1199. Minimum Time to Build Blocks

Problem Description

In this problem, we are given a series of tasks represented by blocks, where each block's value represents the amount of time required to complete that task (blocks[i] = t). Initially, there's only one worker available, capable of performing two types of actions:

  1. Building a block, which completes that task.
  2. Splitting into two workers, which allows for more tasks to be done in parallel.

There is a constraint on the second action: splitting a worker into two costs a fixed amount of time (split), regardless of how many splits occur simultaneously. The goal is to calculate the minimum time required to complete all tasks with the initial single worker, taking into account the time it takes to build each block and the possibility of splitting workers.


The essence of the solution to this problem lies in understanding that it's more efficient to allow workers to build the longest taking blocks simultaneously while splitting workers to multiply their efficiency.

Here's the intuitive step-by-step reasoning for the algorithm:

  1. Use more workers on longer tasks first: Prioritize assigning more workers to build the blocks that take the longest time first, because this minimizes overall waiting time.
  2. Split workers only when necessary: It's often better to split workers early on so that they are available to work on parallel tasks, but doing so should be a calculated decision since each split incurs a fixed time cost.
  3. Utilize a min-heap to organize tasks: By using a min-heap data structure (a binary heap where the parent node is less than or equal to its children), we can efficiently keep track of the smallest (quickest) tasks, which helps in determining when to split workers.

Here's how the algorithm is implemented:

  • The heapq module in Python is used to convert the list of blocks into a min-heap in-place.
  • Then, the process begins of repeatedly taking two elements off the top of the heap, which takes O(log n) time - the smallest element (the one that requires the least amount of time) and the second smallest.
  • The smallest element (fastest task) is discarded, and the second smallest is replaced after adding the split time to it. This simulates a worker that's been split and then finished building the next quickest block.
  • We continue this process until we're left with a single block in the heap, which now contains the minimum time needed to complete all tasks when considering split timing and building of tasks in parallel.
  • Thus, the function returns the value of the sole remaining block in the heap, which is the answer to the problem.

This strategy works because it effectively simulates an optimal way of splitting workers and assigning tasks to them in a way that minimizes the overall completion time.

Solution Approach

The solution approach for this problem makes use of the heap data structure to efficiently manage the blocks while deciding the order in which to build them and when to split the workers. Here's a detailed breakdown:

  1. Heapify the blocks: The heapify function from Python's heapq module is used to transform the list of blocks into a min-heap data structure in-place. A min-heap ensures that the smallest element is always at the root, which is important for efficiently accessing the quickest task to complete.

  2. Processing the heap: A while-loop runs as long as there is more than one block present in the heap. The purpose of this loop is to simulate the worker's actions in building blocks and splitting.

    1while len(blocks) > 1:
  3. Simulate block building and worker split: Inside the loop, the heappop function is called twice. The first heappop removes and returns the smallest element from the heap, which represents the block that would finish the quickest. Since that worker will be split, we don't need that block's time anymore. The second heappop call gets the next smallest block's time which actually gets combined with the split time, simulating the time taken for a worker to split and then complete the next block.

    1heappop(blocks)  # This is the block taken by a worker who decides to split.
    2heappush(blocks, heappop(blocks) + split)  # Next quickest block plus the split time.
  4. Completing the build: The loop continues until there is only one block (or time value) left in the heap. This remaining value represents the minimum time required to complete all tasks, considering all necessary splits and concurrent work.

  5. Return the result: Finally, the last remaining value in the heap (the minimum time required to complete all tasks) is returned as the result.

    1return blocks[0]  # The minimum time to build all the blocks.

By using a heap data structure, this solution approach avoids the need to sort the blocks after each step, which keeps the overall time complexity down. Additionally, by always keeping the smallest elements (quickest tasks) at the front of the heap, it ensures that we're always making the most time-efficient decision for splitting workers or assigning them to build blocks.

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Example Walkthrough

Let's consider an example with a set of tasks with block times expressed in units of time as follows: blocks = [4, 2, 10], and a fixed split time of split = 3.

  1. Heapify the blocks: First, we need to transform the blocks into a min-heap. We end up with a heap that might look like this: [2, 4, 10], where the smallest task (taking 2 time units) is at the root.

    1heapify(blocks)  # blocks becomes [2, 4, 10]
  2. Processing the heap: At the start of our while-loop, our heap has more than one block, so we proceed with the following steps inside the loop.

  3. Simulate block building and worker split: We pop the smallest element, which is 2, from the heap. This is the task that the worker who decides to split can complete.

    1heappop(blocks)  # Removes 2, blocks is now [4, 10]

    The worker splits (adding a split time to the next task), and finishes the next block. We pop the next smallest block, which is 4, add the split time of 3, and push the resulting 7 back onto the heap.

    1heappush(blocks, heappop(blocks) + split)  # Removes 4, adds 3, and pushes 7, blocks is now [7, 10]

    Now we continue with the heap [7, 10].

  4. Continue loop: The while-loop runs again because we still have more than one block in the heap.

    We pop the smallest element, which is 7. Then pop the next smallest element, which is 10, add the split time of 3, and push the result back onto the heap.

    1heappop(blocks)  # Removes 7, blocks is now [10]
    2heappush(blocks, heappop(blocks) + split)  # Removes 10, adds 3, and pushes 13, blocks is now [13]
  5. Completing the build: The loop breaks because we now have a single element in our heap: [13]. This means it will take a minimum of 13 units of time to complete all tasks when considering the necessity for workers to split and the concurrent processing of tasks.

  6. Return the result: Since we have only one element in our heap, it represents the minimum time required to complete all tasks.

    1return blocks[0]  # returns 13, which is the minimum time to build all the blocks

In conclusion, by following this step-by-step approach to simulating the worker splits and block builds, our single worker would be able to complete all the tasks in a minimum of 13 units of time.

Python Solution

1from heapq import heapify, heappop, heappush  # Import required functions from heapq module
3class Solution:
4    def minBuildTime(self, blocks: List[int], split: int) -> int:
5        """
6        Calculate the minimum time to build blocks with a given split time.
8        :param blocks: A list of integers representing the time needed to build each block.
9        :param split: An integer representing the time needed to split a worker into two.
10        :return: The minimum time needed to build all the blocks.
11        """
13        # Transform blocks into a min-heap in place
14        heapify(blocks)
16        # Continue processing until there is only one block left
17        while len(blocks) > 1:
18            # Pop the smallest block (done by one worker) from the heap
19            heappop(blocks)
20            # Pop next smallest block and add the split time
21            # This simulates the split, next job acquisition and pushing it back
22            heappush(blocks, heappop(blocks) + split)
24        # The last element is the total time required to build all blocks
25        return blocks[0]  # Return the minimum time

Java Solution

1class Solution {
2    public int minBuildTime(int[] blocks, int splitTime) {
3        // Create a priority queue to hold the blocks, with the natural order (i.e., min-heap)
4        PriorityQueue<Integer> blockQueue = new PriorityQueue<>();
6        // Add all blocks to the priority queue
7        for (int block : blocks) {
8            blockQueue.offer(block);
9        }
11        // Continue combining blocks until there's only one block left
12        while (blockQueue.size() > 1) {
13            // Remove the smallest two blocks from the queue
14            int fastestWorkerBlock = blockQueue.poll();
15            int secondFastestWorkerBlock = blockQueue.poll();
17            // The two workers team up to split and build the next block;
18            // add the combined time back to the priority queue.
19            // The time is the time of the second worker plus the split time.
20            blockQueue.offer(secondFastestWorkerBlock + splitTime);
21        }
23        // The remaining block in the priority queue requires the longest time to complete.
24        // This will be the total time required to build all the blocks.
25        return blockQueue.poll();
26    }

C++ Solution

1#include <vector>
2#include <queue>
4class Solution {
6    // Finds the minimum time required to build all blocks with a given split time.
7    int minBuildTime(vector<int>& blocks, int split) {
8        // Use a min heap to always process the block that requires the least amount of time.
9        priority_queue<int, vector<int>, greater<int>> minHeap;
11        // Insert all the blocks into the min heap.
12        for (int blockTime : blocks) {
13            minHeap.push(blockTime);
14        }
16        // Continue processing the blocks in the min heap until only one remains.
17        while (minHeap.size() > 1) {
18            // Remove the block with the least building time.
19            minHeap.pop();
21            // Fetch the next block with the least building time.
22            int nextBlockTime = minHeap.top();
23            minHeap.pop();
25            // Combine the two blocks by adding the split time to the next block.
26            // This represents two workers: one splitting and the other building the next block.
27            // Once the split is done, they join forces to finish the next block.
28            minHeap.push(nextBlockTime + split);
29        }
31        // The value remaining in the min heap is the total time needed to build all blocks.
32        return minHeap.top();
33    }

Typescript Solution

1function minHeapify(arr: number[], i: number, n: number) {
2    // Function to heapify a subtree rooted with node i, which is an index in arr[]
3    let smallest = i;
4    let l = 2 * i + 1; // left child
5    let r = 2 * i + 2; // right child
7    // If left child is smaller than root
8    if (l < n && arr[l] < arr[smallest]) {
9        smallest = l;
10    }
12    // If right child is smaller than smallest so far
13    if (r < n && arr[r] < arr[smallest]) {
14        smallest = r;
15    }
17    // If smallest is not root
18    if (smallest !== i) {
19        // Swap
20        [arr[i], arr[smallest]] = [arr[smallest], arr[i]];
22        // Recursive call to heapify the affected sub-tree
23        minHeapify(arr, smallest, n);
24    }
27function buildMinHeap(arr: number[]) {
28    // Function to build a min-heap from a given array
29    let n = arr.length;
31    // Start from the last non-leaf node and heapify each node
32    for (let i = Math.floor(n / 2) - 1; i >= 0; i--) {
33        minHeapify(arr, i, n);
34    }
37function extractMin(arr: number[]): number {
38    // Function to extract the minimum element from the heap
39    let n = arr.length;
41    // The minimum element is at the root of the heap
42    let minElement = arr[0];
44    // Move the last element to the root and reduce heap size
45    arr[0] = arr[n - 1];
46    arr.pop();
48    // Heapify the root node
49    minHeapify(arr, 0, arr.length);
51    return minElement;
54function insertHeap(arr: number[], value: number) {
55    // Function to insert a new value into the heap
56    arr.push(value);
58    // Heapify from the last element up to the root
59    let i = arr.length - 1;
60    while (i != 0 && arr[Math.floor((i - 1) / 2)] > arr[i]) {
61        [arr[i], arr[Math.floor((i - 1) / 2)]] = [arr[Math.floor((i - 1) / 2)], arr[i]];
62        i = Math.floor((i - 1) / 2);
63    }
66// Finds the minimum time required to build all blocks with a given split time
67function minBuildTime(blocks: number[], split: number): number {
68    // Build a min-heap from the given blocks array
69    buildMinHeap(blocks);
71    // Continue processing the blocks until only one remains
72    while (blocks.length > 1) {
73        // Remove the block with the least building time
74        extractMin(blocks);
76        // Fetch the next block with the least building time
77        let nextBlockTime = extractMin(blocks);
79        // Combine the two blocks by adding the split time to the next block
80        // This represents two workers - one splitting and the other building the next block
81        insertHeap(blocks, nextBlockTime + split);
82    }
84    // The value remaining in the heap is the total time needed to build all blocks
85    return blocks[0];

Time and Space Complexity

Time Complexity

The time complexity of the provided code can be analyzed as follows:

  1. The first operation is heapify(blocks), which turns the list into a min-heap. This operation has a time complexity of O(n), where n is the number of elements in the blocks list.
  2. The while loop runs until there is only one block left, so it will run n-1 times, where n is the initial number of elements in blocks.
  3. Inside the while loop, we perform two heap operations for each iteration: heappop(blocks) and heappush(blocks, heappop(blocks) + split).
    • heappop(blocks) has a time complexity of O(log n) for each pop operation because it maintains the heap property after removing the smallest element.
    • heappush(blocks, heappop(blocks) + split) is composed of another heappop(blocks) which is O(log n) and heappush(...) which is O(log n), making the combined operations also O(log n) for each push operation.

Since both heap operations happen each time in the loop, the total time complexity for the while loop is O((n-1) * 2 * log n), which simplifies to O(n log n) considering that n-1 is approximately n for large n.

Because the heapify() operation is O(n) and dominates for smaller n, while the while loop dominates for larger n, the overall time complexity of the code is O(n log n).

Space Complexity

The space complexity of the provided code is analyzed as follows:

  1. The heapify(blocks) operation is done in-place, and does not require additional space, so its space complexity is O(1).
  2. The while loop does not use any extra space other than the space used for the heap operations, which is also done in-place on the blocks list.

Therefore, the overall space complexity of the code is O(1), as no additional space is needed beyond the input list.

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