Minimum Time to Remove All Cars Containing Illegal Goods

Minimum Time to Remove All Cars Containing Illegal Goods - Problem

You are a train conductor tasked with removing all cars containing illegal goods from your train. The train is represented as a binary string s where:

'0' represents a car with legal goods
'1' represents a car with illegal goods

You have three removal operations available:

Remove from left end: Remove s[0] - costs 1 unit of time
Remove from right end: Remove s[n-1] - costs 1 unit of time
Remove from middle: Remove any car from anywhere - costs 2 units of time

Your goal is to find the minimum total time needed to remove all cars containing illegal goods. Once all illegal cars are removed, the remaining legal cars can stay.

Example: For string "1100101", you might remove cars from the left until you reach the middle, then remove specific middle cars, then remove from the right - whichever combination gives minimum total cost.

Input & Output

example_1.py — Basic Case

$ Input: s = "1100101"

› Output: 5

💡 Note: Remove the first 4 cars from the left (cost=4), then remove the last car from right (cost=1). Total = 4+1 = 5. This removes all illegal cars: positions 0,1,4,6.

example_2.py — All Illegal

$ Input: s = "111"

› Output: 3

💡 Note: All cars contain illegal goods. Best strategy is to remove all 3 cars from one end, costing 3 units. Alternative would be removing each from middle costing 2×3=6 units.

example_3.py — Edge Case

$ Input: s = "0000"

› Output: 0

💡 Note: No cars contain illegal goods, so no removal needed. Cost = 0.

Constraints

1 ≤ s.length ≤ 2 × 10⁵
s[i] is either '0' or '1'
All removal operations can be performed any number of times
Empty sequence after removals is considered valid

Visualization

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Understanding the Visualization

Identify Problem

You have a train with cars marked 0 (legal) and 1 (illegal). Need to remove all illegal cars.

Consider Operations

Three options: remove from left (1 cost), right (1 cost), or middle (2 cost).

DP Insight

At each position, choose optimally between extending left removal or previous solution.

Track Minimum

Consider all possible right boundary extensions and track the overall minimum cost.

Key Takeaway

🎯 Key Insight: Use dynamic programming to track optimal cost at each position. At position i, choose between removing everything from left (cost i+1) or extending the previous optimal solution. Consider all possible right boundary extensions to find the global minimum. This achieves O(n) time complexity with O(1) space.

Asked in

G Google 45 a Amazon 38 f Meta 25 ⊞ Microsoft 22

This problem requires finding the minimum cost to remove all illegal train cars using three operations: remove from left (cost 1), remove from right (cost 1), or remove from middle (cost 2). The optimal approach uses dynamic programming to track the minimum cost at each position, considering whether to extend left removal or build upon the previous optimal solution. Key insight: at each position, choose between removing everything from the left or extending the previous optimal solution, then consider adding right removals. Time complexity: O(n), Space complexity: O(1).

Common Approaches

Approach	Time	Space	Notes
✓ Dynamic Programming (Optimal)	O(n)	O(1)	Use DP to track minimum cost at each position with optimal substructure

Dynamic Programming (Optimal) — Algorithm Steps

Initialize leftCost[i] = cost to remove everything from left up to position i
For each position, calculate minimum cost considering all options
Track the minimum of: remove all from left, previous optimal + remove current from middle
The answer considers all possible right boundary extensions
Use the recurrence: dp[i] = min(i+1, dp[i-1] + 2*s[i])

Visualization

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Step-by-Step Walkthrough

Initialize

Start with leftCost=0, consider each position

At each position i

Choose min of: remove all from left (i+1) or extend previous solution

Consider right removals

For each position, also try removing everything to the right

Track minimum

Keep track of overall minimum across all combinations

Code -

solution.c — C

#include <stdio.h>
#include <string.h>

int min(int a, int b) {
    return a < b ? a : b;
}

int minimumTime(char* s) {
    int n = strlen(s);
    if (n == 0) return 0;
    
    int leftCost = 0;
    int minTime = n;
    
    for (int i = 0; i < n; i++) {
        int leftRemoveAll = i + 1;
        
        if (s[i] == '1') {
            leftCost = min(leftRemoveAll, leftCost + 2);
        } else {
            leftCost = min(leftRemoveAll, leftCost);
        }
        
        int rightRemoveAll = n - 1 - i;
        int totalCost = leftCost + rightRemoveAll;
        minTime = min(minTime, totalCost);
    }
    
    return minTime;
}

int main() {
    char s[100001];
    scanf("%s", s);
    
    int len = strlen(s);
    if (s[0] == '"' && s[len-1] == '"') {
        memmove(s, s+1, len-2);
        s[len-2] = '\0';
    }
    
    printf("%d\n", minimumTime(s));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through the string, constant work per position

✓ Linear Growth

Space Complexity

O(1)

Only storing a few variables, no additional arrays needed

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Dynamic Programming (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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