Leetcode #2335: Minimum Amount of Time to Fill Cups
In this guide, we solve Leetcode #2335 Minimum Amount of Time to Fill Cups 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 have a water dispenser that can dispense cold, warm, and hot water. Every second, you can either fill up 2 cups with different types of water, or 1 cup of any type of water.
Quick Facts
- Difficulty: Easy
- Premium: No
- Tags: Greedy, Array, Sorting, Heap (Priority Queue)
Intuition
A locally optimal choice leads to a globally optimal result for this structure.
That means we can commit to decisions as we scan without backtracking.
Approach
Sort or preprocess if needed, then repeatedly take the best available local choice.
Maintain the minimal state necessary to validate the greedy decision.
Steps:
- Sort or preprocess as needed.
- Iterate and pick the best local option.
- Track the current solution.
Example
Input: amount = [1,4,2]
Output: 4
Explanation: One way to fill up the cups is:
Second 1: Fill up a cold cup and a warm cup.
Second 2: Fill up a warm cup and a hot cup.
Second 3: Fill up a warm cup and a hot cup.
Second 4: Fill up a warm cup.
It can be proven that 4 is the minimum number of seconds needed.
Python Solution
class Solution:
def fillCups(self, amount: List[int]) -> int:
ans = 0
while sum(amount):
amount.sort()
ans += 1
amount[2] -= 1
amount[1] = max(0, amount[1] - 1)
return ans
Complexity
The time complexity is O(n log n). The space complexity is O(1) to O(n).
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.