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Minimum Time to Remove All Cars Containing Illegal Goods

Number: 2286

Difficulty: Hard

Paid? No

Companies: Google

Problem Description

Given a binary string s where s[i] == '1' indicates that the iᵗʰ train car contains illegal goods (and '0' otherwise), we can perform three types of operations any number of times:

  1. Remove the leftmost car (cost 1).
  2. Remove the rightmost car (cost 1).
  3. Remove any car (cost 2).

The goal is to remove all cars containing illegal goods at minimum total time.

Key Insights

  • We are allowed to remove from the ends at cost 1 and from anywhere at cost 2.
  • Removing from the ends “sweeps” a whole block of cars even if some of them do not contain illegal goods.
  • For each illegal car, we face a choice: either “expose” it to an end removal by removing enough cars from one end or remove it directly (at cost 2).
  • We can think of a strategy that removes some of the illegal cars via left removals, some via right removals, and the remaining ones (in the middle) directly.
  • Let pos be the sorted list of indices where s[i]=='1'. If we remove k illegal cars from the left (by removing the first pos[k–1]+1 cars) and r illegal cars from the right (by removing the last n – pos[m–r] cars), then the remaining m – (k + r) illegal cars must be removed one‐by‐one at cost 2 each.
  • The task reduces to choosing integers k and r (with k, r ≥ 0 and k + r ≤ m, where m is the total number of illegal cars) to minimize:   cost = (if k > 0 then pos[k–1] + 1 else 0) + (if r > 0 then n – pos[m–r] else 0) + 2 * (m – (k + r))
  • Because the “boundary” removal costs (from left and right) are monotonic (increasing for left and decreasing for right), we can combine them efficiently. We define two auxiliary arrays:   • prefix[k] = cost to remove k cars from the left = (if k > 0 then pos[k–1] + 1 else 0)   • suffix[r] = cost to remove r cars from the right = (if r > 0 then n – pos[m–r] else 0)
  • Then for a fixed total t = k + r (with t from 0 to m), the total cost is:   total_cost = (2m – 2t) + min₀≤k≤t { prefix[k] + suffix[t – k] }
  • Because prefix is increasing and suffix is decreasing the function is unimodal in k for fixed t. We can “slide” the best k along as t increases.
  • Finally, the answer is the minimum total_cost over all possible t.

Space and Time Complexity

Time Complexity: O(n + m) where n = length of s and m = number of illegal cars. (Finding positions takes O(n) and iterating over t from 0 to m is O(m).) Space Complexity: O(m) for storing the positions and the prefix/suffix arrays.

Solution

We first extract the indices of all illegal cars from the string s (i.e. where s[i]=='1'). Let m be the number of such indices. If m == 0, return 0. We precompute two arrays:   • prefix[k]: the cost to remove the first k cars (if k > 0, it equals pos[k–1]+1; otherwise 0).   • suffix[r]: the cost to remove the last r cars (if r > 0, it equals n – pos[m–r]; otherwise 0). Next, for each t from 0 to m (where t = k + r) we want to choose an optimal split between k (illegal cars removed by left removals) and r = t – k (illegal cars removed by right removals). The remaining m – t illegal cars will be directly removed at a cost of 2 units each. Because prefix is increasing and suffix is decreasing, for each t the function f(k) = prefix[k] + suffix[t–k] is convex in k. We can maintain a pointer (best_k) that moves forward as t increases to obtain the minimal value of f(k). The answer is the minimum over t of:   total_cost = (2m – 2t) + (prefix[best_k] + suffix[t – best_k]) This formulation considers every possible combination:

  • t = 0 corresponds to removing all illegal cars individually (at cost 2 each).
  • t = m corresponds to removing all illegal cars via boundary removals.
  • Values in between mix these approaches.

Code Solutions

Below are code solutions in Python, JavaScript, C++, and Java with line‐by‐line comments.

# Python solution
def minimumTime(s: str) -> int:
    n = len(s)
    # Collect indices of cars with illegal goods ('1')
    pos = [i for i, ch in enumerate(s) if ch == '1']
    m = len(pos)
    if m == 0:
        return 0  # no illegal car

    # Precompute prefix and suffix cost arrays
    prefix = [0] * (m + 1)  # prefix[0] = 0, prefix[k] = cost to remove left k cars
    suffix = [0] * (m + 1)  # suffix[0] = 0, suffix[r] = cost to remove right r cars
    for k in range(1, m + 1):
        prefix[k] = pos[k - 1] + 1
    for r in range(1, m + 1):
        suffix[r] = n - pos[m - r]
    
    ans = float('inf')
    best_k = 0  # optimal k for current total t = k + r
    # t represents total illegal cars removed by boundary operations
    for t in range(0, m + 1):
        # For current t, valid k are in the range [0, t]
        # Adjust best_k pointer to maintain minimal prefix[k] + suffix[t - k]
        while best_k < t:
            # Compare value at best_k+1 and best_k; note that suffix index is t - (k)
            current = prefix[best_k] + suffix[t - best_k]
            candidate = prefix[best_k + 1] + suffix[t - best_k - 1]
            if candidate <= current:
                best_k += 1
            else:
                break
        total_cost = (2 * m - 2 * t) + (prefix[best_k] + suffix[t - best_k])
        ans = min(ans, total_cost)
    return ans

# Example usage:
s = "1100101"
print(minimumTime(s))  # Output should be 5
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