How do you solve Minimum Amount of Time to Fill Cups on LeetCode?

Minimum Amount of Time to Fill Cups (LeetCode #2335) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Maximizing the number of cups filled each second reduces total time.

What is the time complexity of Minimum Amount of Time to Fill Cups?

The optimal solution for Minimum Amount of Time to Fill Cups runs in O(n log n) time and uses O(n) space. The complexity is O(n log n) due to the heap operations for maintaining the max-heap, which allows us to efficiently get the two largest amounts.

What is the brute force solution for Minimum Amount of Time to Fill Cups?

The brute force approach for Minimum Amount of Time to Fill Cups is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach tries to fill cups one by one, checking all possible combinations of filling two cups at a time. This is straightforward but inefficient as it doesn't optimize the filling process. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Minimum Amount of Time to Fill Cups?

Minimum Amount of Time to Fill Cups uses the following concepts: Array, Greedy, Sorting, Heap (Priority Queue). The recommended patterns to study are: Greedy Algorithms, Priority Queue.

What are the interview tips for Minimum Amount of Time to Fill Cups?

Always think about how to maximize efficiency in filling tasks. Consider edge cases where one type of cup is much larger than the others. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Minimum Amount of Time to Fill Cups?

Not considering the optimal way to fill cups leads to longer times. Overcomplicating the logic instead of using simple greedy strategies.

#2335

Minimum Amount of Time to Fill Cups

Easy

ArrayGreedySortingHeap (Priority Queue)Greedy AlgorithmsPriority Queue

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

💡

Intuition

Time O(n log n)Space O(n)

The optimal solution uses a greedy approach to always fill the two cups with the most water left. This minimizes the total time by maximizing the number of cups filled each second.

⚙️

Algorithm

3 steps

1Step 1: Sort the amount array in descending order.
2Step 2: While the largest amount is greater than 0, fill the two largest cups.
3Step 3: If only one cup remains, fill it until it's empty.

solution.py16 lines

1# Full working Python code
2import heapq
3amount = [1, 4, 2]
4seconds = 0
5amount = [-x for x in amount]  # Use max-heap
6heapq.heapify(amount)
7while amount:
8    first = -heapq.heappop(amount)
9    if amount:
10        second = -heapq.heappop(amount)
11        if second > 1:
12            heapq.heappush(amount, -(second - 1))
13    seconds += 1
14    if first > 1:
15        heapq.heappush(amount, -(first - 1))
16return seconds

ℹ

Complexity note: The complexity is O(n log n) due to the heap operations for maintaining the max-heap, which allows us to efficiently get the two largest amounts.

1Maximizing the number of cups filled each second reduces total time.
2Using a greedy approach with a priority queue allows efficient selection of the largest amounts.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.