LeetCode 2403 - Minimum Time to Kill All Monsters

The problem is asking us to compute the minimum number of days required to defeat all monsters in an array power, where power[i] represents the strength of the i-th monster. You start with zero mana and gain mana daily, with the initial daily gain of 1.

leetcodehardarraydynamic-programmingbit-manipulationbitmask

LeetCode Problem 2403

Difficulty: 🔴 Hard
Topics: Array, Dynamic Programming, Bit Manipulation, Bitmask

Solution

Problem Understanding

The problem is asking us to compute the minimum number of days required to defeat all monsters in an array power, where power[i] represents the strength of the i-th monster. You start with zero mana and gain mana daily, with the initial daily gain of 1. After defeating a monster, your mana resets to zero and the daily gain increases by 1. The challenge is to schedule which monsters to defeat in which order to minimize the total days.

The input is an integer array with length up to 17 and values up to 10^9. This small array length suggests that a solution that explores subsets of monsters or uses bitmasking is feasible, despite large individual values. The output is a single integer representing the minimum number of days required.

Important edge cases include having monsters with very high power early on, multiple monsters with the same power, and arrays with only one monster. The problem guarantees at least one monster, so we do not need to handle an empty input.

Approaches

Brute Force

A naive approach would try all permutations of monster defeat orders. For each order, we simulate each day, accumulating mana and checking if it is enough to defeat the current monster. While this approach guarantees correctness, the number of permutations is n!, which is infeasible for n up to 17.

Optimal Approach

The key insight is that the problem can be reduced to a bitmask dynamic programming problem. Each subset of monsters can be represented by a bitmask, and we can compute the minimum days to defeat all monsters in that subset.

Let dp[mask] be the minimum days required to defeat monsters represented by mask. For each state, we iterate over monsters not yet defeated in the mask, simulate the number of days required to defeat that monster with the current gain, and update the dp state for the new mask. Since n <= 17, there are at most 2^17 ≈ 131072 subsets, which is feasible.

Approach Time Complexity Space Complexity Notes
Brute Force O(n!) O(n) Try all permutations of monsters; too slow for n=17
Optimal O(n * 2^n) O(2^n) Bitmask DP; iterate over all subsets of monsters efficiently

Algorithm Walkthrough

  1. Initialize a DP array dp of size 2^n with infinity, except dp[0] = 0 representing zero monsters defeated.

  2. Iterate through all masks from 0 to (1 << n) - 1. For each mask:

  3. Count the number of monsters already defeated, which determines the current gain.

  4. For each monster i not in the mask:

  5. Compute the number of days needed to accumulate enough mana to defeat monster i with current gain using ceil(power[i] / gain).

  6. Update dp[new_mask] as the minimum of its current value and dp[mask] + days_needed.

  7. Return dp[(1 << n) - 1] which contains the minimum days for defeating all monsters.

Why it works: The DP invariant ensures that for every subset of defeated monsters, dp[mask] stores the minimal days required. By iterating through all possible states and always taking the minimum, we guarantee the global optimum.

Python Solution

from typing import List
import math

class Solution:
    def minimumTime(self, power: List[int]) -> int:
        n = len(power)
        dp = [math.inf] * (1 << n)
        dp[0] = 0

        for mask in range(1 << n):
            defeated = bin(mask).count("1")
            gain = defeated + 1
            for i in range(n):
                if not (mask & (1 << i)):
                    days_needed = (power[i] + gain - 1) // gain  # ceil division
                    new_mask = mask | (1 << i)
                    dp[new_mask] = min(dp[new_mask], dp[mask] + days_needed)

        return dp[(1 << n) - 1]

Explanation: We use bitmasking to represent defeated monsters. For each state, we determine the current daily gain, calculate days needed to defeat any remaining monster, and update the DP for the new mask. The final answer is stored in the DP value representing all monsters defeated.

Go Solution

package main

import "math"

func minimumTime(power []int) int64 {
    n := len(power)
    size := 1 << n
    dp := make([]int64, size)
    for i := 1; i < size; i++ {
        dp[i] = math.MaxInt64
    }

    for mask := 0; mask < size; mask++ {
        defeated := 0
        for j := 0; j < n; j++ {
            if mask&(1<<j) != 0 {
                defeated++
            }
        }
        gain := int64(defeated + 1)
        for i := 0; i < n; i++ {
            if mask&(1<<i) == 0 {
                daysNeeded := (int64(power[i]) + gain - 1) / gain
                newMask := mask | (1 << i)
                if dp[mask]+daysNeeded < dp[newMask] {
                    dp[newMask] = dp[mask] + daysNeeded
                }
            }
        }
    }
    return dp[size-1]
}

Go Notes: Go does not have built-in ceil for integers, so we use (a + b - 1) / b for ceiling division. We also use int64 to prevent integer overflow for large power values.

Worked Examples

Example 1: power = [3,1,4]

Day Gain Mana Monster defeated DP mask
1 1 1 monster 1 (power 1) 010
2 2 2 - 010
3 2 4 monster 3 (power 4) 110
4 3 3 monster 0 (power 3) 111

Minimum days = 4

Example 2: power = [1,1,4]

Day Gain Mana Monster defeated DP mask
1 1 1 monster 0 001
2 2 2 monster 1 011
3 3 3 - 011
4 3 6 monster 2 111

Minimum days = 4

Complexity Analysis

Measure Complexity Explanation
Time O(n * 2^n) We iterate through 2^n masks and n monsters for each mask
Space O(2^n) DP array stores minimum days for each mask

The exponential space and time is acceptable because n <= 17 yields at most 131072 states.

Test Cases

# Provided examples
assert Solution().minimumTime([3,1,4]) == 4  # Example 1
assert Solution().minimumTime([1,1,4]) == 4  # Example 2
assert Solution().minimumTime([1,2,4,9]) == 6  # Example 3

# Edge cases
assert Solution().minimumTime([1]) == 1  # Single monster
assert Solution().minimumTime([10**9]) == 1000000000  # Max power, single monster
assert Solution().minimumTime([1]*17) == 17  # All monsters same power
assert Solution().minimumTime([i for i in range(1,18)]) == 17  # Increasing powers
Test Why
[3,1,4] Verifies basic example with varied powers
[1] Single monster case
[10^9] Very large power to test integer handling
[1]*17 Max size array with equal powers
[1..17] Increasing powers to test gain accumulation

Edge Cases

  1. Single Monster: With power = [x], the minimum days is simply ceil(x / 1). This checks if the algorithm handles the smallest input correctly. The DP handles this because the only mask update occurs from 0 to 1.
  2. Maximum Power: If power[i] is extremely large (up to 10^9), we need to ensure integer arithmetic does not overflow. In Python, integers are arbitrary precision, while in Go we use int64 to safely handle large numbers.
  3. All Monsters Same Power: When all monsters have equal power, the order of defeating them may not matter, but the DP ensures we calculate days accurately with increasing gain. This case tests if the DP properly accumulates gain after each defeat.