Minimum Time to Remove All Cars Containing Illegal Goods

Minimum Time to Remove All Cars Containing Illegal Goods - Problem

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: An empty sequence of cars is considered to have no cars containing illegal goods.

Input & Output

Example 1 — Basic Mixed Case

$ Input: s = "1100101"

› Output: 5

💡 Note: Remove the first 2 cars from left (2 units), remove the middle '1' at index 4 individually (2 units), and remove the last car from right (1 unit). Total: 2 + 2 + 1 = 5.

Example 2 — All Illegal

$ Input: s = "11111"

› Output: 5

💡 Note: Remove all 5 cars from either left or right end, each taking 1 unit of time. Total: 5 units.

Example 3 — No Illegal Cars

$ Input: s = "0000"

› Output: 0

💡 Note: No cars contain illegal goods, so no removal is needed. Total: 0 units.

Constraints

1 ≤ s.length ≤ 10⁵
s[i] is either '0' or '1'

Visualization

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Understanding the Visualization

Input

Binary string representing train cars (1=illegal, 0=legal)

Operations

Remove from left (1 unit), right (1 unit), or middle (2 units)

Goal

Find minimum time to remove all illegal cars

Key Takeaway

🎯 Key Insight: It's often cheaper to remove entire segments from the ends rather than individual cars from the middle

Asked in

G Google 12 f Meta 8 a Amazon 6 M Microsoft 4

The key insight is that removing cars from the ends costs 1 unit each, while removing from middle costs 2 units. We can use dynamic programming to find the optimal combination of left-end removal, individual middle removal, and right-end removal. Time: O(n), Space: O(n)

Common Approaches

Approach	Time	Space	Notes
✓ Try All Partitions	O(n³)	O(1)	Try every possible way to divide the string into left removal, middle removal, and right removal segments
Dynamic Programming	O(n)	O(n)	Use DP to track minimum cost to remove all '1's up to each position
Prefix Sum Optimization	O(n²)	O(n)	Use prefix sums to quickly calculate the cost of removing middle segments

Try All Partitions — Algorithm Steps

For each left partition point i
For each right partition point j
Calculate total cost: i + 2*(count of '1's between i and j) + (n-j)
Track minimum cost

Visualization

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Step-by-Step Walkthrough

Try Partition

Test removing first i cars from left, last j cars from right

Count Middle

Count '1' cars in middle segment (cost 2 each)

Calculate Total

Total cost = left + 2*middle_ones + right

Code -

solution.c — C

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

int solution(char* s) {
    int n = strlen(s);
    int minTime = INT_MAX;
    
    // Try all possible partitions
    for (int left = 0; left <= n; left++) {
        for (int right = 0; right <= n; right++) {
            if (left + right > n) continue;
            
            // Calculate cost for middle segment
            int middleCost = 0;
            for (int i = left; i < n - right; i++) {
                if (s[i] == '1') {
                    middleCost += 2;
                }
            }
            
            int totalCost = left + middleCost + right;
            if (totalCost < minTime) {
                minTime = totalCost;
            }
        }
    }
    
    return minTime;
}

int main() {
    char s[100001];
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;
    int result = solution(s);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n³)

Three nested loops to try all partition combinations

⚠ Quadratic Growth

Space Complexity

O(1)

Only using constant extra variables

✓ Linear Space

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Try All Partitions — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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