LeetCode #1547 — HARD

Minimum Cost to Cut a Stick

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

Given a wooden stick of length n units. The stick is labelled from 0 to n. For example, a stick of length 6 is labelled as follows:

Given an integer array cuts where cuts[i] denotes a position you should perform a cut at.

You should perform the cuts in order, you can change the order of the cuts as you wish.

The cost of one cut is the length of the stick to be cut, the total cost is the sum of costs of all cuts. When you cut a stick, it will be split into two smaller sticks (i.e. the sum of their lengths is the length of the stick before the cut). Please refer to the first example for a better explanation.

Return the minimum total cost of the cuts.

Example 1:

Input: n = 7, cuts = [1,3,4,5]
Output: 16
Explanation: Using cuts order = [1, 3, 4, 5] as in the input leads to the following scenario:

The first cut is done to a rod of length 7 so the cost is 7. The second cut is done to a rod of length 6 (i.e. the second part of the first cut), the third is done to a rod of length 4 and the last cut is to a rod of length 3. The total cost is 7 + 6 + 4 + 3 = 20.
Rearranging the cuts to be [3, 5, 1, 4] for example will lead to a scenario with total cost = 16 (as shown in the example photo 7 + 4 + 3 + 2 = 16).

Example 2:

Input: n = 9, cuts = [5,6,1,4,2]
Output: 22
Explanation: If you try the given cuts ordering the cost will be 25.
There are much ordering with total cost <= 25, for example, the order [4, 6, 5, 2, 1] has total cost = 22 which is the minimum possible.

Constraints:

2 <= n <= 10⁶
1 <= cuts.length <= min(n - 1, 100)
1 <= cuts[i] <= n - 1
All the integers in cuts array are distinct.

Step 01

Brute Force Baseline

Problem summary: Given a wooden stick of length n units. The stick is labelled from 0 to n. For example, a stick of length 6 is labelled as follows: Given an integer array cuts where cuts[i] denotes a position you should perform a cut at. You should perform the cuts in order, you can change the order of the cuts as you wish. The cost of one cut is the length of the stick to be cut, the total cost is the sum of costs of all cuts. When you cut a stick, it will be split into two smaller sticks (i.e. the sum of their lengths is the length of the stick before the cut). Please refer to the first example for a better explanation. Return the minimum total cost of the cuts.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming

Example 1

7
[1,3,4,5]

Example 2

9
[5,6,1,4,2]

Core Insight

What unlocks the optimal approach

Build a dp array where dp[i][j] is the minimum cost to achieve all the cuts between i and j.
When you try to get the minimum cost between i and j, try all possible cuts k between them, dp[i][j] = min(dp[i][k] + dp[k][j]) + (j - i) for all possible cuts k between them.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1547: Minimum Cost to Cut a Stick
class Solution {
    public int minCost(int n, int[] cuts) {
        List<Integer> nums = new ArrayList<>();
        for (int x : cuts) {
            nums.add(x);
        }
        nums.add(0);
        nums.add(n);
        Collections.sort(nums);
        int m = nums.size();
        int[][] f = new int[m][m];
        for (int l = 2; l < m; ++l) {
            for (int i = 0; i + l < m; ++i) {
                int j = i + l;
                f[i][j] = 1 << 30;
                for (int k = i + 1; k < j; ++k) {
                    f[i][j] = Math.min(f[i][j], f[i][k] + f[k][j] + nums.get(j) - nums.get(i));
                }
            }
        }
        return f[0][m - 1];
    }
}

// Accepted solution for LeetCode #1547: Minimum Cost to Cut a Stick
func minCost(n int, cuts []int) int {
	cuts = append(cuts, []int{0, n}...)
	sort.Ints(cuts)
	m := len(cuts)
	f := make([][]int, m)
	for i := range f {
		f[i] = make([]int, m)
	}
	for l := 2; l < m; l++ {
		for i := 0; i+l < m; i++ {
			j := i + l
			f[i][j] = 1 << 30
			for k := i + 1; k < j; k++ {
				f[i][j] = min(f[i][j], f[i][k]+f[k][j]+cuts[j]-cuts[i])
			}
		}
	}
	return f[0][m-1]
}

# Accepted solution for LeetCode #1547: Minimum Cost to Cut a Stick
class Solution:
    def minCost(self, n: int, cuts: List[int]) -> int:
        cuts.extend([0, n])
        cuts.sort()
        m = len(cuts)
        f = [[0] * m for _ in range(m)]
        for l in range(2, m):
            for i in range(m - l):
                j = i + l
                f[i][j] = inf
                for k in range(i + 1, j):
                    f[i][j] = min(f[i][j], f[i][k] + f[k][j] + cuts[j] - cuts[i])
        return f[0][-1]

// Accepted solution for LeetCode #1547: Minimum Cost to Cut a Stick
struct Solution;

use std::collections::HashMap;

impl Solution {
    fn min_cost(n: i32, mut cuts: Vec<i32>) -> i32 {
        cuts.sort_unstable();
        let mut memo: HashMap<(i32, i32), i32> = HashMap::new();
        Self::dp(0, n, &mut memo, &cuts)
    }

    fn dp(left: i32, right: i32, memo: &mut HashMap<(i32, i32), i32>, cuts: &[i32]) -> i32 {
        if let Some(&res) = memo.get(&(left, right)) {
            return res;
        }
        let mut cuts_inside = vec![];
        for &x in cuts {
            if x > left && x < right {
                cuts_inside.push(x);
            }
        }
        let res = if !cuts_inside.is_empty() {
            let mut min = std::i32::MAX;
            for x in cuts_inside {
                min = min.min(Self::dp(left, x, memo, cuts) + Self::dp(x, right, memo, cuts));
            }
            min + right - left
        } else {
            0
        };
        memo.insert((left, right), res);
        res
    }
}

#[test]
fn test() {
    let n = 7;
    let cuts = vec![1, 3, 4, 5];
    let res = 16;
    assert_eq!(Solution::min_cost(n, cuts), res);
    let n = 9;
    let cuts = vec![5, 6, 1, 4, 2];
    let res = 22;
    assert_eq!(Solution::min_cost(n, cuts), res);
}

// Accepted solution for LeetCode #1547: Minimum Cost to Cut a Stick
function minCost(n: number, cuts: number[]): number {
    cuts.push(0, n);
    cuts.sort((a, b) => a - b);
    const m = cuts.length;
    const f: number[][] = Array.from({ length: m }, () => Array(m).fill(0));
    for (let l = 2; l < m; l++) {
        for (let i = 0; i < m - l; i++) {
            const j = i + l;
            f[i][j] = Infinity;
            for (let k = i + 1; k < j; k++) {
                f[i][j] = Math.min(f[i][j], f[i][k] + f[k][j] + cuts[j] - cuts[i]);
            }
        }
    }
    return f[0][m - 1];
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(m^3)

Space

O(m^2)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.