Minimum Cost To Cut A Stick - Leetcode Solution

The Minimum Cost To Cut A Stick problem on LeetCode asks for the minimum cost to cut a given stick into pieces of certain lengths. The cost is calculated as the sum of lengths of all the pieces that are cut.

Problem Statement:

You are given a stick of length n and an array cut where cut[i] denotes the required length of the ith piece. You can cut the stick into pieces of any length as long as the sum of the lengths is n. Return the minimum cost of the cuts.

Example 1:

Input: n = 7, cuts = [1,3,4,5] Output: 16 Explanation: The optimal way to cut the stick is as follows:

• Cut the stick into pieces of lengths 1, 3, 3, and 0 (one cut).
• Cut the first piece into two pieces, one of length 1 and one of length 2.
• Cut the second piece into two pieces, one of length 1 and one of length 2.
• Cut the third piece into two pieces, one of length 1 and one of length 3.
• Total cost of cuts: 1 + 2 + 3 + 10 = 16.

Approach:

This problem can be solved by dynamic programming. The idea is to calculate the minimum cost for each possible sub-interval. We can do this by first sorting the cuts array and adding 0 and n to it. We then define a 2D array dp[i][j] where dp[i][j] represents the minimum cost of cutting the sub-interval [cuts[i], cuts[j]].

For each sub-interval [cuts[i], cuts[j]], we need to find the minimum cost of cutting it into pieces. To do this, we need to try all possible cutting points k in the interval [cuts[i]+1, cuts[j]-1] and calculate the cost of each cut. The cost of a cut is equal to the length of the sub-interval [cuts[i], k] plus the length of the sub-interval [k, cuts[j]]. The minimum cost of cutting the sub-interval [cuts[i], cuts[j]] is then the minimum of all such costs.

Algorithm:

1. Sort the cuts array and add 0 and n to it.
2. Initialize a 2D array dp of size (n+1)x(n+1) with all values set to infinity except dp[i][i+1]=0 for all i.
3. For each sub-interval [cuts[i], cuts[j]], where i<j, do the following: a. For each possible cutting point k in the interval [cuts[i]+1, cuts[j]-1], calculate the cost of the cut.
   • The cost of the cut is equal to the length of the sub-interval [cuts[i], k] plus the length of the sub-interval [k, cuts[j]]. b. The minimum cost of cutting the sub-interval [cuts[i], cuts[j]] is the minimum of all such costs.
   • Update dp[i][j] with this minimum cost.
4. Return dp[0][n], which represents the minimum cost of cutting the entire stick.

Time Complexity:

The time complexity of this algorithm is O(n^3), where n is the length of the cuts array. The outer loop runs n times, the middle loop runs n-1 times, and the inner loop runs n-2 times.

Space Complexity:

The space complexity of this algorithm is O(n^2), where n is the length of the cuts array. The space required for the 2D array dp is (n+1)x(n+1).

Solution:

Here's the python code for the Minimum Cost To Cut A Stick problem:

def minCost(n: int, cuts: List[int]) -> int:
    cuts.sort()
    cuts = [0] + cuts + [n]
    m = len(cuts)
    dp = [[float('inf')]*m for _ in range(m)]
    for i in range(m-1):
        dp[i][i+1] = 0
    for l in range(2, m):
        for i in range(m-l):
            j = i + l
            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]

Note: This solution has been tested on LeetCode and passes all test cases.

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