Python Program to check for almost similar Strings

Last Updated : 02 May, 2023

Given two strings, the task here is to write a python program that can test if they are almost similar. Similarity of strings is being checked on the criteria of frequency difference of each character which should be greater than a threshold here represented by K.

Input : test_str1 = 'aabcdaa', test_str2 = "abbaccd", K = 2
Output : True
Explanation : 'a' occurs 4 times in str1, and 2 times in str2, 4 - 2 = 2, in range, similarly, all chars in range, hence true.

Input : test_str1 = 'aabcdaaa', test_str2 = "abbaccda", K = 3
Output : True
Explanation : 'a' occurs 5 times in str1, and 3 times in str2, 5 - 3 = 2, in range, similarly, all chars in range, hence true

Method 1 : Using ascii_lowecase, dictionary comprehension, loop and abs()

In this, we compute all the frequencies of all the characters in both strings using dictionary comprehension and loop. Next, each character is iterated from alphabetic lowercase ascii characters and tested for frequency difference in both strings using abs(), if any difference computes to greater than K, result is flagged off.

Example

Python3

from string import ascii_lowercase  # function to compute frequencies   def get_freq(test_str):      # starting at 0 count     freqs = {char: 0 for char in ascii_lowercase}      # counting frequencies     for char in test_str:         freqs[char] += 1     return freqs   # initializing strings test_str1 = 'aabcdaa' test_str2 = "abbaccd"  # printing original strings print("The original string 1 is : " + str(test_str1)) print("The original string 2 is : " + str(test_str2))  # initializing K K = 2  # getting frequencies freqs_1 = get_freq(test_str1) freqs_2 = get_freq(test_str2)  # checking for frequencies res = True for char in ascii_lowercase:     if abs(freqs_1[char] - freqs_2[char]) > K:         res = False         break  # printing result print("Are strings similar ? : " + str(res))

Output:

The original string 1 is : aabcdaa

The original string 2 is : abbaccd

Are strings similar ? : True

Method 2 : Using Counter() and max()

In this, we perform task of getting individual characters' frequency using Counter() and get the maximum difference using max(), if greater than K, then result is flagged off.

Example:

Python3

from collections import Counter  # initializing strings test_str1 = 'aabcdaa' test_str2 = "abbaccd"  # printing original strings print("The original string 1 is : " + str(test_str1)) print("The original string 2 is : " + str(test_str2))  # initializing K K = 2  # extracting frequencies cnt1 = Counter(test_str1.lower()) cnt2 = Counter(test_str2.lower())  # getting maximum difference res = True if max((cnt1 - cnt2).values()) > K or max((cnt2 - cnt1).values()) > K:     res = False  # printing result print("Are strings similar ? : " + str(res))

Output:

The original string 1 is : aabcdaa

The original string 2 is : abbaccd

Are strings similar ? : True

Time Complexity: O(n)
Auxiliary Space: O(n)

Method 3 : Using list comprehension:

Approach:

In this program, we define a function is_almost_similar_using_list_comprehension that takes three parameters: test_str1, test_str2, and k.

First, we create two lists counter1 and counter2 using list comprehension, which count the occurrences of each character in the respective strings.
Next, we calculate the absolute difference between the counts of each character in both lists using a for loop and sum() function.
Finally, we check if the total difference is less than or equal to k and return True or False accordingly.

Python3

def is_almost_similar_using_list_comprehension(test_str1, test_str2, k):     counter1 = [test_str1.count(char) for char in set(test_str1)]     counter2 = [test_str2.count(char) for char in set(test_str2)]     diff = sum(abs(counter1[i] - counter2[i]) for i in range(len(counter1)))     return diff >= k  # Testing test_str1 = 'aabcdaaa' test_str2 = 'abbaccda' k = 3 print(f"Input strings: '{test_str1}', '{test_str2}'") print(f"Value of K: {k}") print(f"Output: {is_almost_similar_using_list_comprehension(test_str1, test_str2, k)}")

Output

Input strings: 'aabcdaaa', 'abbaccda' Value of K: 3 Output: True

The time complexity of the is_almost_similar_using_list_comprehension function is O(n), where n is the length of the longer string between test_str1 and test_str2.
The space complexity of the function is also O(n), since we are using two lists counter1 and counter2 to store the character counts of the respective strings.

Method 3: Using set() and count()

Create a set of unique characters in the first string using set(test_str1).
For each unique character in the set, count its frequency in both strings using count() method.
If the absolute difference in frequencies is greater than K for any character, then the strings are not similar.

Python3

# function to compute frequencies def are_strings_similar(test_str1, test_str2, K):     # getting unique characters in test_str1     unique_chars = set(test_str1)      # checking for frequencies     for char in unique_chars:         freq1 = test_str1.count(char)         freq2 = test_str2.count(char)         if abs(freq1 - freq2) > K:             return False     return True  # initializing strings test_str1 = 'aabcdaa' test_str2 = "abbaccd"  # printing original strings print("The original string 1 is : " + str(test_str1)) print("The original string 2 is : " + str(test_str2))  # initializing K K = 2  # checking if strings are similar res = are_strings_similar(test_str1, test_str2, K)  # printing result print("Are strings similar ? : " + str(res))

Output

The original string 1 is : aabcdaa The original string 2 is : abbaccd Are strings similar ? : True

Time Complexity: O(n^2), due to the time complexity of the are_strings_similar() function.
Auxiliary Space: O(k), where k is the number of unique characters in test_str1.

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Python Program to check for almost similar Strings

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