Minimum Time to Build Blocks — LeetCode 1199 Python Solution
HardLeetCode PremiumGreedyArrayMathHeap (Priority Queue)
- Problem
- #1199
- Pattern
- Heap / Priority Queue
- Reading time
- 2 min
- Source
- leetcode.com
The problem
You are given a list of blocks, where blocks[i] = t means that the i-th block needs t units of time to be built. A block can only be built by exactly one worker.
Example
- Input
- blocks = [1], split = 1
- Output
- 1
- Explanation
- We use 1 worker to build 1 block in 1 time unit.
Python solution
Python
class Solution:
def minBuildTime(self, blocks: List[int], split: int) -> int:
heapify(blocks)
while len(blocks) > 1:
heappop(blocks)
heappush(blocks, heappop(blocks) + split)
return blocks[0]Complexity
| Measure | Complexity |
|---|---|
| Time | O(n \times \log n) |
| Space | O(n) auxiliary |
Pattern: Heap / Priority Queue
Keep only the best k elements, or always pull the smallest, in log time. LeetCode 1199. Minimum Time to Build Blocks is filed here because LeetCode tags it Heap (Priority Queue), which is the vocabulary this hub collects.
The heap / priority queue guide has the Python template for the pattern and the 163 LeetCode problems that use it.
Related problems
Frequently asked questions
- How hard is LeetCode 1199. Minimum Time to Build Blocks?
- LeetCode 1199. Minimum Time to Build Blocks is rated Hard on LeetCode.
- What is the time complexity of LeetCode 1199. Minimum Time to Build Blocks?
- The Python solution on this page runs in O(n \times \log n).
- What is the space complexity of LeetCode 1199. Minimum Time to Build Blocks?
- The Python solution on this page uses O(n) auxiliary space.
- What topics does LeetCode 1199. Minimum Time to Build Blocks cover?
- LeetCode 1199. Minimum Time to Build Blocks is tagged Greedy, Array, Math and Heap (Priority Queue) on LeetCode.
- Is LeetCode 1199. Minimum Time to Build Blocks a premium problem?
- Yes. LeetCode 1199. Minimum Time to Build Blocks is a LeetCode Premium problem, so the full statement and test cases require a paid LeetCode subscription.