CODEWITHSUNDEEP
pythonalgorithmdata-structurestime-complexitygreedy
Adel
Adel

Reputation: 301

How can I enhance the time complexity of arranging bricks into stacks based on specific conditions?

I’m working on a problem where I need to arrange N bricks of varying lengths into the least number of stacks. The rules for stacking bricks are that a brick 𝑖 can be placed on top of brick 𝑗 only if the condition 𝐴𝑖 + 𝑥 ≤ 𝐴𝑗 holds true, where 𝑥 is a given integer.

Here is my current implementation in Python:

def arrange_bricks(N, x, bricks):
    # Sort the bricks in descending order
    bricks.sort(reverse=True)

    # List to hold stacks
    stacks = []

    for brick in bricks:
        placed = False
        for stack in stacks:
            if brick + x <= stack[-1]:
                stack.append(brick)
                placed = True
                break
        
        if not placed:
            stacks.append([brick])

    print(len(stacks))
    for stack in stacks:
        print(len(stack), ' '.join(map(str, stack)))

N, x = map(int, input().split())
bricks = list(map(int, input().split()))
arrange_bricks(N, x, bricks)

Problem: Time Complexity: Currently, the time complexity of my solution is 𝑂(𝑁^2) in the worst case, as each brick potentially checks against all existing stacks. This can be inefficient, especially when 𝑁 is large (up to 10^6).

Desired Outcome: I want to enhance the time complexity to be more efficient, preferably around 𝑂(𝑁log𝑁)if possible.

Constraints:

• 1 ≤ 𝑁 ≤ 10^6

• 1 ≤ 𝑥 ≤ 10^9

• 1 ≤ 𝐴𝑖 ≤ 10^9

Questions: How can I modify my approach to use a more efficient data structure or algorithm to reduce the time complexity? Are there specific algorithms or techniques I can apply to solve this problem more effectively? Thank you for your help!

Upvotes: 1

Views: 176

Answers (2)

no comment
no comment

Reputation: 10122

You can optimize your solution to O(N) after the initial sorting of the bricks (so it's O(N log N) overall).

Instead of trying the current brick on all stacks, try just the one with the longest top brick. Keep the stacks sorted by top brick, from longest top brick to shortest top brick. This way you just try the first stack. If you do place the current brick onto it, then that stack now has the shortest top brick, so move that stack to the end.

So import deque from collections and change your

    # List to hold stacks
    stacks = []

    for brick in bricks:
        placed = False
        for stack in stacks:
            if brick + x <= stack[-1]:
                stack.append(brick)
                placed = True
                break
        
        if not placed:
            stacks.append([brick])

to this:

    # Deque to hold stacks
    stacks = deque()

    for brick in bricks:
        if stacks and brick + x <= stacks[0][-1]:
            stacks[0].append(brick)
            stacks.rotate(-1)
        else:
            stacks.append([brick])

Attempt This Online!

Upvotes: 2

Y.T.
Y.T.

Reputation: 2730

A simple greedy algorithm does the trick.

First, sort the bricks into descending order, a1 >= a2 >= ... >= an.

Observation 1: For a stack, we only need to store the length of the top brick.

We'll put the bricks one by one and maintain the sorted length of the top brick of the current stacks in a BST S. For each brick ai, we find the shortest top brick in S such that ai can be put on top (si >= ai + x) if such a top brick exists and put ai on top. Otherwise, we need to start a new stack and put ai as the base brick. Searching the shortest top brick such that ai can be put on top can be done via binary search on the BST S.

Observation 2: This is optimal. It can be proven by the exchange argument.

The outline of the exchange argument goes as follows:

Assume an optimal solution uses n stacks T = {stack_1, stack_2, ..., stack_n} and our algorithm uses m stacks S = {stack_1, stack_2, ..., stack_m}.

We identify the location of a1 in T, say stack_i. a1 must be at the bottom of stack_i.

Now, if a2 + x <= a1 but a2 is not on top of a1 in T, a2 must be at the base of another stack stack_j where i != j. We can swap {stack_i minus a_1} with {stack_j} and the resulted set T' will also be a valid solution with n stacks where the first step of putting a2 matches with our algorithm. We repeat the exchange process recursively by ignoring a1 from T and focuses on {a2, ..., an}

Otherwise, if a2 + x > a1, then a2 must be at the bottom of some stack_j such that i != j. This matches with our algorithm. We repeat the exchange process recursively by ignoring a1 from T and focuses on {a2, ..., an}.

At the very end, the set T will be transformed to S while keeping the number of stacks unchanged, proving that |S| = |T|.

Upvotes: 0

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