Date: 2026-06-07
Time: 04:52
This file solves LeetCode 2383: Minimum Hours of Training to Win a Competition. It computes the minimum number of training hours (each hour adds +1 to either energy or experience) needed so a player can defeat every opponent in sequence. The player must have strictly greater energy and experience than each opponent at the time of the fight.
mintraininghours(initialEnergy, initialExperience, energy, experience) -> intThe single exported function. It splits the problem into two independent subproblems and sums the results:
Energy subproblem (line 22): Energy is consumed additively — every fight costs energy[i] and the player must survive all fights. This reduces to a single arithmetic check: total energy needed is sum(energy) + 1, and the training gap is max(0, sum(energy) + 1 - initialEnergy).
Experience subproblem (lines 25–31): Experience is gained after each win — the player gets experience[i] added after defeating opponent i. Because gains compound, the deficit must be resolved opponent-by-opponent via simulation. When curexp <= exp, the function computes the exact gap (exp + 1 - curexp), adds it to both the running total and the current experience, then adds the opponent's experience as the win reward.
Imports: typing.List — used only for type annotations.
Imported by: minimum-hours-of-training-to-win-a-competition/test_solution.py (the "Imported By" list in the prompt is misleading — it lists all test files across the repo that import from their own solution.py, not files that import from *this* solution).
1. Compute total energy cost in O(n), derive energy training hours in O(1).
2. Walk the experience array left-to-right, tracking curexp. At each opponent, if curexp <= exp, bridge the gap with training hours. After each opponent, gain their experience.
3. Return the sum of energy and experience training hours.
Time complexity: O(n) — one pass for sum(energy), one pass for the experience simulation.
Space complexity: O(1) — only scalar accumulators.
+ 1 in both sum(energy) + 1 (line 22) and exp + 1 - cur_exp (line 28).max(0, ...) on line 22 and the if cur_exp <= exp guard on line 27 ensure training hours are never negative.None. The function trusts its inputs match the LeetCode contract (equal-length lists, non-negative values). No validation, no exceptions.
minimum-hours-of-training-to-win-a-competition/test_solution.py — See what edge cases (zero initial stats, single opponent, already-sufficient stats) are testedminimum-hours-of-training-to-win-a-competition/review.md — Read the code review for quality observations and potential issuesgreedy-simulation-pattern — Many LeetCode solutions in this repo use greedy single-pass simulation; compare with diet-plan-performance or lemonade-changebest-time-to-buy-and-sell-stock/solution.py:maxProfit — Another example of decomposing a sequential problem into a single-pass greedy scanenergy-experience-independence — Energy and experience training hours are computed independently and summed; no interaction between the two subproblemsexperience-requires-simulation — Unlike energy (which reduces to a sum), experience must be simulated sequentially because wins compound the player's experiencestrict-greater-than-enforced — The + 1 in both the energy and experience calculations enforces the strict-greater-than requirement from the problem statementlinear-time-constant-space — The solution runs in O(n) time and O(1) space with no auxiliary data structures