Minimum Time to Build Blocks

class Solution {
public:
    int minimumBuildTime(vector<int>& cells, int divisionTime) {
        // Sort cells in descending order
        sort(cells.begin(), cells.end(), greater<int>());

        // Helper function to check feasible time limit
        auto canBuildWithin = [&](int timeLimit) {
            int workers = 1;

            for (int i = 0; i < cells.size(); i++) {
                int currentBlockTime = cells[i];
                if (workers == 0 || currentBlockTime > timeLimit) {
                    return false;
                }

                while (currentBlockTime + divisionTime <= timeLimit) {
                    timeLimit -= divisionTime;
                    workers <<= 1;

                    if (workers >= cells.size() - i) {
                        return true;
                    }
                }

                workers--;
            }

            return true;
        };

        int minimum = cells[0];
        int maximum = divisionTime * int(ceil(log2(cells.size()))) + cells[0];
        while (minimum < maximum) {
            int middle = minimum + (maximum - minimum) / 2;
            if (canBuildWithin(middle)) {
                maximum = middle;
            } else {
                minimum = middle + 1;
            }
        }

        return maximum;
    }
};

This C++ solution efficiently determines the minimum time required to build blocks, given a list of individual build times (cells) and a division time (divisionTime). The high-level approach uses a binary search strategy combined with a greedy algorithm.

The strategy involves:

• Sorting the block times in descending order.
• Using a binary search to find the minimum possible time to complete all builds.
• Utilizing a helper function canBuildWithin to check if all blocks can be built within a proposed time limit. This function:
  • Initiates a counter for the number of workers.
  • Iterates through the block times and determines if the currently available number of workers can complete the construction within the proposed time by repeatedly dividing and doubling the number of workers based on the division time until either all blocks are assigned to workers or the time limit exceeds.

The core of the solution lies in dynamically adjusting the number of workers in response to the remaining time left for building within each iteration of the binary search. This approach efficiently narrows down the minimum required time.

The cells vector is modified in-place for sorting but the adjusted code efficiently manages the indexed access to support the binary and greedy function's requirements.

The function returns the minimum time calculated via binary search, ensuring all blocks are built within this time while maximizing the utility of each worker's contribution adjusted by the division time.

class Solution {
    public int minimumBuildTime(int[] blocks, int splitCost) {
        // Reverse sorting of arrays
        Arrays.sort(blocks);
        for (int i = 0; i < blocks.length / 2; i++) {
            int swap = blocks[i];
            blocks[i] = blocks[blocks.length - i - 1];
            blocks[blocks.length - i - 1] = swap;
        }
        
        // Check if it's possible to build within 'timeLimit'
        Predicate<Integer> canBuild = (timeLimit) -> {
            int workers = 1;

            for (int blockTime : blocks) {
                if (workers == 0 || blockTime > timeLimit) return false;

                // Increase workers while possible within current time limit
                while (blockTime + splitCost <= timeLimit) {
                    timeLimit -= splitCost;
                    workers <<= 1;
                    
                    if (workers >= blocks.length - Arrays.binarySearch(blocks, blockTime))
                        return true;
                }
                
                workers--;
            }
            
            return true;
        };
        
        // Determine the minimum time using binary search
        int lowerBound = blocks[0];
        int upperBound = splitCost * (int) Math.ceil(Math.log(blocks.length) / Math.log(2)) + blocks[0];
        while (lowerBound < upperBound) {
            int middle = lowerBound + (upperBound - lowerBound) / 2;
            if (canBuild.test(middle)) {
                upperBound = middle;
            } else {
                lowerBound = middle + 1;
            }
        }
        
        return upperBound;
    }
}

The solution to the "Minimum Time to Build Blocks" problem involves calculating the smallest amount of time required to build all the given blocks, considering the time cost both to split the workers and to actually work on the blocks. The Java solution provided uses a combination of techniques: reverse sorting of the array, predicate-based checking, and binary search, encapsulated in a class method.

• Reverse Sorting: Reverse sorting is applied to the array of block times to facilitate the processing of blocks in descending order. This is achieved by first sorting the array using Arrays.sort(blocks) followed by a manual reversal.

• Predicate for Time Feasibility Check: A lambda function canBuild defines a predicate to determine if it's possible to build all blocks within a given timeLimit. Here, workers double every time the accumulated time including the split would not exceed the current timeLimit, indicating efficiency gain by splitting workers. The check (blockTime > timeLimit) ensures that at no point does the time required to build a block exceed the permitted time, maintaining feasibility.

• Binary Search for Optimal Time: To efficiently determine the minimum time, a binary search is executed between a calculated lower bound (blocks[0], the fastest single block time after sorting) and an upper bound (related to the split cost and the logarithmic scale of block count). The binary search helps in pinpointing the smallest timeLimit where canBuild returns true.

Execute this method with your given block times and split cost to determine the minimum time required to build all blocks. The approach ensures that the solution is computationally efficient, leveraging both sorting and binary search for performance optimization.

class Solution:
    def constructTime(self, bricks: List[int], timeToSplit: int) -> int:        
        bricks.sort(reverse=True)

        def canFinishWithin(targetTime):           
            activeWorkers = 1

            for idx, brickTimeRequired in enumerate(bricks):
                if activeWorkers <= 0 or brickTimeRequired > targetTime:
                    return False
                
                while brickTimeRequired + timeToSplit <= targetTime:
                    targetTime -= timeToSplit
                    activeWorkers *= 2

                    if activeWorkers >= len(bricks) - idx:
                        return True
                
                activeWorkers -= 1

            return True

        leftLimit = bricks[0]
        rightLimit = math.ceil(log2(len(bricks))) * timeToSplit + bricks[0]
        while leftLimit < rightLimit:
            middle = (leftLimit + rightLimit) // 2
            if canFinishWithin(middle):
                rightLimit = middle
            else:
                leftLimit = middle + 1
        
        return int(rightLimit)

This Python3 solution addresses the problem of determining the minimum time required to build blocks, given a list of blocks and a constant time cost for each block-splitting operation.

The strategy involves sorting the block-build times in descending order and attempting to minimize the total time spent given the constraints of each brick's build time and the time it takes to split workers to handle multiple bricks simultaneously.

Here's the step-by-step breakdown of the code:

1. Sort the list of brick times in descending order.
2. Implement a helper function canFinishWithin(targetTime) to determine if it's feasible to finish constructing all bricks within a given period targetTime using the current number of workers.
3. Start with an initial worker. Iterate through each brickTimeRequired:
   • If the current brick's required time is greater than targetTime, or if there are no available workers left, it's impossible to finish within targetTime.
   • Adjust available workers by considering the split time. If at any step, available workers are enough to finish the remaining bricks, conclude that it is possible.
   • Continue this until all bricks are considered, adjusting the count of active workers suitively.
4. Establish search bounds for binary search (leftLimit and rightLimit) based on the largest brick time and a theoretical upper time limit derived from the logarithmic relationship between the number of bricks and the split time.
5. Execute a binary search to find the minimal targetTime for which canFinishWithin returns true.
6. Return the optimized minimal time as the solution.

This approach leverages sorting, greedy decisions (in worker management), and binary search to efficiently determine the minimal required time, making it a robust solution for the problem.