Problem Description

You have a list of blocks where blocks[i] = t means the i-th block requires t units of time to build. Each block must be built by exactly one worker.

Starting with just one worker, you have two options at any point:

1. Split a worker: A worker can split into two workers, which takes split units of time. If multiple workers split simultaneously, they do so in parallel (the time cost is still just split, not multiplied).
2. Build a block: A worker can build one block and then go home (that worker is no longer available).

Your goal is to find the minimum time needed to build all blocks.

Example scenario:

• If you have blocks [1, 2] and split = 1:
  • The single worker first splits (takes 1 time unit), creating 2 workers
  • Then both workers build their blocks in parallel
  • Total time = 1 (split) + max(1, 2) = 3

The challenge is determining the optimal strategy for when to split workers versus when to assign them to build blocks, minimizing the total time required to complete all blocks.

Intuition

The key insight is to think about this problem in reverse. Instead of thinking forward about splitting workers, we can think backward about merging blocks.

Consider what happens when we have multiple blocks to build:

• With one block: No splitting needed, just build it directly
• With two blocks: We must split once to get 2 workers, then they build in parallel. Total time = split + max(block1, block2)
• With more blocks: The pattern becomes complex if we think forward

The reverse perspective reveals something elegant: When we split a worker to build two blocks, the total time is split + max(block_i, block_j). This is equivalent to "merging" two blocks into one super-block that takes split + max(block_i, block_j) time to build.

Why does greedy selection work here? The blocks with longer build times should participate in as few merges as possible because each merge adds a split cost. By always merging the two smallest blocks first:

1. We minimize the impact of the split overhead on larger blocks
2. Larger blocks get merged later (or not at all if they're the last one), reducing their participation in costly merge operations

Think of it like a tournament bracket in reverse - we want the "strongest" (longest) blocks to have the shortest path to the final, while the "weakest" (shortest) blocks can afford to go through multiple rounds of merging.

Using a min heap allows us to efficiently extract the two smallest blocks at each step, merge them into a new block with time split + max(block_i, block_j) (which simplifies to split + block_j since block_j is the larger one after popping from the heap), and continue until only one block remains.

Learn more about Greedy, Math and Heap (Priority Queue) patterns.

Solution Approach

The implementation uses a greedy algorithm with a min heap to efficiently merge blocks from smallest to largest.

Step-by-step implementation:

1. Initialize the min heap: Convert the blocks list into a min heap using heapify(blocks). This operation takes O(n) time and transforms the list in-place into a heap structure where the smallest element is always at the root.

2. Merge blocks iteratively: While there's more than one block in the heap:

   • Extract the smallest block using heappop(blocks) - this is discarded since it gets absorbed into the merge
   • Extract the second smallest block using heappop(blocks)
   • Create a new merged block with time second_smallest + split
   • Push this new block back into the heap using heappush(blocks, second_smallest + split)

3. Return the final block: After all merges, only one block remains in the heap - this represents the minimum total time needed.

Why the algorithm works:

• When we pop two blocks and only keep the larger one plus split, we're simulating the scenario where one worker splits and two workers build these blocks in parallel
• The time for parallel execution is determined by the slower (larger) block
• By always merging the smallest blocks first, we ensure larger blocks participate in fewer merges, minimizing the accumulation of split costs

Time Complexity: O(n log n) where n is the number of blocks

• Initial heapify: O(n)
• We perform n-1 merge operations
• Each merge involves 2 pops and 1 push, each taking O(log n)

Space Complexity: O(1) as we modify the input list in-place (the heap is built on the original list)

Example Walkthrough

Let's walk through the solution with blocks = [3, 1, 4, 2] and split = 2.

Initial Setup:

• Convert the array into a min heap: [1, 2, 4, 3] (heap representation)
• Visual: We start with 4 blocks that need to be built

Iteration 1:

• Pop smallest: 1 (discarded)
• Pop second smallest: 2
• New block time: 2 + 2 (split) = 4
• Push 4 back to heap
• Heap now: [3, 4, 4]
• Interpretation: One worker splits (2 time), creating two workers who build blocks of size 1 and 2 in parallel. Total time for this group: 2 + max(1,2) = 4

Iteration 2:

• Pop smallest: 3 (discarded)
• Pop second smallest: 4
• New block time: 4 + 2 (split) = 6
• Push 6 back to heap
• Heap now: [4, 6]
• Interpretation: A worker handling the merged block (size 4) splits, and the two resulting workers build blocks of size 3 and 4 in parallel. Total time: 2 + max(3,4) = 6

Iteration 3:

• Pop smallest: 4 (discarded)
• Pop second smallest: 6
• New block time: 6 + 2 (split) = 8
• Push 8 back to heap
• Heap now: [8]
• Interpretation: The initial worker splits, with one branch eventually handling the block of size 4, and the other handling the complex merged block of size 6

Result: The final block has time 8, which is the minimum time needed to build all blocks.

This represents the strategy where:

1. First split creates 2 workers (time: 2)
2. One worker builds block 4, the other splits again (time: 2)
3. After second split, one worker splits again while other builds block 3 (time: 2)
4. After third split, workers build blocks 1 and 2 in parallel (time: max(1,2) = 2)
5. Total: Following the critical path = 2 + 2 + 2 + 2 = 8

Solution Implementation

Time and Space Complexity

The time complexity is O(n × log n), where n is the number of blocks.

• The heapify(blocks) operation takes O(n) time to build a min-heap from the list.
• The while loop runs n - 1 iterations (until only one element remains in the heap).
• In each iteration:
  • heappop(blocks) is called twice, each taking O(log n) time
  • heappush(blocks, ...) is called once, taking O(log n) time
  • Total per iteration: O(log n)
• Overall time complexity: O(n) + O((n-1) × log n) = O(n × log n)

The space complexity is O(n), where n is the number of blocks.

• The heap is created in-place by modifying the input list blocks, so no additional space proportional to n is needed for a separate data structure.
• However, the heap operations themselves use O(log n) space for the call stack during the heapify and heap operations.
• Since the input list itself takes O(n) space and we're considering the total space used, the space complexity is O(n).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

Pitfall 1: Misunderstanding the Parallel Execution Model

The Problem: A common misconception is thinking that when workers split and build blocks in parallel, the total time should be split + block1 + block2. This leads to incorrect implementations that sum all block times.

Why It's Wrong: When multiple workers operate in parallel, they work simultaneously, not sequentially. The total time for parallel operations is determined by the slowest worker, not the sum of all work times.

Incorrect Implementation Example:

# WRONG: This adds both block times
merged_time = first_smallest + second_smallest + split

Correct Understanding: When two workers build blocks in parallel after a split, the time is split + max(block1, block2). Since we always pop the smallest first and discard it, keeping only the larger block time plus split achieves this.

Pitfall 2: Not Handling Edge Cases Properly

The Problem: Failing to handle special cases like empty block lists or single blocks can cause runtime errors or incorrect results.

Solution: Add proper validation:

def minBuildTime(self, blocks: List[int], split: int) -> int:
    # Handle edge cases
    if not blocks:
        return 0
    if len(blocks) == 1:
        return blocks[0]
  
    # Continue with main algorithm...
    heapq.heapify(blocks)
    # ... rest of the code

Pitfall 3: Modifying the Input List Without Consideration

The Problem: The current implementation modifies the input blocks list in-place. If the caller needs the original list preserved, this causes unexpected side effects.

Solution: Create a copy if preserving the original is important:

def minBuildTime(self, blocks: List[int], split: int) -> int:
    # Create a copy to avoid modifying the original
    blocks_heap = blocks.copy()  # or blocks[:] 
    heapq.heapify(blocks_heap)
  
    while len(blocks_heap) > 1:
        heapq.heappop(blocks_heap)
        second_smallest = heapq.heappop(blocks_heap)
        heapq.heappush(blocks_heap, second_smallest + split)
  
    return blocks_heap[0]

Pitfall 4: Incorrect Greedy Strategy

The Problem: Some might think pairing the smallest with the largest block is optimal (to "balance" the work), leading to implementations using both min and max values.

Why It's Wrong: The optimal strategy is to always merge the two smallest values. This ensures that larger block times (which take longer) participate in fewer merge operations, minimizing the accumulation of split costs.

Visual Example:

• Blocks: [1, 1, 3, 5], split = 1
• Correct (merge smallest): Total time = 8
• Wrong (merge min-max): Would result in suboptimal time > 8