Minimum Cost to Cut a Stick

Hard

Amazon

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Topics:

Dynamic Programming

Given a wooden stick of length n units. The stick is labelled from 0 to n. You are also given an integer array cuts where cuts[i] denotes a position you should perform a cut at. You should perform the cuts in any order you want. The cost of one cut is the length of the stick to be cut, and the total cost is the sum of costs of all cuts. When you cut a stick, it will be split into two smaller sticks.

Return the minimum total cost of the cuts.

For example:

If n = 7 and cuts = [1, 3, 4, 5], one possible cut order is [1, 3, 4, 5]. This leads to the following scenario:

The first cut is done to a rod of length 7, so the cost is 7.
The second cut is done to a rod of length 6 (the second part of the first cut), the third is done to a rod of length 4, and the last cut is to a rod of length 3.
The total cost is 7 + 6 + 4 + 3 = 20.

However, rearranging the cuts to be [3, 5, 1, 4] leads to a total cost of 16.

What is the most efficient way to determine the minimum total cost?

Minimum Cost to Cut a Stick

Hard

Amazon

1 view

Topics:

Dynamic Programming

Given a wooden stick of length n units. The stick is labelled from 0 to n. You are also given an integer array cuts where cuts[i] denotes a position you should perform a cut at. You should perform the cuts in any order you want. The cost of one cut is the length of the stick to be cut, and the total cost is the sum of costs of all cuts. When you cut a stick, it will be split into two smaller sticks.

Return the minimum total cost of the cuts.

For example:

If n = 7 and cuts = [1, 3, 4, 5], one possible cut order is [1, 3, 4, 5]. This leads to the following scenario:

The first cut is done to a rod of length 7, so the cost is 7.
The second cut is done to a rod of length 6 (the second part of the first cut), the third is done to a rod of length 4, and the last cut is to a rod of length 3.
The total cost is 7 + 6 + 4 + 3 = 20.

However, rearranging the cuts to be [3, 5, 1, 4] leads to a total cost of 16.

What is the most efficient way to determine the minimum total cost?

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 are the possible values and data types for the elements in the `cuts` array and the `n` integer? Specifically, can the cuts be zero, negative, or equal to n?
Is the `cuts` array guaranteed to be sorted, and are there any duplicate cut positions allowed?
If the `cuts` array is empty, what should be returned?
Can you provide an example input and the corresponding optimal cut sequence to clarify the cost calculation?
Is there a specific maximum value for 'n', and what is the expected order of magnitude for the number of cuts, to help guide my thinking on potential time complexity bottlenecks?

Brute Force Solution

Approach

We want to find the cheapest way to cut a stick at specific locations. The brute force method explores every single possible order in which we could make those cuts, like trying all the routes through a maze.

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

Consider all possible sequences of cuts.
For each sequence, calculate the cost of making the cuts in that specific order. Remember that the cost of a cut is equal to the length of the remaining stick before the cut.
Once you have the cost for every possible sequence of cuts, compare them all.
Select the sequence of cuts that has the lowest total cost. This is the optimal solution found by checking every possibility.

Code Implementation

def min_cost_cut_stick_brute_force(stick_length, cuts):
    import itertools

    min_total_cost = float('inf')

    # Iterate through all possible permutations of the cuts
    for cut_permutation in itertools.permutations(cuts):
        total_cost_for_permutation = 0
        current_stick_length = stick_length
        left = 0

        for cut_position in cut_permutation:
            total_cost_for_permutation += current_stick_length
            current_stick_length = cut_position - left
            left = cut_position
        total_cost_for_permutation += current_stick_length

        # Update minimum total cost if the current permutation's cost is lower
        min_total_cost = min(min_total_cost, total_cost_for_permutation)

    # Return the minimum cost found
    return min_total_cost

Big(O) Analysis

Time Complexity

O(n!) – The brute force approach considers all possible orderings of the cuts. Given 'n' cuts, there are n! (n factorial) possible sequences in which these cuts can be made. Calculating the cost for each sequence involves 'n' cut operations. Therefore, the total time complexity is dominated by generating all the permutations, leading to a time complexity of O(n!).

Space Complexity

O(1) – The brute force approach described considers all possible sequences of cuts, calculates the cost for each sequence, and selects the sequence with the minimum cost. This iterative process inherently does not maintain any data structure that grows proportionally with the input (N which represents the number of cuts). It only requires a few variables to track the minimum cost found so far, and the cost of the sequence being evaluated; the storage requirements for these variables are constant with respect to the input size. Therefore, the auxiliary space complexity of the brute force approach is O(1).

Optimal Solution

Approach

This is a problem about finding the cheapest way to cut a stick at specific points. The core idea is to break down the big problem into smaller, overlapping subproblems and solve them in an efficient order to find the overall minimum cost.

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

Imagine the cuts as dividing the stick into sections. Think of each section between two cuts as a smaller 'stick' that needs to be cut further.
Start by considering the smallest possible sections of the stick. What's the cheapest way to cut those? This will involve figuring out the cost of a single cut within that small section.
Once you know the cost of cutting the small sections, use that information to figure out the cost of cutting slightly larger sections.
Keep building up, considering larger and larger sections. For each section, try every possible place to make the first cut, and then use the previously calculated costs for the smaller sections created by that cut.
Always choose the cut that results in the lowest total cost for that section.
Continue this process until you've considered the entire stick. The cost you calculated for the entire stick is the minimum cost to cut it at all the given points.

Code Implementation

def minCost(stick_length, cuts):    cuts = [0] + sorted(cuts) + [stick_length]
    number_of_cuts = len(cuts)
    dp = [[0] * number_of_cuts for _ in range(number_of_cuts)]
    
    for length in range(2, number_of_cuts): 
        for left in range(number_of_cuts - length):            right = left + length
            dp[left][right] = float('inf')
            
            # Try each cut to find the min cost
            for cut_index in range(left + 1, right):                dp[left][right] = min(dp[left][right], dp[left][cut_index] + dp[cut_index][right] + cuts[right] - cuts[left])
                
    # The min cost for the entire stick
    return dp[0][number_of_cuts - 1]

Big(O) Analysis

Time Complexity

O(n^3) – The algorithm iterates through all possible sub-sticks. There are O(n^2) such sub-sticks, defined by the start and end cut indices after padding the cuts array. For each sub-stick, the algorithm iterates through all possible cut locations within that sub-stick to find the optimal cut. This inner loop iterates at most n times, where n represents the number of cuts. Therefore, the overall time complexity is O(n^2) * O(n) = O(n^3).

Space Complexity

O(n^2) – The dynamic programming approach described in the plain English explanation uses a table (often a 2D array) to store the minimum costs of cutting various sub-sections of the stick. If there are n cuts (after including the stick's ends), this table will have dimensions roughly n x n to store the results of all possible subproblems. Therefore, the auxiliary space required is proportional to n squared, where n is the number of cuts plus two (for the start and end of the stick). This dominates the space complexity, leading to O(n^2).

Edge Cases

Case	How to Handle
cuts is null or empty	If cuts is null or empty, the cost is 0; return 0.
stick length is zero or negative	If stick length is zero, return 0; if it is negative, throw an IllegalArgumentException.
cuts contains duplicate values	The algorithm should still function correctly with duplicates, effectively cutting at the same position multiple times, which increases cost.
cuts contains negative values	Throw an IllegalArgumentException if cuts contains negative values, as cutting at negative positions is invalid.
cuts contains values greater than stick length	Throw an IllegalArgumentException if cuts contains values greater than the stick length, as cutting beyond the stick is invalid.
cuts is sorted in ascending or descending order	The dynamic programming approach handles any order of cuts correctly; sorting might improve performance slightly, but isn't required for correctness.
Integer overflow during cost calculation	Use long data type for storing intermediate cost values to prevent potential integer overflow.
Maximum number of cuts (scalability/memory)	The dynamic programming table size is O(n^2), where n is the number of cuts, so consider the memory constraints and potential for stack overflow with recursion for very large inputs.