Leetcode #2403: Minimum Time to Kill All Monsters
In this guide, we solve Leetcode #2403 Minimum Time to Kill All Monsters in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given an integer array power where power[i] is the power of the ith monster. You start with 0 mana points, and each day you increase your mana points by gain where gain initially is equal to 1.
Quick Facts
- Difficulty: Hard
- Premium: Yes
- Tags: Bit Manipulation, Array, Dynamic Programming, Bitmask
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
Input: power = [3,1,4]
Output: 4
Explanation: The optimal way to beat all the monsters is to:
- Day 1: Gain 1 mana point to get a total of 1 mana point. Spend all mana points to kill the 2nd monster.
- Day 2: Gain 2 mana points to get a total of 2 mana points.
- Day 3: Gain 2 mana points to get a total of 4 mana points. Spend all mana points to kill the 3rd monster.
- Day 4: Gain 3 mana points to get a total of 3 mana points. Spend all mana points to kill the 1st monster.
It can be proven that 4 is the minimum number of days needed.
Python Solution
class Solution:
def minimumTime(self, power: List[int]) -> int:
def dfs(mask: int) -> int:
if mask == 0:
return 0
ans = inf
gain = 1 + (n - mask.bit_count())
for i, x in enumerate(power):
if mask >> i & 1:
ans = min(ans, dfs(mask ^ (1 << i)) + (x + gain - 1) // gain)
return ans
n = len(power)
return dfs((1 << n) - 1)
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.