Notes on Minimum Cost to Cut the Stick Problem

Overview

The video discusses a dynamic programming problem known as "Minimum Cost to Cut the Stick." The problem involves determining the minimum cost required to cut a stick of a given length at specified positions. The cost of each cut is equal to the length of the stick being cut at that moment.

Problem Statement

Input:
- A stick of length n (e.g., 7).
- An array of cut positions (e.g., [1, 3, 4, 5]).
Output:
- The minimum cost to cut the stick at the specified positions.
Cost Calculation:
- The cost of a cut is equal to the current length of the stick being cut.

Example Walkthrough

Initial Setup:
- Stick length: 7
- Cuts: [1, 3, 4, 5]
Cutting Process:
- If cuts are made in the order [1, 3, 4, 5]:
  - Cut at 1: Cost = 7 (remaining sticks: lengths 1 and 6)
  - Cut at 3: Cost = 6 (remaining sticks: lengths 1, 2, and 4)
  - Cut at 4: Cost = 4 (remaining sticks: lengths 1, 2, and 3)
  - Cut at 5: Cost = 3 (remaining sticks: lengths 1, 1, and 2)
  - Total Cost = 20
Optimal Order:
- By rearranging cuts to [3, 5, 1, 4], the total cost can be minimized to 16.

Key Concepts

Dynamic Programming (DP):
- The problem can be solved using a DP approach by breaking it down into subproblems.
Sorting Cuts:
- It is essential to sort the cuts to ensure that subproblems are independent.
Cost Calculation Formula:
- When making a cut at position i, the cost incurred is the length of the stick being cut, calculated as: [ \text{Cost} = \text{length of stick} = \text{cuts}[j + 1] - \text{cuts}[i - 1] ]

Approach to Solve the Problem

Define the DP Function:
- Use a recursive function f(i, j) where i and j are the indices of the cuts array.
- Base case: If i > j, return 0 (no cuts to make).
Recursive Case:
- For each cut position between i and j, calculate the cost and recursively solve for the left and right segments: [ \text{cost} = \text{cuts}[j + 1] - \text{cuts}[i - 1] + f(i, k - 1) + f(k + 1, j) ]
- Keep track of the minimum cost across all possible cuts.
Memoization:
- Store results of subproblems to avoid redundant calculations.
Tabulation (Optional Optimization):
- Convert the recursive solution to an iterative one using a table to store results.

Complexity Analysis

Time Complexity:
- The time complexity is (O(m^3)) where (m) is the number of cuts due to the nested loops for each cut position.
Space Complexity:
- The space complexity is (O(m^2)) for the DP table.

Conclusion

The "Minimum Cost to Cut the Stick" problem is a classic example of dynamic programming that illustrates the importance of breaking down problems into manageable subproblems and optimizing through memoization or tabulation. The key takeaway is to sort the cuts to ensure independence of subproblems, which simplifies the solution process.

DP 50. Minimum Cost to Cut the Stick