How do you solve Check Whether Two Strings are Almost Equivalent on LeetCode?

Check Whether Two Strings are Almost Equivalent (LeetCode #2068) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Using frequency counts allows for efficient comparison of character occurrences.

What is the brute force solution for Check Whether Two Strings are Almost Equivalent?

The brute force approach for Check Whether Two Strings are Almost Equivalent is Brute Force, with O(n²) time complexity and O(1) space complexity. We can compare the frequency of each character in both strings by counting them individually. This method is straightforward but inefficient as it checks each character multiple times. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Check Whether Two Strings are Almost Equivalent?

Check Whether Two Strings are Almost Equivalent uses the following concepts: Hash Table, String, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Check Whether Two Strings are Almost Equivalent?

Always clarify the problem statement and edge cases with the interviewer. Think about the data structures that can simplify counting and comparison tasks. Explain your thought process and reasoning as you write the code.

What are common mistakes when solving Check Whether Two Strings are Almost Equivalent?

Not considering characters that are present in one string but not the other. Failing to account for all characters from 'a' to 'z' when checking differences.

#2068

Check Whether Two Strings are Almost Equivalent

Easy

Hash TableStringCountingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

By using a frequency count with a single pass for each string, we can efficiently determine the differences without redundant checks.

⚙️

Algorithm

3 steps

1Step 1: Use a frequency array of size 26 to count occurrences of each character in both strings.
2Step 2: For each character, calculate the absolute difference in frequency counts.
3Step 3: If any difference exceeds 3, return false. If all differences are within 3, return true.

solution.py11 lines

1# Full working Python code
2def areAlmostEquivalent(word1, word2):
3    count = [0] * 26
4    for c in word1:
5        count[ord(c) - ord('a')] += 1
6    for c in word2:
7        count[ord(c) - ord('a')] -= 1
8    for freq in count:
9        if abs(freq) > 3:
10            return False
11    return True

ℹ

Complexity note: The time complexity is O(n) since we only traverse each string once. The space complexity is O(1) because the frequency array size is constant (26 for the alphabet).

1Using frequency counts allows for efficient comparison of character occurrences.
2The absolute difference check is crucial for determining equivalence.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.