Minimum Time to Remove All Cars Containing Illegal Goods

Minimum Time to Remove All Cars Containing Illegal Goods

What is this problem about?

The Minimum Time to Remove All Cars Containing Illegal Goods problem gives you a binary string where '1' represents a car with illegal goods. You can remove cars from either end (cost 1 each) or from the middle (cost 2 each). Find the minimum total cost to remove all cars containing illegal goods. This hard Minimum Time to Remove All Cars Containing Illegal Goods coding problem uses prefix/suffix DP to compute optimal removal costs from both ends.

Why is this asked in interviews?

Google asks this because it requires recognizing that the optimal strategy always removes a contiguous prefix from the left, a contiguous suffix from the right, and uses middle removal for the remaining illegal cars in between. DP from both ends captures the optimal costs efficiently. The string and dynamic programming interview pattern is central, and the problem rewards candidates who can model multi-directional removals cleanly.

Algorithmic pattern used

Prefix DP + Suffix DP.

• left[i] = minimum cost to remove all '1's in the prefix s[0..i] (can remove from left end or in middle).
• right[i] = minimum cost to remove all '1's in the suffix s[i..n-1]. Transition for left[i]: if s[i]='0', left[i] = left[i-1]; if s[i]='1', left[i] = min(left[i-1] + 2, i+1) (middle removal vs removing entire prefix). Final answer: min over all split points i of (left[i] + right[i+1]).

Example explanation

String: "1100101".

• left[]: [1, 2, 2, 2, 3, 3, 4] (computing optimal prefix removal costs).
• right[]: [4, 3, 3, 2, 2, 1, 1] (computing optimal suffix removal costs, scanning from right).
• For each split point, sum left[i] + right[i+1]. Minimum = 3 (e.g., remove first 2 from left, remove last from right, and middle '1' is handled).

Common mistakes candidates make

• Trying to simulate removal operations directly instead of precomputing DP arrays.
• Using only left DP without the right DP (missing suffix removals).
• Incorrect transition: using cost 2 for prefix removal (should be i+1 — the cost of removing the whole prefix up to i).
• Not iterating over all split points after building both DP arrays.

Interview preparation tip

Two-sided prefix/suffix DP problems have a consistent structure: precompute the optimal cost from the left, precompute from the right, then merge by trying all split points. This technique appears in multiple hard interview problems involving left/right segment optimization. Build strong intuition for the left[i] and right[i] transition rules before tackling combined DP. Practice problems like "minimum cost to remove/paint/change elements from both ends" to master this pattern.