Rod Cutting

Problem Overview

The problem "Minimum Cost to Cut a Stick" is an interesting variant of the dynamic programming problems typically seen in competitive programming and programming interviews. The problem stems from an everyday scenario where one needs to figure out the minimum cost of making cuts on a stick at specified positions. Here's a brief breakdown:

• Objective: Return the minimum cost to perform a sequence of cuts on a stick. The cost of making a cut is determined by the length of the stick at the time of the cut.
• Input: An integer n representing the length of a stick and an array cuts representing the positions to cut the stick.
• Constraints: Cuts are non-negative and strictly less than n.

Understanding the Problem

In this problem, you are given a stick of length n which you need to cut at specific positions provided in the array cuts. The cost of making a cut is equal to the length of the stick at the time you make the cut. Thus, we need to cleverly arrange the sequence of cuts to minimize the total cost.

For example, if n = 9 and cuts = [5, 6, 1, 4, 2], the task involves finding the optimal sequence of making these cuts such that the total cost, i.e., the sum of lengths of smaller sticks at each cut is minimized.

Key DSA Concepts and Theory

Dynamic Programming

Dynamic programming (DP) is a method for solving complex problems by breaking them down into simpler subproblems. It is applicable in problems exhibiting the properties of overlapping subproblems and optimal substructure. For this problem, DP helps in exploring all possible ways to cut the stick at given positions and use the results of solved subproblems to determine what's optimal for solving the complete problem.

Problem Decomposition

We can consider each stick segment as an independent problem, and recursively find the minimum cost for each segment using a DP table. We will maintain a 2D DP array where dp[i][j] will denote the minimum cost to cut a stick from position cuts[i] to cuts[j].

Solution Approach

Step-by-step Solution

1. Sorting: Begin by sorting the array cuts. This helps in determining segments easily.

2. Extend Cuts Array: Originally, cuts does not include the starting (0) and ending (n) positions, which will be needed for computation. Extend the array to include 0 as the start and n as the end of the stick.

3. Initialize DP Table: Define a 2D DP array dp such that dp[i][j] represents the minimum cost to cut the segment between cuts[i] and cuts[j].

4. Bottom-Up Calculation:

   • Iterate over the range of segments, starting from smaller segments and expanding to the whole stick.
   • For each segment (i, j), calculate the cost of every cut k between cuts[i] and cuts[j].
   • Update dp[i][j] to be the minimum possible cost obtained by adding the segment cost to the sum of the costs to cut the left and right segments recursively.

5. Final Answer: The solution to the problem will be stored at dp[0][m+1], where m is the number of cuts.

Code Examples

C++

#include <vector>
#include <algorithm>
#include <climits>

class Solution {
public:
    int minCost(int n, std::vector<int>& cuts) {
        // Add the two ends to the cuts
        cuts.push_back(0);
        cuts.push_back(n);
        std::sort(cuts.begin(), cuts.end());
        
        int m = cuts.size();
        std::vector<std::vector<int>> dp(m, std::vector<int>(m, 0));
        
        // Fill the DP table
        for (int length = 2; length < m; ++length) {
            for (int i = 0; i < m - length; ++i) {
                int j = i + length;
                dp[i][j] = INT_MAX;
                // Find the minimum cost for cuts between i and j
                for (int k = i + 1; k < j; ++k) {
                    dp[i][j] = std::min(dp[i][j], dp[i][k] + dp[k][j] + cuts[j] - cuts[i]);
                }
            }
        }
        
        return dp[0][m-1];
    }
};

Java

import java.util.Arrays;

public class Solution {
    public int minCost(int n, int[] cuts) {
        int m = cuts.length;
        int[] newCuts = new int[m + 2];
        newCuts[0] = 0;
        newCuts[m + 1] = n;
        System.arraycopy(cuts, 0, newCuts, 1, m);
        Arrays.sort(newCuts);
        
        int[][] dp = new int[m + 2][m + 2];
        
        for (int length = 2; length < m + 2; ++length) {
            for (int i = 0; i < m + 2 - length; ++i) {
                int j = i + length;
                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] + newCuts[j] - newCuts[i]);
                }
            }
        }
        
        return dp[0][m+1];
    }
}

Time and Space Complexity Analysis

• Time Complexity: The time complexity is O(m^3), where m is the number of cuts. This is due to the triple nested loops iterating over the range and number of cuts.

• Space Complexity: The space complexity is O(m^2) for storing the DP table.

Common Mistakes to Avoid

1. Not Sorting the Cuts Array: Failing to sort the cuts array before computation might lead to incorrect cost calculations.
2. Excluding the Stick Start and End in Cuts: Ensure you include the start (0) and end (n) of the stick in the cuts list for accurate segment calculation.

Similar Problems on LeetCode

1. Burst Balloons (LeetCode #312)
2. Stone Game VII (LeetCode #1690)
3. Paint House (LeetCode #256)

Additional Resources and References

• Dynamic Programming Pattern on LeetCode
• GeeksForGeeks - Dynamic Programming Introduction
• YouTube Tutorials on DP Problems

This article provides a thorough understanding of solving the "Minimum Cost to Cut a Stick" problem using a dynamic programming approach, along with comprehensive theoretical background and practical coding solutions in C++ and Java.