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Formatted question description: https://leetcode.ca/all/2383.html

2383. Minimum Hours of Training to Win a Competition

  • Difficulty: Easy.
  • Related Topics: Array, Greedy.
  • Similar Questions: .

Problem

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

Solution (Java, C++, Python)

  • class Solution {
    public int minNumberOfHours(
            int initialEnergy, int initialExperience, int[] energy, int[] experience) {
        int totalEnergy = 0;
        for (int e : energy) {
            totalEnergy += e;
        }
        int result = Math.max(0, totalEnergy - initialEnergy + 1);
        int currentExp = initialExperience;
        int exp;
        for (int i = 0; i < experience.length - 1; ++i) {
            exp = experience[i];
            if (currentExp <= exp) {
                result += exp - currentExp + 1;
                currentExp = exp + 1;
            }
            currentExp += exp;
        }
        int last = experience[experience.length - 1];
        if (currentExp <= last) {
            result += last - currentExp + 1;
        }
        return result;
    }
}

############

class Solution {
    public int minNumberOfHours(
        int initialEnergy, int initialExperience, int[] energy, int[] experience) {
        int ans = 0;
        for (int i = 0; i < energy.length; ++i) {
            int a = energy[i], b = experience[i];
            if (initialEnergy <= a) {
                ans += a - initialEnergy + 1;
                initialEnergy = a + 1;
            }
            if (initialExperience <= b) {
                ans += b - initialExperience + 1;
                initialExperience = b + 1;
            }
            initialEnergy -= a;
            initialExperience += b;
        }
        return ans;
    }
}

  • class Solution:
    def minNumberOfHours(
        self,
        initialEnergy: int,
        initialExperience: int,
        energy: List[int],
        experience: List[int],
    ) -> int:
        ans = 0
        for a, b in zip(energy, experience):
            if initialEnergy <= a:
                ans += a - initialEnergy + 1
                initialEnergy = a + 1
            if initialExperience <= b:
                ans += b - initialExperience + 1
                initialExperience = b + 1
            initialEnergy -= a
            initialExperience += b
        return ans

############

# 2383. Minimum Hours of Training to Win a Competition
# https://leetcode.com/problems/minimum-hours-of-training-to-win-a-competition/

class Solution:
    def minNumberOfHours(self, A: int, B: int, energy: List[int], experience: List[int]) -> int:
        res = 0
        
        for a, b in zip(energy, experience):
            if a >= A:
                res += a - A + 1
                A += a - A + 1
            
            if b >= B:
                res += b - B + 1
                B += b - B + 1
                
            A -= a
            B += b
        
        return res


  • class Solution {
public:
    int minNumberOfHours(int initialEnergy, int initialExperience, vector<int>& energy, vector<int>& experience) {
        int ans = 0;
        for (int i = 0; i < energy.size(); ++i) {
            int a = energy[i], b = experience[i];
            if (initialEnergy <= a) {
                ans += a - initialEnergy + 1;
                initialEnergy = a + 1;
            }
            if (initialExperience <= b) {
                ans += b - initialExperience + 1;
                initialExperience = b + 1;
            }
            initialEnergy -= a;
            initialExperience += b;
        }
        return ans;
    }
};

  • func minNumberOfHours(initialEnergy int, initialExperience int, energy []int, experience []int) int {
	ans := 0
	for i, a := range energy {
		b := experience[i]
		if initialEnergy <= a {
			ans += a - initialEnergy + 1
			initialEnergy = a + 1
		}
		if initialExperience <= b {
			ans += b - initialExperience + 1
			initialExperience = b + 1
		}
		initialEnergy -= a
		initialExperience += b
	}
	return ans
}

  • function minNumberOfHours(
    initialEnergy: number,
    initialExperience: number,
    energy: number[],
    experience: number[],
): number {
    const n = energy.length;
    let ans = 0;
    for (let i = 0; i < n; i++) {
        const minEnergy = energy[i];
        const minExperience = experience[i];
        if (initialEnergy <= minEnergy) {
            const need = minEnergy - initialEnergy + 1;
            ans += need;
            initialEnergy += need;
        }
        if (initialExperience <= minExperience) {
            const need = minExperience - initialExperience + 1;
            ans += need;
            initialExperience += need;
        }
        initialEnergy -= minEnergy;
        initialExperience += minExperience;
    }
    return ans;
}


  • impl Solution {
    pub fn min_number_of_hours(
        mut initial_energy: i32,
        mut initial_experience: i32,
        energy: Vec<i32>,
        experience: Vec<i32>,
    ) -> i32 {
        let n = energy.len();
        let mut res = 0;
        for i in 0..n {
            if initial_energy <= energy[i] {
                res += energy[i] - initial_energy + 1;
                initial_energy = energy[i] + 1;
            }
            if initial_experience <= experience[i] {
                res += experience[i] - initial_experience + 1;
                initial_experience = experience[i] + 1;
            }
            initial_energy -= energy[i];
            initial_experience += experience[i];
        }
        res
    }
}

Explain:

nope.

Complexity:

  • Time complexity : O(n).
  • Space complexity : O(n).

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