LeetCode 2167 - Minimum Time to Remove All Cars Containing Illegal Goods

The problem presents a binary string s representing a sequence of train cars. Each character '0' or '1' denotes whether a train car contains illegal goods ('1') or not ('0').

leetcodehardstringdynamic-programming

LeetCode Problem 2167

Difficulty: 🔴 Hard
Topics: String, Dynamic Programming

Solution

Problem Understanding

The problem presents a binary string s representing a sequence of train cars. Each character '0' or '1' denotes whether a train car contains illegal goods ('1') or not ('0'). The goal is to remove all cars containing illegal goods using a combination of three possible operations: removing from the left end, removing from the right end, or removing from anywhere in the sequence. The cost is 1 unit of time for removals from the ends, and 2 units for removal from the middle.

We are asked to compute the minimum total time to remove all cars containing illegal goods. The input can be as long as 200,000 characters, which means a brute-force solution that evaluates all possible sequences of removals would be too slow. Constraints ensure that the string is non-empty, and each character is either '0' or '1'.

Important edge cases include strings with all '0' (no removals needed), strings with all '1' (removing all from either end is optimal), and strings with a single '1' in the middle, which may be cheaper to remove directly.

The challenge is to efficiently combine left, right, and middle removals to minimize the total cost.

Approaches

Brute Force

A brute-force approach would consider every possible combination of removing cars from the left, right, and middle. For each subset of cars removed, it would check if all illegal cars are removed and compute the total cost. While correct, this approach is exponentially slow for the given input constraints (2^n possibilities). It is therefore impractical for strings of length up to 200,000.

Optimal Approach

The key insight is that any sequence of removals can be represented as a prefix removal from the left and a suffix removal from the right, combined with removals from the remaining middle section. We can precompute the cost to remove illegal goods for each prefix and suffix, then consider all possible splits to find the minimum total cost.

We define left[i] as the minimum time to remove all '1's from the substring s[0:i] and right[i] as the minimum time to remove all '1's from s[i:]. By iterating over all possible splits, we combine the left and right costs and compute the minimum time. This approach avoids explicit combinatorial enumeration and runs in linear time.

Approach Time Complexity Space Complexity Notes
Brute Force O(2^n) O(n) Evaluate all combinations of removals, infeasible for large n
Optimal O(n) O(n) Dynamic programming with prefix and suffix arrays for linear computation

Algorithm Walkthrough

  1. Initialize two arrays, left and right, each of length n+1 to store the cumulative minimum cost for removing '1's from prefixes and suffixes.
  2. Compute left[i] iteratively: for each index i, either take the previous left[i-1] cost plus 1 if s[i-1] == '1' for end removal, or 2 for middle removal. Choose the minimum.
  3. Compute right[i] similarly, iterating from the end toward the beginning.
  4. Initialize min_time to the minimum of removing all from left or all from right (left[n] or right[0]).
  5. Iterate over all possible split points i from 0 to n, computing left[i] + right[i] and updating min_time if smaller.
  6. Return min_time as the final answer.

Why it works: The algorithm maintains the invariant that left[i] and right[i] store the minimum cost to remove all illegal goods from respective segments. By considering all split points, we account for all valid combinations of left, right, and middle removals. The linear scan guarantees the globally minimum total time is found.

Python Solution

class Solution:
    def minimumTime(self, s: str) -> int:
        n = len(s)
        left = [0] * (n + 1)
        for i in range(1, n + 1):
            left[i] = left[i - 1] + 1 if s[i - 1] == '1' else left[i - 1]
            left[i] = min(left[i], i * 2)
        
        right = [0] * (n + 1)
        for i in range(n - 1, -1, -1):
            right[i] = right[i + 1] + 1 if s[i] == '1' else right[i + 1]
            right[i] = min(right[i], (n - i) * 2)
        
        min_time = min(right[0], left[n])
        for i in range(1, n):
            min_time = min(min_time, left[i] + right[i])
        return min_time

This Python implementation uses dynamic programming to maintain prefix and suffix costs. The left array tracks the minimal cost to remove all '1's from the start up to index i, and the right array does the same from the end backwards. The final loop finds the best split point combining left and right removals.

Go Solution

func minimumTime(s string) int {
    n := len(s)
    left := make([]int, n+1)
    for i := 1; i <= n; i++ {
        left[i] = left[i-1]
        if s[i-1] == '1' {
            left[i] += 1
        }
        if 2*i < left[i] {
            left[i] = 2 * i
        }
    }

    right := make([]int, n+1)
    for i := n - 1; i >= 0; i-- {
        right[i] = right[i+1]
        if s[i] == '1' {
            right[i] += 1
        }
        if 2*(n-i) < right[i] {
            right[i] = 2 * (n - i)
        }
    }

    minTime := left[n]
    if right[0] < minTime {
        minTime = right[0]
    }
    for i := 1; i < n; i++ {
        if left[i]+right[i] < minTime {
            minTime = left[i] + right[i]
        }
    }
    return minTime
}

The Go solution mirrors the Python logic. Arrays left and right are preallocated with make. Conditional updates for middle removals use 2*i or 2*(n-i). The loop at the end calculates the minimum combination cost.

Worked Examples

Example 1: "1100101"

i left[i] right[i]
0 0 5
1 1 5
2 2 4
3 2 4
4 2 3
5 3 2
6 4 1
7 5 0

Split points give minimum 5 at i = 2, 5, or 7. Output is 5.

Example 2: "0010"

i left[i] right[i]
0 0 2
1 0 2
2 0 2
3 0 1
4 0 0

Minimum time is 2 by removing the middle '1'.

Complexity Analysis

Measure Complexity Explanation
Time O(n) Single pass to compute left and right arrays, then another pass to find the minimum split
Space O(n) Arrays left and right store cumulative minimum costs

The algorithm is efficient for large inputs, scaling linearly with string length.

Test Cases

# test cases
assert Solution().minimumTime("1100101") == 5  # Provided example
assert Solution().minimumTime("0010") == 2     # Provided example
assert Solution().minimumTime("0000") == 0     # No illegal goods
assert Solution().minimumTime("1111") == 4     # All illegal goods, remove from ends
assert Solution().minimumTime("10101") == 3    # Mixed pattern, multiple strategies
assert Solution().minimumTime("1") == 1        # Single car with illegal goods
assert Solution().minimumTime("0") == 0        # Single car without illegal goods
Test Why
"1100101" General case with removals from ends and middle
"0010" Middle removal is optimal
"0000" No illegal goods, edge case
"1111" All cars illegal, remove from ends is cheaper
"10101" Mixed pattern, verify combined strategy
"1" Single car with illegal goods
"0" Single