Count the Number of matching characters in a pair of strings
Given a pair of non-empty strings str1 and str2, the task is to count the number of matching characters in these strings. Consider the single count for the character which have duplicates in the strings.
Examples:
Input: str1 = “abcdef”, str2 = “defghia”
Output: 4
Matching characters are: a, d, e, fInput: str1 = “aabcddekll12”, str2 = “bb22ll@55k”
Output: 5
Matching characters are: b, 1, 2, @, k
Approach:
- Initialize a counter variable with 0.
- Iterate over the first string from the starting character to ending character.
- If the character extracted from the first string is found in the second string, then increment the value of the counter by 1.
- The final answer will be count/2 as the duplicates are not being considered.
- Output the value of counter
Below is the implementation of the above approach.
C++
// C++ code to count number of matching // characters in a pair of strings #include <bits/stdc++.h> using namespace std; // Function to count the matching characters void count(string str1, string str2) { int c = 0, j = 0; // Traverse the string 1 char by char for (int i = 0; i < str1.length(); i++) { // This will check if str1[i] // is present in str2 or not // str2.find(str1[i]) returns -1 if not found // otherwise it returns the starting occurrence // index of that character in str2 if (str2.find(str1[i]) >= 0 and j == str1.find(str1[i])) c += 1; j += 1; } cout << "No. of matching characters are: " << c / 2; } // Driver code int main() { string str1 = "aabcddekll12@"; string str2 = "bb2211@55k"; count(str1, str2); } |
chevron_right
filter_none
Java
// Java code to count number of matching // characters in a pair of strings class GFG { // Function to count the matching characters static void count(String str1, String str2) { int c = 0, j = 0; // Traverse the string 1 char by char for (int i = 0; i < str1.length(); i++) { // This will check if str1[i] // is present in str2 or not // str2.find(str1[i]) returns -1 if not found // otherwise it returns the starting occurrence // index of that character in str2 if (str2. indexOf(str1.charAt(i)) >= 0) { c += 1; } } System.out.println("No. of matching characters are: " + c); } // Driver code public static void main (String[] args) { String str1 = "aabcddekll12@"; String str2 = "bb2211@55k"; count(str1, str2); } } // This code is contributed by AnkitRai01 |
chevron_right
filter_none
Python3
# Python3 code to count number of matching # characters in a pair of strings # Function to count the matching characters def count(str1, str2) : c = 0; j = 0; # Traverse the string 1 char by char for i in range(len(str1)) : # This will check if str1[i] # is present in str2 or not # str2.find(str1[i]) returns -1 if not found # otherwise it returns the starting occurrence # index of that character in str2 if str1[i] in str2 : c += 1; #print(str1[i]) j += 1; print("No. of matching characters are: ", c ); # Driver code if __name__ == "__main__" : str1 = "aabcddekll12@"; str2 = "bb2211@55k"; count(str1, str2); # This code is contributed by AnkitRai01 |
chevron_right
filter_none
C#
// C# code to count number of matching // characters in a pair of strings using System; class GFG { // Function to count the matching characters static void count(string str1, string str2) { int c = 0, j = 0; // Traverse the string 1 char by char for (int i = 0; i < str1.Length; i++) { // This will check if str1[i] // is present in str2 or not // str2.find(str1[i]) returns -1 if not found // otherwise it returns the starting occurrence // index of that character in str2 if (str2.IndexOf(str1[i]) >= 0) { c += 1; } } Console.WriteLine("No. of matching characters are: " + c); } // Driver code public static void Main() { string str1 = "aabcddekll12@"; string str2 = "bb2211@55k"; count(str1, str2); } } // This code is contributed by AnkitRai01 |
chevron_right
filter_none
Output:
No. of matching characters are: 5
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.
