Minimum Cost to Cut a Stick

class Solution {
    int[][] dpTable;
    int[] cutsModified;
    
    private int computeCost(int start, int end) {
        if (dpTable[start][end] != -1) {
            return dpTable[start][end];
        }
        if (end - start == 1) {
            return 0;
        }
        int minimumCost = Integer.MAX_VALUE;
        for (int mid = start + 1; mid < end; mid++) {
            int splitCost = computeCost(start, mid) + computeCost(mid, end) + cutsModified[end] - cutsModified[start];
            minimumCost = Math.min(minimumCost, splitCost);
        }
        dpTable[start][end] = minimumCost;
        return minimumCost;
    }
    
    public int minCost(int n, int[] cuts) {
        int len = cuts.length;
        cutsModified = new int[len + 2];
        System.arraycopy(cuts, 0, cutsModified, 1, len);
        cutsModified[len + 1] = n;
        Arrays.sort(cutsModified);
        
        dpTable = new int[len + 2][len + 2];
        for (int i = 0; i <= len + 1; ++i) {
            Arrays.fill(dpTable[i], -1);
        }
        
        return computeCost(0, cutsModified.length - 1);
    }    
}

This solution utilizes dynamic programming to solve the problem of minimizing the cost to cut a stick. Defined in Java, the class named Solution contains methods to compute and return the minimal cutting cost.

• Initialize an array cutsModified and a 2D array dpTable.

  • cutsModified holds the modified positions of cuts including the ends of the stick,
  • dpTable is used to store the cost of cutting the stick between indices. Each element in dpTable is initialized to -1 to denote that no calculation has been made yet.

• The core logic is handled in the computeCost method, which computes the minimum cost of making cuts between two indices.

  • Base Case: If there's only one segment (difference between the end and start indices is 1), return a cost of 0.
  • Recursive Case: Iteratively check the cost of making a cut at every possible position between start and end. Calculate the total cost by adding the cost of left and right segments along with the cost of the current cut. Update minimumCost if a cheaper option is found.
  • Memoization: Store the computed minimum cost in dpTable to avoid recalculating costs for the same indices.

• In the minCost method:

  • Populate cutsModified with the provided cuts and the stick's total length n.
  • Sort cutsModified for a sequenced approach in cutting.
  • Call computeCost passing the whole range to find the minimum cost for all cuts.

This implementation ensures efficiency by leveraging memoization, preventing recalculation by storing the results of subproblems. Arrays are handled efficiently and only necessary data is computed, which is well suited for larger inputs and constraints.

class Solution:
    def minimumCost(self, total_length: int, cut_positions: List[int]) -> int:
        calculated_costs = {}
        cut_positions = [0] + sorted(cut_positions) + [total_length]

        def calculate_cost(start, end):
            if (start, end) in calculated_costs:
                return calculated_costs[(start, end)]
            if end - start == 1:
                return 0
            minimum_cost = min(calculate_cost(start, mid) + calculate_cost(mid, end) + cut_positions[end] - cut_positions[start] for mid in range(start + 1, end))
            calculated_costs[(start, end)] = minimum_cost
            return minimum_cost

        return calculate_cost(0, len(cut_positions) - 1)

The given Python code defines a solution to determine the minimum cost required to cut a stick into predefined parts. The program uses a memoized recursion for efficient cost calculation.

• You define a solution class with a method minimumCost that takes the total length of the stick and a list of positions where the cuts are to be made.
• Start by initializing a dictionary calculated_costs to store previously calculated subproblems, which helps in reducing unnecessary calculations (memoization).
• Amend the cut_positions list to include the start (0) and the end (total_length) of the stick.
• The calculate_cost function is defined inside minimumCost. It calculates the minimum cost to cut the stick between two positions, start and end.
  • If the cost has been calculated before, it returns the stored value.
  • If start and end are consecutive, no cut is required, and the cost is 0.
  • It calculates the cost by iterating from start + 1 to end - 1 to find the optimal cut position mid that minimizes the cost of cutting the stick into two parts and adds the cost of making the cut.
• The entry point of recursion is calculate_cost(0, len(cut_positions)-1), which calculates the cost for the whole stick.
• The final return from the minimumCost method provides the minimum cost to completely separate the stick according to the cut positions provided.