Leetcode #1547: Minimum Cost to Cut a Stick
In this guide, we solve Leetcode #1547 Minimum Cost to Cut a Stick in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Given a wooden stick of length n units. The stick is labelled from 0 to n.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Array, Dynamic Programming, Sorting
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
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).
Python Solution
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]
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.