Introduction

In this blog post, we will explore how to solve the LeetCode problem "1547" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Given a wooden stick of length n units. The stick is labelled from 0 to n. For example, a stick of length 6 is labelled as follows: Given an integer array cuts where cuts[i] denotes a position you should perform a cut at. You should perform the cuts in order, you can change the order of the cuts as you wish. The cost of one cut is the length of the stick to be cut, the total cost is the sum of costs of all cuts. When you cut a stick, it will be split into two smaller sticks (i.e. the sum of their lengths is the length of the stick before the cut). Please refer to the first example for a better explanation. Return the minimum total cost of the cuts. Example 1: Input: n = 7, cuts = [1,3,4,5] Output: 16 Explanation: 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 Explanation: 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.

Explanation

Here's the solution to the optimal cutting stick problem:

• High-Level Approach: The problem can be solved using dynamic programming. We can consider the cuts as splitting the stick into smaller subproblems. The optimal solution can be found by considering all possible orders of performing the cuts and choosing the one that minimizes the total cost. We use memoization to avoid recomputing overlapping subproblems.

• Complexity:

  • Runtime Complexity: O(m^3), where m is the number of cuts.
  • Storage Complexity: O(m^2), where m is the number of cuts.

Code

    def minCost(n: int, cuts: list[int]) -> int:
    """
    Calculates the minimum cost to cut a stick of length n at given cut positions.

    Args:
        n: The length of the stick.
        cuts: A list of positions where the stick should be cut.

    Returns:
        The minimum total cost of the cuts.
    """

    cuts = sorted(cuts)
    cuts = [0] + cuts + [n]  # Add endpoints

    m = len(cuts)
    dp = [[0] * m for _ in range(m)]

    for length in range(2, m):
        for i in range(m - length):
            j = i + length
            dp[i][j] = float('inf')
            for k in range(i + 1, j):
                dp[i][j] = min(dp[i][j], dp[i][k] + dp[k][j] + cuts[j] - cuts[i])

    return dp[0][m - 1]