1547. Minimum Cost to Cut a Stick

Problem Description

Given a wooden stick of length n units and an array cuts containing distinct positions where cuts should be made, find the minimum total cost to cut the stick. The cost of a cut is the length of the stick being cut. The order of cuts can be chosen to minimize the total cost.

Solution: Dynamic Programming (Interval DP)

This problem can be efficiently solved using dynamic programming with an interval DP approach.

1. Preprocessing:

• Add 0 and n to the cuts array, representing the beginning and end of the stick.
• Sort the cuts array in ascending order. This allows us to easily work with intervals.

2. DP State:

Let dp[i][j] represent the minimum cost to cut the stick within the interval [cuts[i], cuts[j]].

3. Base Case:

• dp[i][i+1] = 0 for all i, as no cuts are needed for a single interval.

4. Recurrence Relation:

For an interval [cuts[i], cuts[j]], we need to make at least one cut. Let's say we make a cut at cuts[k], where i < k < j. This divides the interval into two subintervals: [cuts[i], cuts[k]] and [cuts[k], cuts[j]]. The cost of this cut is cuts[j] - cuts[i]. Therefore, the minimum cost for the interval [cuts[i], cuts[j]] is:

dp[i][j] = min(dp[i][k] + dp[k][j] + cuts[j] - cuts[i]) for all k such that i < k < j

5. Algorithm:

• Iterate through intervals of increasing length.
• For each interval, consider all possible cut positions within that interval.
• Use the recurrence relation to compute the minimum cost for that interval.

6. Time and Space Complexity:

• Time Complexity: O(m³), where m is the length of the cuts array after adding 0 and n. This is because we have three nested loops: one for the length of the interval, one for the starting index of the interval, and one for iterating through possible cut positions.
• Space Complexity: O(m²), due to the dp table.

Code (Python)

def minCost(n: int, cuts: List[int]) -> int:
    cuts = [0] + cuts + [n]  # Add 0 and n to cuts
    cuts.sort()
    m = len(cuts)
    dp = [[0] * m for _ in range(m)]  # Initialize DP table
 
    for length in range(2, m):  # Iterate through interval lengths
        for i in range(m - length):
            j = i + length
            dp[i][j] = float('inf')  # Initialize with infinity
            for k in range(i + 1, j):  # Iterate through possible cut positions
                dp[i][j] = min(dp[i][j], dp[i][k] + dp[k][j] + cuts[j] - cuts[i])
 
    return dp[0][m - 1]  # Result is in dp[0][m-1]
 

Code (Java)

import java.util.*;
 
class Solution {
    public int minCost(int n, int[] cuts) {
        List<Integer> cutList = new ArrayList<>();
        for (int cut : cuts) {
            cutList.add(cut);
        }
        cutList.add(0);
        cutList.add(n);
        Collections.sort(cutList);
        int m = cutList.size();
        int[][] dp = new int[m][m];
 
        for (int len = 2; len < m; len++) {
            for (int i = 0; i < m - len; i++) {
                int j = i + len;
                dp[i][j] = Integer.MAX_VALUE;
                for (int k = i + 1; k < j; k++) {
                    dp[i][j] = Math.min(dp[i][j], dp[i][k] + dp[k][j] + cutList.get(j) - cutList.get(i));
                }
            }
        }
        return dp[0][m - 1];
    }
}

Other languages (C++, Go, TypeScript) would follow a similar structure, using their respective data structures and syntax. The core algorithmic idea remains the same.