Check Whether Two Strings are Almost Equivalent

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Two strings word1 and word2 are considered almost equivalent if the differences between the frequencies of each letter from 'a' to 'z' between word1 and word2 is at most 3.

Given two strings word1 and word2, each of length n, return true if word1 and word2 are almost equivalent, or false otherwise.

The frequency of a letter x is the number of times it occurs in the string.

Example 1:

Input: word1 = "aaaa", word2 = "bccb"
Output: false
Explanation: There are 4 'a's in "aaaa" but 0 'a's in "bccb".
The difference is 4, which is more than the allowed 3.

Example 2:

Input: word1 = "abcdeef", word2 = "abaaacc"
Output: true
Explanation: The differences between the frequencies of each letter in word1 and word2 are at most 3:
- 'a' appears 1 time in word1 and 4 times in word2. The difference is 3.
- 'b' appears 1 time in word1 and 1 time in word2. The difference is 0.
- 'c' appears 1 time in word1 and 2 times in word2. The difference is 1.
- 'd' appears 1 time in word1 and 0 times in word2. The difference is 1.
- 'e' appears 2 times in word1 and 0 times in word2. The difference is 2.
- 'f' appears 1 time in word1 and 0 times in word2. The difference is 1.

Example 3:

Input: word1 = "cccddabba", word2 = "babababab"
Output: true
Explanation: The differences between the frequencies of each letter in word1 and word2 are at most 3:
- 'a' appears 2 times in word1 and 4 times in word2. The difference is 2.
- 'b' appears 2 times in word1 and 5 times in word2. The difference is 3.
- 'c' appears 3 times in word1 and 0 times in word2. The difference is 3.
- 'd' appears 2 times in word1 and 0 times in word2. The difference is 2.

Constraints:

n == word1.length == word2.length
1 <= n <= 100
word1 and word2 consist only of lowercase English letters.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

Are the input strings `word1` and `word2` guaranteed to only contain lowercase English letters?
Are the input strings always of equal length, or should I handle cases where they are not?
Is there a specific return value I should use if the input strings are null or empty?
Could you provide a few examples of strings that are considered almost equivalent and some that are not?
Is there a memory constraint that I should be aware of while solving this problem?

Brute Force Solution

Approach

The core idea is to count how many times each letter appears in both phrases. Then, for each letter, we compare its counts and see if the difference ever goes beyond a certain limit. If it does, the phrases are too different to be almost the same.

Here's how the algorithm would work step-by-step:

First, count how many times each letter of the alphabet shows up in the first phrase.
Next, do the same thing and count how many times each letter appears in the second phrase.
Now, go through the alphabet, letter by letter. For each letter, subtract its count in the second phrase from its count in the first phrase.
Check if the result of the subtraction is more than a certain number or less than a certain number. These numbers define how much difference is acceptable.
If at any point this difference is too big, then the two phrases are not considered almost the same. You can stop looking at other letters.
If you go through all the letters and none of the differences are too big, then the two phrases are almost equivalent.

Code Implementation

def are_strings_almost_equivalent(first_word, second_word):
    first_word_letter_counts = {}
    second_word_letter_counts = {}
    
    for letter in first_word:
        first_word_letter_counts[letter] = first_word_letter_counts.get(letter, 0) + 1
    
    for letter in second_word:
        second_word_letter_counts[letter] = second_word_letter_counts.get(letter, 0) + 1

    #Iterate through all 26 lowercase alphabet characters
    for ascii_value in range(ord('a'), ord('z') + 1):
        letter = chr(ascii_value)

        #If the letter does not exist, default count is zero
        first_word_count = first_word_letter_counts.get(letter, 0)
        second_word_count = second_word_letter_counts.get(letter, 0)

        # Check if the absolute difference is greater than the limit
        if abs(first_word_count - second_word_count) > 3:
            return False

    return True

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the characters of both strings once to count their occurrences in frequency arrays. This takes O(n) time, where n is the length of the strings. Then, it iterates through the alphabet (26 characters), performing constant-time operations for each letter. Since the alphabet size is constant, this step takes O(1) time. Therefore, the overall time complexity is dominated by the initial counting step, resulting in O(n).

Space Complexity

O(1) – The provided steps describe counting character frequencies. This can be done using a fixed-size array or hash map (e.g., of size 26 for lowercase English letters). This auxiliary space used to store character counts is independent of the input string lengths. Thus, the space complexity is constant.

Optimal Solution

Approach

The efficient way to solve this problem involves counting how many times each letter appears in both strings. Then, we check if the difference between the counts of each letter in the two strings is small enough.

Here's how the algorithm would work step-by-step:

First, we need a way to keep track of how many times each letter shows up in each string. Imagine having separate containers for each string where we'll count the letters.
Go through the first string, and for each letter we find, increase its count in the first container.
Do the same thing for the second string: go through it and increase the count of each letter in the second container.
Now, for every possible letter of the alphabet, find the difference between its count in the first container and its count in the second container.
If any of these differences is bigger than the allowed threshold, we know the strings are not almost equivalent.
If we go through all the letters and their differences are all within the limit, the strings are almost equivalent.

Code Implementation

def are_almost_equivalent(string1, string2):
    first_string_counts = {}
    second_string_counts = {}

    for char in string1:
        first_string_counts[char] = first_string_counts.get(char, 0) + 1

    for char in string2:
        second_string_counts[char] = second_string_counts.get(char, 0) + 1

    # Iterate through all lowercase letters to check counts.
    for letter_index in range(ord('a'), ord('z') + 1):
        character = chr(letter_index)

        first_string_count = first_string_counts.get(character, 0)
        second_string_count = second_string_counts.get(character, 0)

        #Check if difference exceeds threshold.
        if abs(first_string_count - second_string_count) > 3:
            return False

    return True

Big(O) Analysis

Time Complexity

O(n) – We iterate through the first string of length n to count character frequencies. Then we iterate through the second string of length n to count character frequencies. Finally, we iterate through the alphabet, which has a constant size (26), to compare the frequency differences. The dominant operation is iterating through the strings, each taking O(n) time. The comparison of character frequencies takes O(1) time as the alphabet size is fixed. Thus, the overall time complexity is O(n) + O(n) + O(1) = O(n).

Space Complexity

O(1) – The algorithm uses two containers (likely arrays or hash maps) to store the counts of each letter for both strings. Since we are dealing with the English alphabet, each container will have a fixed size of 26, regardless of the input string lengths. Therefore, the auxiliary space used is constant. The space does not scale with the input size N (where N is the length of the input strings).

Edge Cases

Case	How to Handle
Null or empty input strings	Return true if both are null or empty, or false if one is null/empty and the other isn't, and throw an exception if null checks are not desired.
Strings of length 1 or 2	The frequency check should work correctly for short strings; the core logic remains the same.
Strings with maximum length	The solution should use a frequency map approach for efficiency to handle large strings within reasonable memory constraints and avoid excessive iterations.
Strings where all characters are identical	The frequency difference for the repeated character will be within the limit, likely zero, hence returning true.
Strings with highly skewed character distribution (e.g., one character dominates)	The frequency comparison will accurately reflect the difference even if one character is much more frequent than others.
Strings with characters outside 'a' to 'z'	The solution should either ignore or throw an error if it encounters characters other than lowercase English letters, depending on requirements.
Strings with equal length but differing characters that result in frequencies exceeding the threshold	The frequency difference for at least one letter will be greater than 3, resulting in the function returning false.
Language-specific memory constraints with extremely long strings	Consider using an iterative solution with constant space for frequency counting instead of creating new strings, optimizing memory usage.

Null or empty input strings

How to Handle:

Return true if both are null or empty, or false if one is null/empty and the other isn't, and throw an exception if null checks are not desired.

Strings of length 1 or 2

How to Handle:

The frequency check should work correctly for short strings; the core logic remains the same.

Strings with maximum length

How to Handle:

The solution should use a frequency map approach for efficiency to handle large strings within reasonable memory constraints and avoid excessive iterations.

Strings where all characters are identical

How to Handle:

The frequency difference for the repeated character will be within the limit, likely zero, hence returning true.

Strings with highly skewed character distribution (e.g., one character dominates)

How to Handle:

The frequency comparison will accurately reflect the difference even if one character is much more frequent than others.

Strings with characters outside 'a' to 'z'

How to Handle:

The solution should either ignore or throw an error if it encounters characters other than lowercase English letters, depending on requirements.

Strings with equal length but differing characters that result in frequencies exceeding the threshold

How to Handle:

The frequency difference for at least one letter will be greater than 3, resulting in the function returning false.

Language-specific memory constraints with extremely long strings

How to Handle:

Consider using an iterative solution with constant space for frequency counting instead of creating new strings, optimizing memory usage.