Given a pair of non-empty strings. Count the number of matching characters in those strings (consider the single count for the character which have duplicates in the strings).
Examples:
Input : str1 = 'abcdef'
str2 = 'defghia'
Output : 4
(i.e. matching characters :- a, d, e, f)
Input : str1 = 'aabcddekll12@'
str2 = 'bb22ll@55k'
Output : 5
(i.e. matching characters :- b, 1, 2, @, k)
Approach 1: 1. Initialize a counter variable with 0. 2. Iterate over the first string from the starting character to ending character. 3. If the character extracted from first string is found in the second string and also first occurrence index of that extracted character in first string is same as that of index of current extracted character then increment the value of counter by 1.
Note: For this, use string.find(character) in python. This returns the first occurrence index of character in string, if found, otherwise return -1. For example : str=’abcdedde’ str.find(‘d’) –> 3 str.find(‘e’) –> 4 str.find(‘g’) –> -1
4. Output the value of counter. Below is the implementation of above approach.
Python
// 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 charactersvoid 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 codeint main(){ string str1 = "aabcddekll12@" string str2 = "bb2211@55k" count(str1, str2)} |
Output :
No. of matching characters are : 5
Approach 2: 1.In this approach set() is used to remove duplicate on a given string. 2.After that concept of set(intersection) is used on given string. 3.After that we find a length using len() method.
Python3
# Python code to find no. of common characters between two strings#first covert both string to set and then find common character between twodef commonfun(str1,str2): return(len((set(str1)).intersection(set(str2))))#string1=input("Enter String 1 : ")string1="VISHAV"string2="VANSHIKA"#string2=input("Enter String 2 : ")no_of_common_character=commonfun(string1.lower(),string2.lower())print("NO. OF COMMON CHRACTERS ARE : ",no_of_common_character) #this code is contributed by VISHAV PRATAP SINGH RANA(UIET KURUKSHETRA) |
Output :
NO. OF COMMON CHRACTERS ARE : 5
Approach 3:
- Initialize a dictionary to keep count of each character in the first string.
- Iterate over the first string and update the count of each character in the dictionary.
- Initialize a counter variable with 0.
- Iterate over the second string and check if the character is present in the dictionary and its count is greater than 0. If yes, increment the counter and decrement the count of the character in the dictionary.
- Print the final value of the counter variable.
Python3
def count(str1, str2): # Initialize an empty dictionary to keep track of the count of each character in str1. char_count = {} # Iterate over each character in str1. for char in str1: # If the character is already in the dictionary, increment its count. if char in char_count: char_count[char] += 1 # Otherwise, add the character to the dictionary with a count of 1. else: char_count[char] = 1 # Initialize a counter variable to 0. counter = 0 # Iterate over each character in str2. for char in str2: # If the character is in the char_count dictionary and its count is greater than 0, increment the counter and decrement the count. if char in char_count and char_count[char] > 0: counter += 1 char_count[char] -= 1 # Print the number of matching characters. print("No. of matching characters are: " + str(counter))# Driver codeif __name__ == "__main__": # Define two strings to compare. str1 = 'aabcddekll12@' str2 = 'bb2211@55k' # Call the count function with the two strings. count(str1, str2) |
No. of matching characters are: 5
Time complexity: O(m+n), where m is the length of str1 and n is the length of str2. This is because we are iterating through each character in both strings once, and the time it takes to look up a character in a dictionary is constant.
Auxiliary space: O(k), where k is the number of distinct characters in str1. This is because we are creating a dictionary to store the count of each character in str1
