#2383

Easy

auto_awesomeGreedy choice plus invariant validation

LeetCode Problem Workspace

Minimum Hours of Training to Win a Competition

Find the minimum hours of training needed to beat all opponents in a competition based on energy and experience.

Problem Statement

You are entering a competition with initial energy and experience levels. You are given two arrays, energy and experience, where each element in these arrays represents the energy and experience required to defeat an opponent in sequence. To win, you need to have strictly greater energy and experience than each opponent in order.

Your task is to determine the minimum total hours of training required for you to defeat all opponents. Each hour of training can increase either your energy or experience. You must find the optimal combination of training to minimize the total hours spent.

Examples

Example 1

Input: initialEnergy = 5, initialExperience = 3, energy = [1,4,3,2], experience = [2,6,3,1]

Output: 8

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.

Example 2

Input: initialEnergy = 2, initialExperience = 4, energy = [1], experience = [3]

Output: 0

You do not need any additional energy or experience to win the competition, so we return 0.

Constraints

n == energy.length == experience.length
1 <= n <= 100
1 <= initialEnergy, initialExperience, energy[i], experience[i] <= 100

Solution Approach

Greedy Strategy for Energy and Experience

The problem can be broken into two parts: increasing energy and experience. Apply a greedy approach where you prioritize increasing energy or experience based on which one is lagging behind the opponent’s requirement. Ensure that after each opponent, your energy and experience are updated accordingly.

Simulate the Training Process

For each opponent, check if your current energy and experience are sufficient to defeat them. If not, calculate the minimum training hours needed to make them sufficient and accumulate the hours. The order of facing the opponents matters, as defeating one opponent increases your experience and energy, which may help in facing the next.

Efficient Calculation

To optimize the training time, compute the minimum hours of training for both energy and experience separately. Combine the total hours required for both energy and experience, ensuring that you meet both conditions as you face each opponent.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time and space complexity of this problem depends on the specific approach used. A straightforward simulation approach with greedy logic will have a time complexity of O(n), where n is the number of opponents, as we are iterating through the array of opponents once and performing constant time operations for each. The space complexity is O(1), as no additional space is required apart from a few variables to track the current energy, experience, and training hours.

What Interviewers Usually Probe

The candidate uses a greedy strategy to balance energy and experience.
The candidate demonstrates an understanding of simulation techniques for tracking and updating energy/experience.
The candidate applies a clear approach for calculating minimum training hours efficiently.

Common Pitfalls or Variants

Common pitfalls

Forgetting to update both energy and experience after each opponent.
Overestimating the amount of training needed—ensure that you only train as much as necessary to defeat the next opponent.
Ignoring the fact that increasing experience may help with defeating future opponents more easily.

Follow-up variants

What if the energy and experience arrays were sorted in descending order of difficulty? This would change the approach slightly, as the order of opponents could affect the minimum training hours.
What if you could train in increments other than 1 hour at a time? This could complicate the calculation of training hours, requiring a different greedy method or approach.
What if you could train energy and experience together instead of separately? This would change how you calculate and balance the minimum training required.

FAQ

How do I calculate the minimum training hours for each opponent?

You need to compare your current energy and experience with the opponent’s requirements and calculate how much additional training is needed to exceed both values.

What is the greedy strategy for this problem?

The greedy strategy involves increasing either energy or experience first based on which one is lagging behind the opponent’s requirements, ensuring you defeat opponents with minimal training.

What is the importance of updating both energy and experience?

After each opponent, you need to update both energy and experience as defeating an opponent increases your experience and energy, which can help in defeating future opponents.

How does the training process affect future opponents?

Each opponent you defeat increases your energy and experience, which makes it easier to defeat subsequent opponents. So, managing the training effectively is key to minimizing the total hours.

Can this problem be solved without a greedy approach?

While the greedy approach is the most efficient, it is possible to approach this problem with a more exhaustive simulation. However, that would result in a higher time complexity.

terminal

Solution

Solution 1: Greedy + Simulation

Let's denote the current energy as $x$ and the current experience as $y$.

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

Continue Practicing

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