Minimum Time to Build Blocks

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Greedy Algorithms

You are given a list of blocks, where each block is represented by its length. Each block needs to be built sequentially such that each block i needs to be built before block i+1.

In the beginning, you have a team of one builder. You can split the team into two or keep the team as is. The cost of splitting the team into two is given by split.

Each builder in the team can work on one block independently and in parallel. Given a list of block lengths blocks and an integer split, return the minimum time to build all blocks.

Initially, you have a team of one builder.

Example 1:

Input: blocks = [1], split = 1
Output: 1
Explanation: We use 1 builder to build the only block in 1 unit of time.

Example 2:

Input: blocks = [1,2], split = 5
Output: 7
Explanation:
We split the team into two, which costs 5.
Then, the two builders build the blocks simultaneously for a total of max(1, 2) = 2 units of time.
The total cost is 5 + max(1, 2) = 7

Example 3:

Input: blocks = [1,2,3], split = 1
Output: 4
Explanation:
We split the team into two, which costs 1.
Then, the two builders build the blocks simultaneously for a total of max(1, 2) = 2 units of time.
Then, we use 1 builder to build the last block in 3 units of time.
The total cost is 1 + max(1, 2) + 3 = 1 + 2 + 1 = 4.

Constraints:

1 <= blocks.length <= 1000
1 <= blocks[i] <= 1000
0 <= split <= 100

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the maximum value for `blocks` and the maximum value for an element within the `splitMins` array?
Can the `splitMins` array ever be empty, or can any of its elements be zero or negative?
If it's impossible to build all the blocks, what value should I return?
Are the number of blocks always greater than zero, or could it be zero indicating an immediate completion?
If multiple build schedules result in the same minimum time, is any of them acceptable?

Brute Force Solution

Approach

The goal is to find the fastest way to build all the blocks. The brute force approach tries every possible combination of assigning workers to blocks and calculates the total time for each combination to find the minimum.

Here's how the algorithm would work step-by-step:

Consider each possible assignment of workers to blocks.
For a given assignment, figure out how long each worker would take to build the blocks they are assigned.
Calculate the total time it would take to build all blocks given that specific assignment by finding the maximum time any worker takes.
Repeat this process for every possible worker assignment.
Compare the total time of each assignment and choose the one that resulted in the smallest total time. This is the minimum time to build the blocks.

Code Implementation

def min_time_to_build_blocks_brute_force(blocks, workers_abilities):
    number_of_blocks = len(blocks)
    number_of_workers = len(workers_abilities)
    min_total_time = float('inf')

    # Iterate through all possible worker assignments
    for i in range(number_of_workers ** number_of_blocks):
        worker_assignments = []
        temp_value = i
        for _ in range(number_of_blocks):
            worker_assignments.append(temp_value % number_of_workers)
            temp_value //= number_of_workers

        worker_times = [0] * number_of_workers

        # Calculate the time each worker takes to complete assigned blocks
        for block_index in range(number_of_blocks):
            worker_index = worker_assignments[block_index]
            worker_times[worker_index] += blocks[block_index] /\
                                            workers_abilities[worker_index]

        # Find the maximum time taken by any worker
        max_time = 0
        for worker_index in range(number_of_workers):
            max_time = max(max_time, worker_times[worker_index])

        # Track the overall min time. 
        min_total_time = min(min_total_time, max_time)

    # Brute force explores all possibilities
    return min_total_time

Big(O) Analysis

Time Complexity

O(m^n) – The algorithm considers every possible assignment of n workers to m blocks. Each worker can be assigned to any of the m blocks, resulting in m choices for each worker. Since there are n workers, the total number of possible assignments is m^n. For each assignment, the time to build all blocks is calculated, which takes O(n) to iterate over all workers' build times. Therefore, the overall time complexity is O(m^n * n). Since m^n dominates n, the Big O complexity is O(m^n).

Space Complexity

O(1) – The brute force approach, as described, doesn't explicitly use any auxiliary data structures that scale with the input size. Although it explores every possible assignment of workers to blocks, the plain English explanation doesn't indicate any temporary data structures used to store these assignments or intermediate results. Therefore, the space complexity remains constant, regardless of the number of blocks or workers. This means the space used does not increase with an increase in the input size, N.

Optimal Solution

Approach

The fastest way to determine the minimum time is to use binary search. Binary search allows us to efficiently guess a possible time and then see if we can complete all tasks within that time. If it is possible, we can try a lower time, otherwise, we increase the time to search within.

Here's how the algorithm would work step-by-step:

Recognize that the answer must be between the shortest build time and the sum of all build times.
Pick a time in the middle of the range between the shortest and longest possible times.
Go through the build times. For each build time, determine how many blocks can be built with that build time within your chosen time.
Add up the number of blocks that can be built across all build times.
If the number of blocks that can be built is equal to or greater than the number of blocks required, it means you might be able to do it in a shorter time, so look at the lower half of your range. If it is less, then you need more time so look at the upper half of the range.
Repeat the process of picking a middle time and checking if it's possible, narrowing the range each time, until you find the smallest possible time needed to build all blocks.

Code Implementation

def min_build_time(blocks, split_cost, number_of_blocks):
    shortest_build_time = min(blocks)
    longest_build_time = sum(blocks)

    while shortest_build_time <= longest_build_time:
        middle_time = (shortest_build_time + longest_build_time) // 2
        blocks_built = 0

        for build_time in blocks:
            blocks_built += middle_time // build_time

        # Check if it's possible to build all blocks within the given time.
        if blocks_built >= number_of_blocks:

            longest_build_time = middle_time - 1
        else:
            # If not, increase the search space.

            shortest_build_time = middle_time + 1

    #After the binary search, shortest_build_time is the minimum time.
    return shortest_build_time

Big(O) Analysis

Time Complexity

O(n log s) – The algorithm uses binary search to find the minimum time. Binary search operates on a range between the shortest build time and the sum of all build times (denoted as 's'). The binary search contributes a factor of O(log s). Inside the binary search loop, we iterate through the 'n' build times to calculate how many blocks can be built within the guessed time. This inner loop contributes a factor of O(n). Therefore, the overall time complexity is O(n log s), where 'n' is the number of build times and 's' is the sum of build times.

Space Complexity

O(1) – The algorithm described utilizes binary search and primarily involves calculations based on the input array of build times. It doesn't create any auxiliary data structures like arrays, hash maps, or lists whose size depends on the input size, N (where N is the number of build times). The space used is limited to a few variables for the binary search (e.g., low, high, mid) and intermediate calculations, resulting in constant extra space.

Edge Cases

Case	How to Handle
Null or empty blocks array	Return 0 immediately as no blocks need building.
Single block with build time 0	Return 0 immediately as the block is already built.
Blocks array with all build times equal to 0	Return 0 immediately as all blocks are already built.
Blocks array with very large build times (potential overflow)	Use long data type to prevent integer overflow during calculations of total time, assuming the language supports it.
Blocks array sorted in ascending order of build time	The algorithm should still process this efficiently as binary search or similar approach is not affected by the order.
Blocks array sorted in descending order of build time	The algorithm should still process this efficiently as binary search or similar approach is not affected by the order.
Maximum number of blocks allowed based on memory constraints	Ensure the algorithm's memory usage does not exceed available memory and potentially crash the program.
Extreme boundary values for time taken to build each block	Check the upper bounds of the time taken to build each block to ensure they are within acceptable limits to prevent unexpected behavior such as integer overflow or underflow.

Null or empty blocks array

How to Handle:

Return 0 immediately as no blocks need building.

Single block with build time 0

How to Handle:

Return 0 immediately as the block is already built.

Blocks array with all build times equal to 0

How to Handle:

Return 0 immediately as all blocks are already built.

Blocks array with very large build times (potential overflow)

How to Handle:

Use long data type to prevent integer overflow during calculations of total time, assuming the language supports it.

Blocks array sorted in ascending order of build time

How to Handle:

The algorithm should still process this efficiently as binary search or similar approach is not affected by the order.

Blocks array sorted in descending order of build time

How to Handle:

The algorithm should still process this efficiently as binary search or similar approach is not affected by the order.

Maximum number of blocks allowed based on memory constraints

How to Handle:

Ensure the algorithm's memory usage does not exceed available memory and potentially crash the program.

Extreme boundary values for time taken to build each block

How to Handle:

Check the upper bounds of the time taken to build each block to ensure they are within acceptable limits to prevent unexpected behavior such as integer overflow or underflow.