#2167

Hard

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Minimum Time to Remove All Cars Containing Illegal Goods

Determine the minimum time to remove all cars with illegal goods using state transition dynamic programming efficiently.

string dynamic programming

Problem Statement

You are given a binary string s representing a sequence of train cars, where s[i] = '1' means the ith car contains illegal goods and s[i] = '0' means it does not. The goal is to remove all cars containing illegal goods in minimum time using available operations.

The allowed operations are removing a car from the left end, removing a car from the right end, or removing a car in the middle that contains illegal goods. Each operation takes 1 unit of time per car removed. Return the minimum total time needed to eliminate all illegal cars.

Examples

Example 1

Input: s = "1100101"

Output: 5

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.

Example 2

Input: s = "0010"

Output: 2

One way to remove all the cars containing illegal goods from the sequence is to

remove a car from the left end 3 times. Time taken is 3 * 1 = 3. This obtains a total time of 3.

Another way to remove all the cars containing illegal goods from the sequence is to

remove the car containing illegal goods found in the middle. Time taken is 2. This obtains a total time of 2.

Another way to remove all the cars containing illegal goods from the sequence is to

remove a car from the right end 2 times. Time taken is 2 * 1 = 2. This obtains a total time of 2.

2 is the minimum time taken to remove all the cars containing illegal goods. There are no other ways to remove them with less time.

Constraints

1 <= s.length <= 2 * 105
s[i] is either '0' or '1'.

Solution Approach

Dynamic Programming with Suffix Array

Construct a DP array withoutFirst where withoutFirst[i] stores the minimum time to remove all illegal goods from the suffix starting at index i. Iterate backwards, updating the state based on removing from left, right, or the current car.

State Transitions

For each position, compute the minimum between removing the current car individually or combining it with the previously computed suffix removal. This captures the optimal decision at each step and prevents redundant calculations.

Final Computation

Once withoutFirst is populated, evaluate the prefix removals in combination with the suffix states to determine the global minimum time. Return the smallest computed total across all split points.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time and space complexity depend on the implementation of the DP array and string iteration, typically O(n) for both if suffix arrays are used.

What Interviewers Usually Probe

Focus on handling each '1' efficiently without redundant removals.
Explain why state transition captures optimal substructure.
Clarify how left, right, and middle operations are combined in DP.

Common Pitfalls or Variants

Common pitfalls

Failing to consider the middle removal cost in DP updates.
Updating DP states incorrectly leading to overcounting time.
Assuming only left or right removals suffice without suffix calculations.

Follow-up variants

Compute minimum time when removal from ends costs different units.
Handle a sequence where illegal goods can appear consecutively in blocks.
Optimize for extremely large strings using rolling DP or prefix sums.

FAQ

What is the best approach for Minimum Time to Remove All Cars Containing Illegal Goods?

Use state transition dynamic programming with a suffix array to compute minimal removal times efficiently.

How do I handle middle car removals in this problem?

Include middle removal as part of the DP state by comparing its cost to left and right removals at each step.

Can I solve this problem with just greedy removal from ends?

Greedy removal alone can fail because optimal sequences may require removing middle cars; DP ensures minimal total time.

What is the time complexity of this DP approach?

It is O(n) since each suffix position is evaluated once with constant-time state updates.

Why use a withoutFirst array in the DP solution?

WithoutFirst stores the minimal time for suffixes, allowing efficient calculation of global minimal time combining left, right, and middle removals.

terminal

Solution

Solution 1

#### Python3

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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:]))

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