The fastest way is to prioritize filling the two largest remaining cup counts each second, reducing the overall number of seconds. Always verify that no single type exceeds half the remaining total, which ensures the greedy choice is valid. By combining max selection with invariant validation, the solution achieves minimal time for any given distribution of cup amounts.
Problem Statement
You are given three types of water cups: cold, warm, and hot. Each second, you can fill either one cup of any type or two cups of different types simultaneously. Determine the minimum number of seconds required to fill all cups given an array amount = [cold, warm, hot].
Implement a function that takes a 0-indexed array amount of length 3 where amount[0], amount[1], and amount[2] represent the number of cups for cold, warm, and hot water respectively. Return the fewest seconds needed to fill all cups under the dispensing rules.
Examples
Example 1
Input: amount = [1,4,2]
Output: 4
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.
Example 2
Input: amount = [5,4,4]
Output: 7
One way to fill up the cups is: Second 1: Fill up a cold cup, and a hot cup. Second 2: Fill up a cold cup, and a warm cup. Second 3: Fill up a cold cup, and a warm cup. Second 4: Fill up a warm cup, and a hot cup. Second 5: Fill up a cold cup, and a hot cup. Second 6: Fill up a cold cup, and a warm cup. Second 7: Fill up a hot cup.
Example 3
Input: amount = [5,0,0]
Output: 5
Every second, we fill up a cold cup.
Constraints
- amount.length == 3
- 0 <= amount[i] <= 100
Solution Approach
Greedy Max Pair Selection
Each second, identify the two cup types with the largest remaining counts and fill one cup of each. This ensures that every second contributes maximally to reducing total cups, minimizing overall time.
Invariant Validation
Check that the largest remaining cup count never exceeds half the sum of all remaining cups. This invariant guarantees that the greedy max-pair choice leads to a valid minimal time solution.
Simple Implementation Using Sorting or Heap
Sort the cup counts each iteration or use a max heap to efficiently select the two largest values. Decrement counts after each step until all cups are filled, accumulating the total seconds.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | Depends on the final approach |
| Space | Depends on the final approach |
Time complexity is O(1) with fixed-length arrays since operations involve only three elements, or O(n log n) if generalized with a heap. Space complexity is O(1) for in-place operations or O(n) if a heap is used.
What Interviewers Usually Probe
- Consider the effect of filling two different types each second versus one type.
- Think about whether the largest cup count ever dictates the minimum total time.
- Watch for off-by-one errors when decrementing cup counts each second.
Common Pitfalls or Variants
Common pitfalls
- Attempting to fill cups in arbitrary order without maximizing each second wastes time.
- Failing to check that the largest cup type does not exceed half of remaining cups may yield incorrect results.
- Neglecting to handle cases where only one type remains can miscount total seconds.
Follow-up variants
- Generalize to more than three cup types, requiring priority queues to select the largest pair each second.
- Allow filling up to k cups per second if types are distinct, adjusting the greedy selection accordingly.
- Introduce unequal fill rates for each type, requiring weighted greedy selection to minimize total time.
How GhostInterview Helps
- Automatically highlights which cup types to pair each second to minimize time based on greedy selection.
- Verifies that the greedy invariant holds, preventing over-allocation of any single cup type.
- Provides step-by-step reasoning and total second calculation for any input distribution.
Topic Pages
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- LeetCode Interview Copilot - Use GhostInterview as a live solver when you want direct help with LeetCode-style coding questions.
- Coding Interview Assistant - See how GhostInterview supports array, string, linked list, graph, and tree interview workflows.
- How GhostInterview Works - Review the screenshot, reasoning, and answer flow before using the solver in a live interview.
FAQ
What is the fastest approach to solve Minimum Amount of Time to Fill Cups?
Use a greedy strategy selecting the two largest remaining cup counts each second and validate that the largest does not exceed half the total remaining.
Can we fill two cups of the same type in one second?
No, the rules allow filling either one cup of any type or two cups of different types per second.
Does the array length always matter?
Yes, the problem specifically uses an array of length 3 representing cold, warm, and hot cups, simplifying selection.
What data structures help implement this efficiently?
Sorting the array each iteration or using a max heap lets you efficiently pick the two largest counts per second.
How does the greedy choice plus invariant pattern apply here?
It ensures each second reduces the largest counts optimally while checking that no single type dominates, guaranteeing minimal total time.
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