How do you solve Minimum Cost to Cut a Stick on LeetCode?

Minimum Cost to Cut a Stick (LeetCode #1547) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n³) time complexity and O(n²) space complexity. Key insight: The order of cuts significantly affects the total cost, so finding the optimal order is crucial.

What is the brute force solution for Minimum Cost to Cut a Stick?

The brute force approach for Minimum Cost to Cut a Stick is Brute Force, with O(n!) time complexity and O(n) space complexity. The brute-force approach tries every possible order of cuts and calculates the total cost for each order. This is straightforward but inefficient as it explores all permutations of cuts. The optimal Optimal Solution approach improves this to O(n³).

What algorithm or data structure is used to solve Minimum Cost to Cut a Stick?

Minimum Cost to Cut a Stick uses the following concepts: Array, Dynamic Programming, Sorting. The recommended patterns to study are: Dynamic Programming, Array Manipulation.

What are the interview tips for Minimum Cost to Cut a Stick?

Explain your thought process clearly and ask clarifying questions about the problem. Start with a brute-force solution to demonstrate understanding before moving to optimization. Practice dynamic programming problems to become familiar with the approach and common patterns.

What are common mistakes when solving Minimum Cost to Cut a Stick?

Students often forget to include the boundaries (0 and n) in the cuts array. Not considering all possible cuts between two indices can lead to incorrect results.

#1547

Minimum Cost to Cut a Stick

Hard

ArrayDynamic ProgrammingSortingDynamic ProgrammingArray Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n!)	O(n³)
Space	O(n)	O(n²)

💡

Intuition

Time O(n³)Space O(n²)

The optimal approach uses dynamic programming to minimize the cost of cuts by considering subproblems. It builds a DP table where each entry represents the minimum cost for a segment of the stick, allowing us to avoid redundant calculations.

⚙️

Algorithm

3 steps

1Step 1: Sort the cuts array and add 0 and n to the cuts for boundaries.
2Step 2: Create a DP table where dp[i][j] represents the minimum cost to cut the stick between cuts[i] and cuts[j].
3Step 3: Fill the DP table using the relation: dp[i][j] = min(dp[i][k] + dp[k][j] + (cuts[j] - cuts[i])) for all k between i and j.

solution.py11 lines

1def minCost(n, cuts):
2    cuts = [0] + sorted(cuts) + [n]
3    m = len(cuts)
4    dp = [[0] * m for _ in range(m)]
5    for length in range(2, m):
6        for i in range(m - length):
7            j = i + length
8            dp[i][j] = float('inf')
9            for k in range(i + 1, j):
10                dp[i][j] = min(dp[i][j], dp[i][k] + dp[k][j] + (cuts[j] - cuts[i]))
11    return dp[0][m - 1]

ℹ

Complexity note: The time complexity is O(n³) due to the three nested loops: one for the length of the segment, one for the starting index, and one for the cut index. The space complexity is O(n²) for the DP table storing minimum costs.

1The order of cuts significantly affects the total cost, so finding the optimal order is crucial.
2Dynamic programming allows us to break down the problem into manageable subproblems, leading to an efficient solution.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.