#1547

Hard

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Minimum Cost to Cut a Stick

Find the minimum cost to cut a stick into segments at specified positions using dynamic programming and sorting.

Problem Statement

You are given a wooden stick of length n and a list of positions where cuts are to be made. The cost of cutting the stick at any position is proportional to the current length of the stick being cut. The task is to find the optimal order of cuts to minimize the total cost. The cuts must be performed in such a way that the total cost is as low as possible.

The problem can be approached using dynamic programming. You need to compute the minimum cost of cutting a stick between any two positions, storing intermediate results in a 2D table to avoid redundant calculations. By evaluating the cost of each cut and considering the subproblems, an optimal solution can be found.

Examples

Example 1

Input: n = 7, cuts = [1,3,4,5]

Output: 16

Using cuts order = [1, 3, 4, 5] as in the input 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 (i.e. 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. Rearranging the cuts to be [3, 5, 1, 4] for example will lead to a scenario with total cost = 16 (as shown in the example photo 7 + 4 + 3 + 2 = 16).

Example 2

Input: n = 9, cuts = [5,6,1,4,2]

Output: 22

If you try the given cuts ordering the cost will be 25. There are much ordering with total cost <= 25, for example, the order [4, 6, 5, 2, 1] has total cost = 22 which is the minimum possible.

Constraints

2 <= n <= 106
1 <= cuts.length <= min(n - 1, 100)
1 <= cuts[i] <= n - 1
All the integers in cuts array are distinct.

Solution Approach

State Transition Dynamic Programming

Use a 2D dynamic programming table dp[i][j] to store the minimum cost of cutting the stick between positions i and j. Start by calculating the cost for smaller subproblems and progressively build up to larger ones. The optimal solution will be found at dp[0][n], where n is the total stick length.

Sorting the Cuts

Sort the cuts to optimize the cost calculation. This allows the dynamic programming approach to efficiently compute the minimum cost as it processes the cuts in a more structured manner.

Minimizing Redundant Calculations

By storing intermediate results in the dynamic programming table, the algorithm avoids recalculating the cost for overlapping subproblems, reducing the overall time complexity.

Complexity Analysis

Metric	Value
Time	O(m^3)
Space	O(m^2)

The time complexity of this solution is O(m^3), where m is the number of cuts. This arises from the need to evaluate all subproblems for each possible cut pair. The space complexity is O(m^2) due to the storage requirements of the 2D dynamic programming table used to store the intermediate results.

What Interviewers Usually Probe

Focus on dynamic programming and state transitions for this problem.
A candidate should be able to optimize the order of cuts and explain how dynamic programming helps with this optimization.
Watch for the candidate's understanding of time and space complexity in relation to dynamic programming solutions.

Common Pitfalls or Variants

Common pitfalls

Failing to sort the cuts optimally, leading to a suboptimal solution.
Not understanding the importance of the 2D dynamic programming table in minimizing redundant calculations.
Not clearly explaining the relationship between the cut positions and the dynamic programming approach.

Follow-up variants

Consider implementing the solution for larger input sizes where the number of cuts increases.
Try optimizing the space complexity by reducing the size of the dynamic programming table.
Explore whether a greedy approach could be used as an approximation for very large input sizes.

FAQ

What is the best way to approach the 'Minimum Cost to Cut a Stick' problem?

The best approach is to use dynamic programming with a state transition table that calculates the minimum cost of cuts between any two positions on the stick.

How does dynamic programming help in minimizing the cutting cost?

Dynamic programming helps by storing intermediate results in a 2D table, preventing redundant calculations and ensuring the optimal cutting order is determined efficiently.

What is the time complexity of the optimal solution for 'Minimum Cost to Cut a Stick'?

The time complexity of the optimal solution is O(m^3), where m is the number of cuts. This comes from evaluating each possible cut pair in the dynamic programming table.

Can a greedy algorithm solve the 'Minimum Cost to Cut a Stick' problem?

A greedy approach might provide a suboptimal solution, but it is not guaranteed to find the minimum cost. Dynamic programming ensures the optimal solution.

What is the role of sorting the cuts in this problem?

Sorting the cuts ensures that the dynamic programming table is filled efficiently, helping minimize the overall cost of cutting the stick.

terminal

Solution

Solution 1: Dynamic Programming (Interval DP)

We can add two elements to the array $\textit{cuts}$, namely $0$ and $n$, representing the two ends of the stick. Then we sort the $\textit{cuts}$ array, so we can divide the entire stick into several intervals, each with two cut points. Let the length of the $\textit{cuts}$ array be $m$.

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class Solution:
    def minCost(self, n: int, cuts: List[int]) -> int:
        cuts.extend([0, n])
        cuts.sort()
        m = len(cuts)
        f = [[0] * m for _ in range(m)]
        for l in range(2, m):
            for i in range(m - l):
                j = i + l
                f[i][j] = inf
                for k in range(i + 1, j):
                    f[i][j] = min(f[i][j], f[i][k] + f[k][j] + cuts[j] - cuts[i])
        return f[0][-1]

Solution 2: Dynamic Programming (Another Enumeration Method)

We can also enumerate $i$ from large to small and $j$ from small to large. This ensures that when calculating $f[i][j]$, the states $f[i][k]$ and $f[k][j]$ have already been computed, where $i \lt k \lt j$.

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class Solution:
    def minCost(self, n: int, cuts: List[int]) -> int:
        cuts.extend([0, n])
        cuts.sort()
        m = len(cuts)
        f = [[0] * m for _ in range(m)]
        for l in range(2, m):
            for i in range(m - l):
                j = i + l
                f[i][j] = inf
                for k in range(i + 1, j):
                    f[i][j] = min(f[i][j], f[i][k] + f[k][j] + cuts[j] - cuts[i])
        return f[0][-1]

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