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Formatted question description: https://leetcode.ca/all/1199.html
1199. Minimum Time to Build Blocks
Level
Hard
Description
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.
A worker can either split into two workers (number of workers increases by one) or build a block then go home. Both decisions cost some time.
The time cost of spliting one worker into two workers is given as an integer split
. Note that if two workers split at the same time, they split in parallel so the cost would be split
.
Output the minimum time needed to build all blocks.
Initially, there is only one worker.
Example 1:
Input: blocks = [1], split = 1
Output: 1
Explanation: We use 1 worker to build 1 block in 1 time unit.
Example 2:
Input: blocks = [1,2], split = 5
Output: 7
Explanation: We split the worker into 2 workers in 5 time units then assign each of them to a block so the cost is 5 + max(1, 2) = 7.
Example 3:
Input: blocks = [1,2,3], split = 1
Output: 4
Explanation: Split 1 worker into 2, then assign the first worker to the last block and split the second worker into 2.
Then, use the two unassigned workers to build the first two blocks.
The cost is 1 + max(3, 1 + max(1, 2)) = 4.
Constraints:
1 <= blocks.length <= 1000
1 <= blocks[i] <= 10^5
1 <= split <= 100
Solution
Splitting a worker into two workers using time split
is equivalent to merging two blocks into one block using time t
, where the block with greater time to be built is remained. Therefore, this problem can be converted into merging blocks so that the total time is minimized. The idea of Huffman Tree can be used here. Each time merge the two blocks with the minimum building time, and the total time after merging is the sum of the two blocks’ building times plus split
. Repeat the process until all the blocks are merged. Finally, return the total time.

class Solution { public int minBuildTime(int[] blocks, int split) { PriorityQueue<Integer> priorityQueue = new PriorityQueue<Integer>(); for (int block : blocks) priorityQueue.offer(block); while (priorityQueue.size() > 1) { priorityQueue.poll(); int secondSmallest = priorityQueue.poll(); int newTime = secondSmallest + split; priorityQueue.offer(newTime); } return priorityQueue.poll(); } }

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]

class Solution { public: int minBuildTime(vector<int>& blocks, int split) { priority_queue<int, vector<int>, greater<int>> pq; for (int v : blocks) pq.push(v); while (pq.size() > 1) { pq.pop(); int x = pq.top(); pq.pop(); pq.push(x + split); } return pq.top(); } };

func minBuildTime(blocks []int, split int) int { q := hp{} for _, v := range blocks { heap.Push(&q, v) } for q.Len() > 1 { heap.Pop(&q) heap.Push(&q, heap.Pop(&q).(int)+split) } return q.IntSlice[0] } type hp struct{ sort.IntSlice } func (h *hp) Push(v interface{}) { h.IntSlice = append(h.IntSlice, v.(int)) } func (h *hp) Pop() interface{} { a := h.IntSlice v := a[len(a)1] h.IntSlice = a[:len(a)1] return v }