Date: 2026-06-06
Time: 19:08
slowest-key/solution.pyThis file solves LeetCode 1629: Slowest Key. It determines which key was held the longest based on an array of cumulative release times and a corresponding string of keys. It's one solution module in a repo of ~500+ LeetCode problems, each following the same solution.py / test_solution.py / review.md convention.
minInteger(releaseTimes, keysPressed) -> str — The sole exported function. Despite the misleading name (minInteger suggests something numeric), it finds the key with the longest press duration, breaking ties by choosing the lexicographically largest key.
Contract:
releaseTimes is a list of strictly increasing integers representing cumulative timestamps. keysPressed is a string of the same length, where keysPressed[i] is the key released at releaseTimes[i].'b' > 'a').Single-pass greedy tracking. The algorithm maintains a running best (bestkey, bestdur) and updates in one linear scan — a standard pattern for "find the max with a secondary comparator" problems. No sorting, no auxiliary data structures.
Implicit first-element handling. The first keypress duration equals releaseTimes[0] (time since 0). Subsequent durations are computed as releaseTimes[i] - releaseTimes[i-1]. The code initializes best_dur = releaseTimes[0] to handle this asymmetry without a special case inside the loop.
typing.List — no external libraries.List from typing, not this module. The actual consumer is slowest-key/test_solution.py.1. Initialize bestkey to the first character and bestdur to releaseTimes[0] (duration of the first keypress, measured from time 0).
2. Iterate i from 1 to len(releaseTimes) - 1:
dur = releaseTimes[i] - releaseTimes[i-1].dur strictly exceeds bestdur, or if dur equals bestdur and the current key is lexicographically greater.3. Return best_key.
releaseTimes is assumed sorted and strictly increasing (per the problem constraints). The code does not validate this.len(releaseTimes) == len(keysPressed) and both are non-empty. No empty-input guard exists.keysPressed[i] > best_key) relies on Python's native character comparison, which is lexicographic by Unicode code point — correct for lowercase English letters.None. If releaseTimes or keysPressed is empty, line 13 (keysPressed[0]) raises an IndexError. This is consistent with the repo's convention of trusting LeetCode's input guarantees.
slowest-key/test_solution.py — See which edge cases are covered (ties, single key, all same duration)slowest-key/review.md — Likely contains a code review with complexity analysis and alternative approachesthe-employee-that-worked-on-the-longest-task/solution.py:minInteger — Structurally identical problem (longest task from cumulative times), good for comparing patternsfunction-naming-conventions — minInteger is a misnomer carried over from LeetCode's class template; worth checking if other solutions have similar naming driftslowest-key/plan.md — Documents the approach chosen before implementationslowest-key-single-pass — The solution runs in O(n) time and O(1) space with exactly one pass over the inputslowest-key-tiebreak-largest — Ties in duration are broken by choosing the lexicographically largest key, not the smallestslowest-key-first-duration-from-zero — The first keypress duration is releaseTimes[0] (implicitly measuring from time 0), not a delta from a previous elementslowest-key-no-input-validation — The function assumes non-empty inputs and strictly increasing release times without any defensive checks