LeetCode #2167 — HARD

Minimum Time to Remove All Cars Containing Illegal Goods

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

The Problem

Problem Statement

You are given a 0-indexed binary string s which represents a sequence of train cars. s[i] = '0' denotes that the i^th car does not contain illegal goods and s[i] = '1' denotes that the i^th car does contain illegal goods.

As the train conductor, you would like to get rid of all the cars containing illegal goods. You can do any of the following three operations any number of times:

Remove a train car from the left end (i.e., remove s[0]) which takes 1 unit of time.
Remove a train car from the right end (i.e., remove s[s.length - 1]) which takes 1 unit of time.
Remove a train car from anywhere in the sequence which takes 2 units of time.

Return the minimum time to remove all the cars containing illegal goods.

Note that an empty sequence of cars is considered to have no cars containing illegal goods.

Example 1:

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.

Example 2:

Input: s = "0010"
Output: 2
Explanation:
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'.

Step 01

Brute Force Baseline

Problem summary: 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. As the train conductor, you would like to get rid of all the cars containing illegal goods. You can do any of the following three operations any number of times: Remove a train car from the left end (i.e., remove s[0]) which takes 1 unit of time. Remove a train car from the right end (i.e., remove s[s.length - 1]) which takes 1 unit of time. Remove a train car from anywhere in the sequence which takes 2 units of time. Return the minimum time to remove all the cars containing illegal goods. Note that an empty sequence of cars is considered to have no cars containing illegal goods.

Baseline thinking

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

Pattern signal: Dynamic Programming

Example 1

"1100101"

Example 2

"0010"

Core Insight

What unlocks the optimal approach

Build an array withoutFirst where withoutFirst[i] stores the minimum time to remove all the cars containing illegal goods from the ‘suffix’ of the sequence starting from the ith car without using any type 1 operations.
Next, build an array onlyFirst where onlyFirst[i] stores the minimum time to remove all the cars containing illegal goods from the ‘prefix’ of the sequence ending on the ith car using only type 1 operations.
Finally, we can compare the best way to split the operations amongst these two types by finding the minimum time across all onlyFirst[i] + withoutFirst[i + 1].

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 #2167: Minimum Time to Remove All Cars Containing Illegal Goods
class Solution {
    public int minimumTime(String s) {
        int n = s.length();
        int[] pre = new int[n + 1];
        int[] suf = new int[n + 1];
        for (int i = 0; i < n; ++i) {
            pre[i + 1] = s.charAt(i) == '0' ? pre[i] : Math.min(pre[i] + 2, i + 1);
        }
        for (int i = n - 1; i >= 0; --i) {
            suf[i] = s.charAt(i) == '0' ? suf[i + 1] : Math.min(suf[i + 1] + 2, n - i);
        }
        int ans = Integer.MAX_VALUE;
        for (int i = 1; i <= n; ++i) {
            ans = Math.min(ans, pre[i] + suf[i]);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2167: Minimum Time to Remove All Cars Containing Illegal Goods
func minimumTime(s string) int {
	n := len(s)
	pre := make([]int, n+1)
	suf := make([]int, n+1)
	for i, c := range s {
		pre[i+1] = pre[i]
		if c == '1' {
			pre[i+1] = min(pre[i]+2, i+1)
		}
	}
	for i := n - 1; i >= 0; i-- {
		suf[i] = suf[i+1]
		if s[i] == '1' {
			suf[i] = min(suf[i+1]+2, n-i)
		}
	}
	ans := 0x3f3f3f3f
	for i := 1; i <= n; i++ {
		ans = min(ans, pre[i]+suf[i])
	}
	return ans
}

# Accepted solution for LeetCode #2167: Minimum Time to Remove All Cars Containing Illegal Goods
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:]))

// Accepted solution for LeetCode #2167: Minimum Time to Remove All Cars Containing Illegal Goods
/**
 * [2167] Minimum Time to Remove All Cars Containing Illegal Goods
 *
 * You are given a 0-indexed binary string s which represents a sequence of train cars. s[i] = '0' denotes that the i^th car does not contain illegal goods and s[i] = '1' denotes that the i^th car does contain illegal goods.
 * As the train conductor, you would like to get rid of all the cars containing illegal goods. You can do any of the following three operations any number of times:
 * <ol>
 * 	Remove a train car from the left end (i.e., remove s[0]) which takes 1 unit of time.
 * 	Remove a train car from the right end (i.e., remove s[s.length - 1]) which takes 1 unit of time.
 * 	Remove a train car from anywhere in the sequence which takes 2 units of time.
 * </ol>
 * Return the minimum time to remove all the cars containing illegal goods.
 * Note that an empty sequence of cars is considered to have no cars containing illegal goods.
 *  
 * Example 1:
 *
 * Input: s = "<u>11</u>00<u>1</u>0<u>1</u>"
 * 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.
 *
 * Example 2:
 *
 * Input: s = "00<u>1</u>0"
 * Output: 2
 * Explanation:
 * 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 * 10^5
 * 	s[i] is either '0' or '1'.
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/minimum-time-to-remove-all-cars-containing-illegal-goods/
// discuss: https://leetcode.com/problems/minimum-time-to-remove-all-cars-containing-illegal-goods/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn minimum_time(s: String) -> i32 {
        0
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    #[ignore]
    fn test_2167_example_1() {
        let s = "1100101".to_string();

        let result = 5;

        assert_eq!(Solution::minimum_time(s), result);
    }

    #[test]
    #[ignore]
    fn test_2167_example_2() {
        let s = "0010".to_string();

        let result = 2;

        assert_eq!(Solution::minimum_time(s), result);
    }
}

// Accepted solution for LeetCode #2167: Minimum Time to Remove All Cars Containing Illegal Goods
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #2167: Minimum Time to Remove All Cars Containing Illegal Goods
// class Solution {
//     public int minimumTime(String s) {
//         int n = s.length();
//         int[] pre = new int[n + 1];
//         int[] suf = new int[n + 1];
//         for (int i = 0; i < n; ++i) {
//             pre[i + 1] = s.charAt(i) == '0' ? pre[i] : Math.min(pre[i] + 2, i + 1);
//         }
//         for (int i = n - 1; i >= 0; --i) {
//             suf[i] = s.charAt(i) == '0' ? suf[i + 1] : Math.min(suf[i + 1] + 2, n - i);
//         }
//         int ans = Integer.MAX_VALUE;
//         for (int i = 1; i <= n; ++i) {
//             ans = Math.min(ans, pre[i] + suf[i]);
//         }
//         return ans;
//     }
// }

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(n × m)

Space

O(n × m)

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.

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.