Check Whether Two Strings are Almost Equivalent - Solution & Explanation

Problem Overview: You receive two equal-length strings word1 and word2. The task is to check whether the difference in frequency of every lowercase letter between the two strings is at most 3. If any character appears more than 3 times more in one string than the other, the strings are not almost equivalent.

The key observation: the exact positions of characters do not matter. Only the frequency of each letter matters. That immediately points toward a hash table or frequency counting solution.

Approach 1: Frequency Counting with Arrays (O(n) time, O(1) space)

Use two fixed-size arrays of length 26 to store counts of each lowercase letter. Iterate through both strings once and increment the appropriate index (c - 'a') for each character. After building the frequency arrays, iterate from 0..25 and compute the absolute difference for each letter. If any difference exceeds 3, return false. Otherwise the strings are almost equivalent.

This works because the alphabet size is constant. The comparison step always checks 26 entries regardless of input length, so the space remains constant. This approach is usually the fastest in practice and avoids hashing overhead.

Approach 2: Using HashMaps (O(n) time, O(k) space)

Store character frequencies using a HashMap for each string. Traverse word1 and word2, updating counts for each character. Then iterate through all lowercase letters or the union of keys in both maps. Compute the absolute difference between the counts from the two maps. If any difference is greater than 3, return false.

This approach is more flexible when the character set is not limited to lowercase letters. Hash maps automatically expand for any character set, making the solution adaptable beyond the typical string problems restricted to 26 letters. The tradeoff is slightly higher constant overhead compared to arrays.

Both methods rely on the same idea: compare per-character frequency differences using a simple counting technique. The difference threshold (3) becomes a straightforward validation step after computing counts.

Recommended for interviews: The array-based frequency counting approach. Interviewers expect candidates to recognize that lowercase letters allow a fixed-size counting array. It runs in O(n) time and O(1) space and demonstrates strong understanding of character frequency techniques. Mentioning the HashMap variant shows awareness of how the solution generalizes when the alphabet size is not fixed.