Leetcode #2383: Minimum Hours of Training to Win a Competition
In this guide, we solve Leetcode #2383 Minimum Hours of Training to Win a Competition in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are entering a competition, and are given two positive integers initialEnergy and initialExperience denoting your initial energy and initial experience respectively. You are also given two 0-indexed integer arrays energy and experience, both of length n.
Quick Facts
- Difficulty: Easy
- Premium: No
- Tags: Greedy, Array
Intuition
A locally optimal choice leads to a globally optimal result for this structure.
That means we can commit to decisions as we scan without backtracking.
Approach
Sort or preprocess if needed, then repeatedly take the best available local choice.
Maintain the minimal state necessary to validate the greedy decision.
Steps:
- Sort or preprocess as needed.
- Iterate and pick the best local option.
- Track the current solution.
Example
Input: initialEnergy = 5, initialExperience = 3, energy = [1,4,3,2], experience = [2,6,3,1]
Output: 8
Explanation: You can increase your energy to 11 after 6 hours of training, and your experience to 5 after 2 hours of training.
You face the opponents in the following order:
- You have more energy and experience than the 0th opponent so you win.
Your energy becomes 11 - 1 = 10, and your experience becomes 5 + 2 = 7.
- You have more energy and experience than the 1st opponent so you win.
Your energy becomes 10 - 4 = 6, and your experience becomes 7 + 6 = 13.
- You have more energy and experience than the 2nd opponent so you win.
Your energy becomes 6 - 3 = 3, and your experience becomes 13 + 3 = 16.
- You have more energy and experience than the 3rd opponent so you win.
Your energy becomes 3 - 2 = 1, and your experience becomes 16 + 1 = 17.
You did a total of 6 + 2 = 8 hours of training before the competition, so we return 8.
It can be proven that no smaller answer exists.
Python Solution
class Solution:
def minNumberOfHours(
self, x: int, y: int, energy: List[int], experience: List[int]
) -> int:
ans = 0
for dx, dy in zip(energy, experience):
if x <= dx:
ans += dx + 1 - x
x = dx + 1
if y <= dy:
ans += dy + 1 - y
y = dy + 1
x -= dx
y += dy
return ans
Complexity
The time complexity is , where is the number of opponents. The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.