Leetcode #2167: Minimum Time to Remove All Cars Containing Illegal Goods
In this guide, we solve Leetcode #2167 Minimum Time to Remove All Cars Containing Illegal Goods 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
You are given a 0-indexed binary string s which represents a sequence of train cars. s[i] = '0' denotes that the ith car does not contain illegal goods and s[i] = '1' denotes that the ith car does contain illegal goods.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: String, Dynamic Programming
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: s = "1100101"
Output: 5
Explanation:
One way to remove all the cars containing illegal goods from the sequence is to
- remove a car from the left end 2 times. Time taken is 2 * 1 = 2.
- remove a car from the right end. Time taken is 1.
- remove the car containing illegal goods found in the middle. Time taken is 2.
This obtains a total time of 2 + 1 + 2 = 5.
An alternative way is to
- remove a car from the left end 2 times. Time taken is 2 * 1 = 2.
- remove a car from the right end 3 times. Time taken is 3 * 1 = 3.
This also obtains a total time of 2 + 3 = 5.
5 is the minimum time taken to remove all the cars containing illegal goods.
There are no other ways to remove them with less time.
Python Solution
class Solution:
def minimumTime(self, s: str) -> int:
n = len(s)
pre = [0] * (n + 1)
suf = [0] * (n + 1)
for i, c in enumerate(s):
pre[i + 1] = pre[i] if c == '0' else min(pre[i] + 2, i + 1)
for i in range(n - 1, -1, -1):
suf[i] = suf[i + 1] if s[i] == '0' else min(suf[i + 1] + 2, n - i)
return min(a + b for a, b in zip(pre[1:], suf[1:]))
Complexity
The time complexity is O(n·m) (typical). The space complexity is O(n·m) or optimized.
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.