Skip to content
Related Articles

Related Articles

Improve Article

Count common characters in two strings

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

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[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:
 

 

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 <bits/stdc++.h>
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[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 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[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 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




<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>
Output: 



8

 

Time Complexity: O(n1 + n2)
Auxiliary Space: O(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
Recommended Articles
Page :