Remove Letter To Equalize Frequency

Remove Letter To Equalize Frequency

What is this problem about?

The Remove Letter To Equalize Frequency interview question asks whether you can remove exactly one character from a given string such that all remaining characters have the same frequency. For example, if removing one 'a' causes every remaining character to appear exactly 3 times, return true; otherwise, return false. This is a frequency analysis problem with careful case enumeration.

Why is this asked in interviews?

This problem is asked at TCS, Walmart Labs, Google, and Bloomberg because it tests a candidate's ability to reason about frequency distributions and enumerate subtle edge cases. Despite appearing simple, there are many tricky scenarios — a string with all identical characters, a string where one character appears once, or a string where one character appears one more time than all others. Careful enumeration of valid conditions is the skill being evaluated.

Algorithmic pattern used

The pattern is hash table counting with case analysis. Count the frequency of each character. Then count the frequency of those frequencies (a "meta-frequency" map). A valid removal exists in these cases:

1. Only one distinct character exists (remove one copy, rest are equal).
2. All characters have frequency 1 (remove any one character).
3. One character has frequency 1 and the rest have the same frequency.
4. One character has frequency f+1 and appears exactly once extra compared to the uniform f.
5. All characters have the same frequency and (total_chars + 1) == 26 (one more character added would still equalize).

Example explanation

String: "aaabbbccc" — each appears 3 times. Remove any 'a' → "aabbbccc", now a=2, b=3, c=3. Not equal. But this input returns false.

String: "aaabbb" — a=3, b=3. Removing one 'a' gives a=2, b=3. Not equal. Returns false.

String: "aabb" — a=2, b=2. Removing one 'a' gives a=1, b=2. Not equal. But only 2 distinct characters — returns false.

String: "aab" — a=2, b=1. Removing 'b' gives a=2. Only 'a' remains, all equal. Returns true.

Common mistakes candidates make

• Not considering the case where removing the single extra character leaves an empty string (which vacuously satisfies the condition).
• Missing the case where all characters appear with the same frequency and removing one reduces a character to 0.
• Overcomplicating by simulating removal for each character rather than reasoning from the frequency-of-frequencies map.

Interview preparation tip

For the Remove Letter To Equalize Frequency coding problem, the hash table and counting interview pattern is the key. Build the frequency map, then build the frequency-of-frequency map. Enumerate valid configurations rather than simulating. Google interviewers appreciate when you list all valid cases before coding — it signals thoroughness and reduces bugs.