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
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)
}
|
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
def commonfun(str1,str2):
return ( len (( set (str1)).intersection( set (str2))))
string1 = "VISHAV"
string2 = "VANSHIKA"
no_of_common_character = commonfun(string1.lower(),string2.lower())
print ( "NO. OF COMMON CHRACTERS ARE : " ,no_of_common_character)
|
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):
char_count = {}
for char in str1:
if char in char_count:
char_count[char] + = 1
else :
char_count[char] = 1
counter = 0
for char in str2:
if char in char_count and char_count[char] > 0 :
counter + = 1
char_count[char] - = 1
print ( "No. of matching characters are: " + str (counter))
if __name__ = = "__main__" :
str1 = 'aabcddekll12@'
str2 = 'bb2211@55k'
count(str1, str2)
|
Output
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
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Last Updated :
02 Aug, 2023
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