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Minimum Time to Build Blocks

Minimum Time to Build Blocks

This problem involves finding the minimum time to build a set of blocks given a certain number of workers and the cost to split workers. A greedy approach using a priority queue (min-heap) provides an efficient solution.

Approach

The core idea is to iteratively merge the blocks with the shortest build times. This is because merging shorter blocks first minimizes the impact of the splitting cost (split).

  1. Initialization: Create a min-heap containing all block build times.

  2. Iteration: While more than one block remains in the heap:

    • Remove the two smallest blocks from the heap.
    • Calculate the time to build a new combined block: split + max(block1, block2).
    • Add this combined block back to the heap.

  3. Result: The final element remaining in the heap represents the minimum total build time.

Time and Space Complexity

  • Time Complexity: O(N log N), dominated by the heap operations (insertion and extraction). N is the number of blocks.
  • Space Complexity: O(N), to store the min-heap.

Code Implementation (Python)

import heapq
 
def minBuildTime(blocks: list[int], split: int) -> int:
    """
    Calculates the minimum time to build blocks.
 
    Args:
        blocks: A list of integers representing the build time for each block.
        split: An integer representing the cost to split a worker.
 
    Returns:
        An integer representing the minimum total build time.
    """
    heapq.heapify(blocks)  # Convert the list to a min-heap in-place
 
    while len(blocks) > 1:
        # Extract the two smallest build times
        block1 = heapq.heappop(blocks)
        block2 = heapq.heappop(blocks)
 
        # Calculate the time for the combined block and re-insert into the heap
        combined_time = max(block1, block2) + split
        heapq.heappush(blocks, combined_time)
 
    return blocks[0] # The last element is the minimum total time

Code Implementation (Java)

import java.util.PriorityQueue;
 
class Solution {
    public int minBuildTime(int[] blocks, int split) {
        PriorityQueue<Integer> pq = new PriorityQueue<>();
        for (int block : blocks) {
            pq.offer(block);
        }
 
        while (pq.size() > 1) {
            int block1 = pq.poll();
            int block2 = pq.poll();
            pq.offer(Math.max(block1, block2) + split);
        }
 
        return pq.poll();
    }
}

Code Implementation (C++)

#include <queue>
#include <algorithm>
#include <vector>
 
using namespace std;
 
class Solution {
public:
    int minBuildTime(vector<int>& blocks, int split) {
        priority_queue<int, vector<int>, greater<int>> pq(blocks.begin(), blocks.end());
 
        while (pq.size() > 1) {
            int block1 = pq.top(); pq.pop();
            int block2 = pq.top(); pq.pop();
            pq.push(max(block1, block2) + split);
        }
 
        return pq.top();
    }
};

The implementations in other languages (Go, TypeScript, Rust) would follow a similar structure, utilizing their respective priority queue implementations. The core algorithm remains the same: a greedy approach using a min-heap to efficiently find the minimum build time.