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Weekly Contest 307 Q1
Greedy
Array

2383. Minimum Hours of Training to Win a Competition

中文文档

Description

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

Solutions

Solution 1: Greedy + Simulation

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

Next, we traverse each opponent. For the $i$-th opponent, let their energy be $dx$ and their experience be $dy$.

  • If $x \leq dx$, then we need to train for $dx + 1 - x$ hours to increase our energy to $dx + 1$.
  • If $y \leq dy$, then we need to train for $dy + 1 - y$ hours to increase our experience to $dy + 1$.
  • Then, we subtract $dx$ from our energy and add $dy$ to our experience.

Finally, return the answer.

The time complexity is $O(n)$, where $n$ is the number of opponents. The space complexity is $O(1)$.

Python3

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

Java

class Solution {
    public int minNumberOfHours(int x, int y, int[] energy, int[] experience) {
        int ans = 0;
        for (int i = 0; i < energy.length; ++i) {
            int dx = energy[i], dy = experience[i];
            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;
    }
}

C++

class Solution {
public:
    int minNumberOfHours(int x, int y, vector<int>& energy, vector<int>& experience) {
        int ans = 0;
        for (int i = 0; i < energy.size(); ++i) {
            int dx = energy[i], dy = experience[i];
            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;
    }
};

Go

func minNumberOfHours(x int, y int, energy []int, experience []int) (ans int) {
	for i, dx := range energy {
		dy := experience[i]
		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
}

TypeScript

function minNumberOfHours(x: number, y: number, energy: number[], experience: number[]): number {
    let ans = 0;
    for (let i = 0; i < energy.length; ++i) {
        const [dx, dy] = [energy[i], experience[i]];
        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;
}

Rust

impl Solution {
    pub fn min_number_of_hours(
        mut x: i32,
        mut y: i32,
        energy: Vec<i32>,
        experience: Vec<i32>,
    ) -> i32 {
        let mut ans = 0;

        for (&dx, &dy) in energy.iter().zip(experience.iter()) {
            if x <= dx {
                ans += dx + 1 - x;
                x = dx + 1;
            }
            if y <= dy {
                ans += dy + 1 - y;
                y = dy + 1;
            }
            x -= dx;
            y += dy;
        }

        ans
    }
}

C

int minNumberOfHours(int x, int y, int* energy, int energySize, int* experience, int experienceSize) {
    int ans = 0;
    for (int i = 0; i < energySize; ++i) {
        int dx = energy[i], dy = experience[i];
        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;
}