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, f
Input: 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++ 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);
} |
// 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 |
# 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 |
// 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 |
<script> // JavaScript code to count number of matching // characters in a pair of strings // Function to count the matching characters function count(str1, str2)
{ var c = 0;
// Traverse the string 1 char by char
for ( var 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.includes(str1[i]))
c += 1;
}
document.write( "No. of matching characters are: " +
+ parseInt(c));
} // Driver code var str1 = "aabcddekll12@" ;
var str2 = "bb2211@55k" ;
count(str1, str2); </script> |
No. of matching characters are: 5
Time Complexity: O(m * (n + m)), where m and n are the length of the given strings str1 and str2 respectively.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Efficient Approach:
Implementation:
#include <bits/stdc++.h> using namespace std;
// Function to count number of matching pairs int countMatchingCharacters(string str1, string str2)
{ unordered_map< char , int > charCount;
int count = 0;
// Count characters in str1
for ( char c : str1) {
charCount++;
}
// Check for matching characters in str2
for ( char c : str2) {
if (charCount > 0) {
count++;
charCount--;
}
}
return count;
} // Driver code int main()
{ string str1 = "abcdef" ;
string str2 = "defghia" ;
// Function call
cout << countMatchingCharacters(str1, str2);
return 0;
} |
import java.util.HashMap;
import java.util.Map;
public class GFG {
// Function to count the number of matching pairs
public static int countMatchingCharacters(String str1, String str2) {
// Create a HashMap to store character frequencies in str1
Map<Character, Integer> charCount = new HashMap<>();
int count = 0 ;
// Count characters in str1
for ( char c : str1.toCharArray()) {
charCount.put(c, charCount.getOrDefault(c, 0 ) + 1 );
}
// Check for matching characters in str2
for ( char c : str2.toCharArray()) {
if (charCount.containsKey(c) && charCount.get(c) > 0 ) {
count++;
charCount.put(c, charCount.get(c) - 1 );
}
}
return count;
}
public static void main(String[] args) {
String str1 = "abcdef" ;
String str2 = "defghia" ;
// Function call
int matchingPairs = countMatchingCharacters(str1, str2);
System.out.println( "Number of matching pairs: " + matchingPairs);
}
} |
# Function to count the number of matching characters def count_matching_characters(str1, str2):
char_count = {}
count = 0
# Count characters in str1
for c in str1:
if c in char_count:
char_count + = 1
else :
char_count = 1
# Check for matching characters in str2
for c in str2:
if c in char_count and char_count > 0 :
count + = 1
char_count - = 1
return count
# Driver code if __name__ = = "__main__" :
str1 = "abcdef"
str2 = "defghia"
# Function call
print (count_matching_characters(str1, str2))
|
using System;
using System.Collections.Generic;
public class GFG
{ // Function to count number of matching pairs
public static int CountMatchingCharacters( string str1, string str2)
{
Dictionary< char , int > charCount = new Dictionary< char , int >();
int count = 0;
// Count characters in str1
foreach ( char c in str1)
{
if (charCount.ContainsKey(c))
{
charCount++;
}
else
{
charCount = 1;
}
}
// Check for matching characters in str2
foreach ( char c in str2)
{
if (charCount.ContainsKey(c) && charCount > 0)
{
count++;
charCount--;
}
}
return count;
}
// Driver code
public static void Main( string [] args)
{
string str1 = "abcdef" ;
string str2 = "defghia" ;
// Function call
Console.WriteLine(CountMatchingCharacters(str1, str2));
}
} |
// Function to count the number of matching pairs function countMatchingCharacters(str1, str2) {
// Create an object to count characters in str1
const charCount = {};
let count = 0;
// Count characters in str1
for (const c of str1) {
if (charCount === undefined) {
charCount = 1;
} else {
charCount++;
}
}
// Check for matching characters in str2
for (const c of str2) {
if (charCount > 0) {
count++;
charCount--;
}
}
return count;
} // Driver code function main() {
const str1 = "abcdef" ;
const str2 = "defghia" ;
// Function call
console.log(countMatchingCharacters(str1, str2));
} // Run the main function main(); |
4
Time Complexity: O(n),where n is the size of the string.
Auxiliary Space: O(K), where K is the number of unique characters in str1.