been paired with s2[0]Input: s1 = “geeksforgeeks”, s2 = “platformforgeeks” Output: 8  
 
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++
Java
Python3
C#
Javascript
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std; // Function to return the count of// valid indices pairsint countPairs(string s1, int n1, string s2, int n2){     // To store the frequencies of characters    // of string s1 and s2    int freq1[26] = { 0 };    int freq2[26] = { 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 codeint main(){    string s1 = "geeksforgeeks", s2 = "platformforgeeks";    int n1 = s1.length(), n2 = s2.length();    cout << countPairs(s1, n1, s2, n2);     return 0;}
// Java implementation of the approachimport java.util.*; class GFG{ // Function to return the count of// valid indices pairsstatic 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 codepublic 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 implementation of the approach # Function to return the count of# valid indices pairsdef 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 codeif __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# implementation of the approachusing System; class GFG{ // Function to return the count of// valid indices pairsstatic 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];    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 codepublic 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
<script>     // JavaScript implementation of the approach         // Function to return the count of    // valid indices pairs    function countPairs(s1, n1, s2, n2)    {         // To store the frequencies of        // characters of string s1 and s2        let freq1 = new Array(26);        let freq2 = new Array(26);        freq1.fill(0);        freq2.fill(0);         // To store the count of valid pairs        let i, count = 0;         // Update the frequencies of        // the characters of string s1        for (i = 0; i < n1; i++)            freq1[s1[i].charCodeAt() - 'a'.charCodeAt()]++;         // Update the frequencies of        // the characters of string s2        for (i = 0; i < n2; i++)            freq2[s2[i].charCodeAt() - 'a'.charCodeAt()]++;         // Find the count of valid pairs        for (i = 0; i < 26; i++)            count += (Math.min(freq1[i], freq2[i]));         return count;    }         let s1 = "geeksforgeeks", s2 = "platformforgeeks";    let n1 = s1.length, n2 = s2.length;    document.write(countPairs(s1, n1, s2, n2));     </script>
8
 
Time Complexity: O(max(n1, n2))Auxiliary Space: O(1)
ankthon
SURENDRA_GANGWAR
Akanksha_Rai
suresh07
pankajsharmagfg
ashutoshsinghgeeksforgeeks
frequency-counting
Algorithms
Competitive Programming
Searching
Strings
Searching
Strings
Algorithms
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