2214. Minimum Health to Beat Game 🔒

DifficultyMedium

Description

You are playing a game that has n levels numbered from 0 to n - 1. You are given a 0-indexed integer array damage where damage[i] is the amount of health you will lose to complete the ith level.

You are also given an integer armor. You may use your armor ability at most once during the game on any level which will protect you from at most armor damage.

You must complete the levels in order and your health must be greater than 0 at all times to beat the game.

Return the minimum health you need to start with to beat the game.

Example 1:

Input: damage = [2,7,4,3], armor = 4
Output: 13
Explanation: One optimal way to beat the game starting at 13 health is:
On round 1, take 2 damage. You have 13 - 2 = 11 health.
On round 2, take 7 damage. You have 11 - 7 = 4 health.
On round 3, use your armor to protect you from 4 damage. You have 4 - 0 = 4 health.
On round 4, take 3 damage. You have 4 - 3 = 1 health.
Note that 13 is the minimum health you need to start with to beat the game.

Example 2:

Input: damage = [2,5,3,4], armor = 7
Output: 10
Explanation: One optimal way to beat the game starting at 10 health is:
On round 1, take 2 damage. You have 10 - 2 = 8 health.
On round 2, use your armor to protect you from 5 damage. You have 8 - 0 = 8 health.
On round 3, take 3 damage. You have 8 - 3 = 5 health.
On round 4, take 4 damage. You have 5 - 4 = 1 health.
Note that 10 is the minimum health you need to start with to beat the game.

Example 3:

Input: damage = [3,3,3], armor = 0
Output: 10
Explanation: One optimal way to beat the game starting at 10 health is:
On round 1, take 3 damage. You have 10 - 3 = 7 health.
On round 2, take 3 damage. You have 7 - 3 = 4 health.
On round 3, take 3 damage. You have 4 - 3 = 1 health.
Note that you did not use your armor ability.

Constraints:

• n == damage.length
• 1 <= n <= 105
• 0 <= damage[i] <= 105
• 0 <= armor <= 105

Solutions

Solution 1: Greedy

We can greedily choose to use the armor skill in the round with the highest damage. Suppose the maximum damage is \(\textit{mx}\), then we can avoid \(\min(\textit{mx}, \textit{armor})\) damage. Therefore, the minimum health required is \(\sum(\textit{damage}) - \min(\textit{mx}, \textit{armor}) + 1\).

The time complexity is \(O(n)\), where \(n\) is the length of the array \(\textit{damage}\). The space complexity is \(O(1)\).

1
2
3

class Solution:
    def minimumHealth(self, damage: List[int], armor: int) -> int:
        return sum(damage) - min(max(damage), armor) + 1

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11

class Solution {
    public long minimumHealth(int[] damage, int armor) {
        long s = 0;
        int mx = damage[0];
        for (int v : damage) {
            s += v;
            mx = Math.max(mx, v);
        }
        return s - Math.min(mx, armor) + 1;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12

class Solution {
public:
    long long minimumHealth(vector<int>& damage, int armor) {
        long long s = 0;
        int mx = damage[0];
        for (int& v : damage) {
            s += v;
            mx = max(mx, v);
        }
        return s - min(mx, armor) + 1;
    }
};

1
2
3
4
5
6
7
8
9

func minimumHealth(damage []int, armor int) int64 {
    var s int64
    var mx int
    for _, v := range damage {
        s += int64(v)
        mx = max(mx, v)
    }
    return s - int64(min(mx, armor)) + 1
}

1
2
3
4
5
6
7
8
9

function minimumHealth(damage: number[], armor: number): number {
    let s = 0;
    let mx = 0;
    for (const v of damage) {
        mx = Math.max(mx, v);
        s += v;
    }
    return s - Math.min(mx, armor) + 1;
}

Comments