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.
You will face n opponents in order. The energy and experience of the ith opponent is denoted by energy[i] and experience[i] respectively. When you face an opponent, you need to have both strictly greater experience and energy to defeat them and move to the next opponent if available.
Defeating the ith opponent increases your experience by experience[i], but decreases your energy by energy[i].
Before starting the competition, you can train for some number of hours. After each hour of training, you can either choose to increase your initial experience by one, or increase your initial energy by one.
Return the minimum number of training hours required to defeat all n opponents.
Example 1:
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.
Example 2:
Input: initialEnergy = 2, initialExperience = 4, energy = [1], experience = [3] Output: 0 Explanation: You do not need any additional energy or experience to win the competition, so we return 0.
Constraints:
n == energy.length == experience.length1 <= n <= 1001 <= initialEnergy, initialExperience, energy[i], experience[i] <= 100This approach starts by ensuring that you have enough energy and experience to defeat each opponent in sequence. If your current energy and experience are insufficient for an opponent, you'll calculate the additional hours of training needed to match the required levels. For each opponent, you'll update your experience after a win, and subtract the opponent's energy from your current energy.
The C implementation initializes the energy and experience levels, checks if they are sufficient to defeat each opponent, calculates any necessary training, adjusts the current experience and energy levels, and continues to the next opponent. The total training hours are summed and returned.
C++
Java
Python
C#
JavaScript
Time Complexity: O(n) since we iterate over each opponent once.
Space Complexity: O(1) since no additional data structures are used beyond the input and a few auxiliary variables.
The unfair way I got good at Leetcode • Dave Burji • 596,394 views views
Watch 9 more video solutions →Practice Minimum Hours of Training to Win a Competition with our built-in code editor and test cases.
Practice on FleetCode