Formatted question description: https://leetcode.ca/all/2214.html

2214. Minimum Health to Beat Game (Medium)

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

Related Topics:
Array, Greedy, Prefix Sum

Similar Questions:

Solution 1. DP

Let yes be the min health needed if we do use armor once.

Let no be the min health needed if we don’t use armor.

Both of them are initialized as 1.

From A[N-1] to A[0], at each step, we get newYes and newNo values:

• newNo is simply no + A[i].
• newYes has two options. It’s the min of these two:
• Use the armor in the current level. The min health needed is no + max(0, A[i] - armor).
• Don’t use the armor in the current level but a previous level. The min health needed is yes + A[i].

In the end, the result is min(yes, no).

// OJ: https://leetcode.com/problems/minimum-health-to-beat-game/
// Time: O()
// Space: O()
class Solution {
public:
long long minimumHealth(vector<int>& A, int armor) {
long yes = 1, no = 1;
for (int i = A.size() - 1; i >= 0; --i) {
long newYes = min(no + max(0, A[i] - armor), yes + A[i]);
long newNo = no + A[i];
yes = newYes, no = newNo;
}
return min(yes, no);
}
};


Solution 2. Greedy

We greedily use the armor at the level with the greatest damage. Without armor, we need 1 + sum(A) min health. The armor can take damange of min(max(A), armor), saving us this amount of health. So in total, we just need 1 + sum(A) - min(max(A), armor).

• 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;
}
}


• // OJ: https://leetcode.com/problems/minimum-health-to-beat-game/
// Time: O(N)
// Space: O(1)
class Solution {
public:
long long minimumHealth(vector<int>& A, int armor) {
return 1 + accumulate(begin(A), end(A), 0L) - min(*max_element(begin(A), end(A)), armor);
}
};

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