Minimum Cost to Cut the Stick (Partition DP)

1. Why Do We Need This? (Problem Statement & Status Quo)

When working with sticks or ropes that need to be cut at multiple specific points, the order in which you make those cuts dramatically impacts the total cost incurred.

Problem Definition: You are given a stick of length $n$ and an array of integer cuts where cuts[i] represents a position on the stick where a cut must be made.
Cost Definition: The cost of making a single cut is equal to the length of the stick currently being cut.
Objective: Find the minimum total cost required to make all the specified cuts. You are allowed to perform the cuts in any arbitrary order.

2. Core Mental Model & Real-Life Analogy

Real-Life Analogy: Cutting a Wooden Plank

Imagine you are a carpenter holding a wooden plank of length $7$ . You need to make notches (cuts) at positions [1, 3, 4, 5].

If you cut at position 1 first, the entire plank of length $7$ is split into a piece of length $1$ and a piece of length $6$ . The cost for this first cut is $7$ .
Subsequent cuts only operate on smaller individual pieces, but their costs depend on the length of whatever specific segment is currently under the saw.

To minimize the total effort (sum of all plank lengths cut), you must choose the optimal sequence of cuts. Trying all permutations manually leads to an exponential time complexity, making Partition Dynamic Programming necessary to find an optimal substructure.

3. The Order Independence Trap & Why Sorting is Mandatory

At first glance, it might seem that you can pick any cut from the cuts array, make it, and split the problem. However, if you make a cut at an arbitrary point, the remaining sub-problems might depend on each other unless the cuts array is sorted.

Why Sorting is Required: If you sort the cuts array beforehand (e.g., [1, 3, 4, 5]), making a cut at a specific index physically divides the stick into completely independent left and right sub-sticks.
Independence: A cut performed on the left sub-stick has zero structural effect or length dependency on the right sub-stick.

4. Step-by-Step Evolution of the Approach

To evaluate partitions cleanly, we augment the cuts array:

Insert 0 at the beginning of the cuts array (representing the left boundary of the main stick).
Insert n (the stick length) at the end of the cuts array (representing the right boundary of the main stick).
Sort the modified cuts array.

Visualizing the Augmented Cuts Array

If original cuts were [1, 3, 4, 5] and length $n = 7$ :

CODE

Augmented Cuts: [0, 1, 3, 4, 5, 7]
Indices:         0  1  2  3  4  5

Now, any partition from index i to j refers to evaluating cuts between cuts[i] and cuts[j].

5. Formulating the Recurrence Relation

When solving a sub-problem for a partition range from index i to j:

We iterate through every possible first cut index k from i up to j.
The cost of making that initial cut at index k is the length of the current stick segment:

\text{Length} = \text{cuts}[j + 1] - \text{cuts}[i - 1]

We then recursively solve the two independent sub-problems:
1. Left Sub-problem: Range from i to k - 1
2. Right Sub-problem: Range from k + 1 to j

Structure & Concept Flow

cuts[i-1] cuts[k] cuts[j+1]

| | |

|<--- Left Sub-problem->|<--- Right Sub-problem->|

(i to k - 1) (k + 1 to j)

Recurrence Equation

\text{f}(i, j) = \min_{k = i}^{j} \Big( (\text{cuts}[j + 1] - \text{cuts}[i - 1]) + \text{f}(i, k - 1) + \text{f}(k + 1, j) \Big)

Base Case: If $i > v$ , there are no cuts to make in this range, so the cost is 0.

6. Complete Recursive & Memoized Code (Python/C++ Paradigm)

Here is how the recurrence translates into clean memoized code using C++:

C++

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

class Solution {
private:
    int solve(int i, int j, vector<int>& cuts, vector<vector<int>>& dp) {
        // Base case: invalid partition range
        if (i > j) return 0;
        
        // Return memoized result if available
        if (dp[i][j] != -1) return dp[i][j];
        
        int mini = INT_MAX;
        
        // Try every possible cut 'ind' between i and j as the first cut
        for (int ind = i; ind <= j; ind++) {
            int cost = (cuts[j + 1] - cuts[i - 1]) 
                     + solve(i, ind - 1, cuts, dp) 
                     + solve(ind + 1, j, cuts, dp);
            
            mini = min(mini, cost);
        }
        
        return dp[i][j] = mini;
    }

public:
    int minCost(int n, vector<int>& cuts) {
        int c = cuts.size();
        
        // Step 1: Add boundary markers (0 and n)
        cuts.push_back(n);
        cuts.insert(cuts.begin(), 0);
        
        // Step 2: Sort to ensure sub-problems are independent
        sort(cuts.begin(), cuts.end());
        
        // Step 3: Initialize DP memoization table
        vector<vector<int>> dp(c + 1, vector<int>(c + 1, -1));
        
        // Step 4: Solve for the full range from 1 to c
        return solve(1, c, cuts, dp);
    }
};

7. Tabulation (Bottom-Up) Approach

To convert the memoized solution into an iterative bottom-up tabulation approach, we reverse the state directions:

i runs backward from c down to 1.
j runs forward from i up to c.

C++

class SolutionTabulation {
public:
    int minCost(int n, vector<int>& cuts) {
        int c = cuts.size();
        
        cuts.push_back(n);
        cuts.insert(cuts.begin(), 0);
        sort(cuts.begin(), cuts.end());
        
        // Allocate extra space (c + 2) to prevent out-of-bounds access during index+1 lookups
        vector<vector<int>> dp(c + 2, vector<int>(c + 2, 0));
        
        // Bottom-up computation
        for (int i = c; i >= 1; i--) {
            for (int j = i; j <= c; j++) {
                int mini = INT_MAX;
                
                for (int ind = i; ind <= j; ind++) {
                    int cost = (cuts[j + 1] - cuts[i - 1]) 
                             + dp[i][ind - 1] 
                             + dp[ind + 1][j];
                    mini = min(mini, cost);
                }
                
                dp[i][j] = mini;
            }
        }
        
        return dp[1][c];
    }
};

8. Complexity Analysis Summary

Approach	Time Complexity	Space Complexity	Notes
Pure Recursion	Exponential ( $O(2^m)$ )	$O(m)$	TLE on large constraints due to overlapping sub-problems.
Memoization (Top-Down)	$O(m^3)$	$O(m^2) + O(m)$ stack space	$m^2$ states, with an $O(m)$ loop per state.
Tabulation (Bottom-Up)	$O(m^3)$	$O(m^2)$	Eliminates recursion stack overhead entirely.

(Note: $m$ represents the size of the initial cuts array).

Summary and Key Takeaways

Problem Classification: This is a classic example of Partition DP, where a problem is broken down into smaller sub-ranges, and we iterate through every possible partition point to find the optimum.
Augmenting Boundaries: Always insert 0 at the start and stick length n at the end of the cuts array to accurately compute segment lengths using cuts[j + 1] - cuts[i - 1].
Sorting is Non-Negotiable: Sorting the cuts array guarantees that left and right sub-problems remain strictly independent of one another.
State Transition: The cost formula combines the length of the current stick segment plus the minimum costs of solving the left sub-problem and right sub-problem.
Optimization Path: Moving from brute-force recursion to memoization, and finally to bottom-up tabulation, avoids redundant calculations and stack overflows.