Date: 2026-06-06
Time: 18:47
remove-letter-to-equalize-frequency/solution.pyThis file solves LeetCode 2423: Remove Letter To Equalize Frequency. It determines whether you can remove exactly one character from a string so that every remaining distinct character appears the same number of times.
canequalfrequency(word: str) -> bool — The sole public function. Takes a lowercase letter string of length 2–100 and returns True if deleting exactly one character makes all remaining character frequencies equal.
Brute-force simulation. Rather than reasoning about frequency counts analytically (which is notoriously error-prone for this problem due to many edge cases), the solution tries every possible single deletion and checks whether the result has uniform frequency. This is the "just simulate it" idiom — trade O(n^2) time for zero risk of missing a corner case.
The uniformity check len(set(counts.values())) == 1 is a standard Python idiom: if all values in a Counter are equal, the set of values has exactly one element.
collections.Counter — used to count character frequencies in each candidate string.remove-letter-to-equalize-frequency/test_solution.py and ~400+ other test files (the "Imported By" list in the prompt is the test harness importing Counter from collections, not this file specifically).1. Iterate i over every index [0, len(word)).
2. For each i, build a new string with index i removed via word[:i] + word[i+1:].
3. Count character frequencies of the shortened string with Counter.
4. Check if all frequencies are identical: len(set(counts.values())) == 1.
5. If any deletion produces uniform frequencies, return True immediately (short-circuit).
6. If no deletion works, return False.
Counter of an empty string is empty, so set(counts.values()) would be an empty set with length 0, not 1. This doesn't arise in practice because len(word) >= 2 per the problem constraints, so the shortened string always has at least one character.None. The function assumes valid input per the LeetCode contract (lowercase English letters, length 2–100). No exceptions are raised or caught.
Counter each use O(n) space.For n ≤ 100 this is trivially fast. An O(n) analytical approach exists but is much harder to get right — this problem is infamous for tricky edge cases (e.g., "aazz", "abc", "aaaa").
remove-letter-to-equalize-frequency/test_solution.py — See which edge cases the test suite covers (single-frequency strings, all-same characters, length-2 inputs)remove-letter-to-equalize-frequency/review.md — Check if the review flags the O(n^2) approach or discusses the analytical alternativecheck-if-all-characters-have-equal-number-of-occurrences/solution.py:areOccurrencesEqual — Related problem (equal frequency without any removal) — contrast the simpler checkbrute-force-vs-analytical-frequency-problems — This problem is a well-known trap where analytical solutions have subtle bugs; understand when brute force is the right callbrute-force-deletion-correctness — The brute-force approach of trying every single-index deletion is correct for all inputs within the problem's constraints (length 2–100) and avoids the edge-case bugs common in O(n) analytical solutions.uniform-frequency-check-idiom — len(set(counter.values())) == 1 is the canonical Python check for whether all character frequencies are equal.quadratic-acceptable-for-n-100 — The O(n^2) time complexity is well within limits for the constraint n ≤ 100, making optimization unnecessary.no-empty-counter-risk — The Counter is never called on an empty string because the problem guarantees len(word) >= 2, so the shortened string always has at least one character.