Minimum Health to Beat Game

Minimum Health to Beat Game - Problem

Array Greedy 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 i-th 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.

Input & Output

Example 1 — Basic Case

$ Input: damage = [2,7,4,3], armor = 4

› Output: 13

💡 Note: Use armor on level 1 (damage 7), reducing it to 3. Total damage becomes [2,3,4,3]. To maintain health > 0 throughout: start with 13, after levels: 13→11→8→4→1. Minimum health needed is 13.

Example 2 — High Armor Value

$ Input: damage = [2,5,3,4], armor = 7

› Output: 10

💡 Note: Use armor on level 1 (damage 5), reducing it to 0. Total damage becomes [2,0,3,4]. Starting with 10 health: 10→8→8→5→1. Minimum health needed is 10.

Example 3 — Small Armor

$ Input: damage = [3,3,3], armor = 1

› Output: 9

💡 Note: Use armor on any level to reduce one 3 to 2. Total damage becomes [2,3,3]. To maintain health > 0 throughout: start with 9, after levels: 9→7→4→1. Minimum health needed is 9.

Constraints

n == damage.length
1 ≤ n ≤ 10⁵
0 ≤ damage[i] ≤ 10⁶
0 ≤ armor ≤ 10⁶

Visualization

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Asked in

a Amazon 15 G Google 12

The key insight is to use armor on the level with maximum damage for optimal benefit. The greedy approach works: find the level where armor provides maximum damage reduction, then simulate the game to calculate minimum starting health. Time: O(n), Space: O(1)

Common Approaches

✓ Brute Force - Try All Armor Positions

⏱️ Time: O(n²) Space: O(1)

For each possible position to use armor (including not using it), simulate the game and calculate the minimum starting health needed. Take the minimum across all possibilities.

Greedy - Use Armor on Maximum Damage

⏱️ Time: O(n) Space: O(1)

Calculate the total damage first. Then find the level with maximum damage where using armor would give maximum benefit. Use armor there and work backwards to find minimum starting health.

Brute Force - Try All Armor Positions — Algorithm Steps

Step 1: Try using armor on each level from 0 to n-1
Step 2: For each armor position, simulate game and find minimum starting health
Step 3: Return the minimum health across all attempts

Visualization

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Step-by-Step Walkthrough

Try Armor at Level 0

Simulate game with armor protecting first level

Try Armor at Level 1

Simulate game with armor protecting second level

Find Minimum

Return the scenario requiring least starting health

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>

long long solution(int* damage, int damageSize, int armor) {
    long long minHealth = LLONG_MAX;
    
    // Try using armor at each position (or not using it)
    for (int armorPos = -1; armorPos < damageSize; armorPos++) { // -1 means don't use armor
        long long currentHealth = 1; // Start with 1 health
        long long maxHealthNeeded = 1;
        
        for (int i = 0; i < damageSize; i++) {
            // Apply damage, reduced by armor if used at this level
            long long actualDamage = damage[i];
            if (i == armorPos) {
                actualDamage = (damage[i] - armor > 0) ? damage[i] - armor : 0;
            }
            
            currentHealth -= actualDamage;
            
            // If health drops to 0 or below, we need more starting health
            if (currentHealth <= 0) {
                long long neededExtra = -currentHealth + 1;
                maxHealthNeeded += neededExtra;
                currentHealth = 1;
            }
        }
        
        if (maxHealthNeeded < minHealth) {
            minHealth = maxHealthNeeded;
        }
    }
    
    return minHealth;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Parse damage array
    int damage[1000];
    int damageSize = 0;
    char* token = strtok(line, "[,]");
    while (token != NULL) {
        if (strlen(token) > 0 && token[0] != '\n') {
            damage[damageSize++] = atoi(token);
        }
        token = strtok(NULL, "[,]");
    }
    
    int armor;
    scanf("%d", &armor);
    
    printf("%lld\n", solution(damage, damageSize, armor));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each of n armor positions, we simulate n levels

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra variables

✓ Linear Space

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Try All Armor Positions — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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