Problem Description

In this problem, we are dealing with a 0-indexed binary string s, where each character represents a train car. A car can either contain illegal goods (denoted by '1') or not (denoted by '0'). The goal is to remove all cars with illegal goods in the least amount of time possible. We have three operations that we can use:

1. Removing a car from the left end of the train takes 1 unit of time.
2. Removing a car from the right end of the train also takes 1 unit of time.
3. Removing a car from any other position takes 2 units of time.

A car removal operation is directed towards a car containing illegal goods, and we want to know the minimum time required to remove all such cars.

Intuition

To solve this problem, we use a dynamic programming approach. The intuition lies in breaking down the problem into simpler sub-problems:

1. The minimum time to remove all 'bad' cars from the beginning (left side) of the train up to any given position.
2. The minimum time to remove all 'bad' cars from the end (right side) of the train from any given position to the end.

The goal is to find out the optimal point (or points) where we switch from removing cars from the left to removing cars from the right, or vice versa, to minimize the total time. We solve for two arrays, pre and suf, where pre[i] is the minimum time to remove all cars with illegal goods from the first i cars, and suf[i] is the minimum time to remove all cars with illegal goods from position i to the end.

For pre:

• If a car at position i does not contain illegal goods ('0'), then pre[i + 1] = pre[i] since no action is needed.
• If it does contain illegal goods ('1'), we check if it's cheaper to remove it individually, taking 2 units of time, or remove all the i+1 cars from the left which take i+1 units of time.

For suf:

• If a car at position i does not contain illegal goods ('0'), then suf[i] = suf[i + 1] as no action is needed here too.
• If it does contain illegal goods ('1'), we check if it's cheaper to remove it individually, taking 2 units of time, or remove all cars from position i to the end which will take n-i units of time (where n is the total number of cars).

Finally, we iterate through the arrays pre and suf to find the minimum combined time taking into account both the left and right removals at each position. The minimum value from this combination gives us the least amount of time needed to remove all the cars containing illegal goods.

Learn more about Dynamic Programming patterns.

Solution Approach

The solution implementation uses a dynamic programming approach with two additional arrays pre and suf to keep track of the optimal removal times from the left and from the right, respectively.

Here's the detailed implementation of the solution step by step:

1. Initialize the pre and suf arrays with zeros, and set their sizes to one more than the length of the string s. This is done to handle the prefix up to i and suffix from i inclusively, easily without running into array index issues.

2. Iterate through the string s from left to right. If you encounter a '0' (no illegal goods), simply copy the value from the previous index of pre. This means no additional time is needed to handle this car. If there is a '1' (illegal goods exist), calculate the minimum time between the time stored in pre at the previous index plus 2 (removing this particular car) and i + 1 (removing all cars from the left up to the current position) and store that minimum value in pre[i + 1].

   for i, c in enumerate(s):
       pre[i + 1] = pre[i] if c == '0' else min(pre[i] + 2, i + 1)

3. Similarly, iterate through the string s from right to left to fill in the suf array. You do the mirror operation of what was done with pre: if a '0' is encountered, carry over the value from the right index of suf. If a '1' is found, calculate the minimum time between the time stored in suf at the next index plus 2 (removing this particular car) and n - i (removing all cars from the current position to the end).

   for i in range(n - 1, -1, -1):
       suf[i] = suf[i + 1] if s[i] == '0' else min(suf[i + 1] + 2, n - i)

4. Once you have the pre and suf arrays constructed, find the point where the sum of removal times from the left and from the right is minimal. Iterate through the combined view of pre (excluding the zeroth position) and suf (excluding the last position) and find the minimum sum of corresponding pairs.

   return min(a + b for a, b in zip(pre[1:], suf[1:]))

In this approach, the dynamic programming pattern helps avoid redundant calculations by building up the solution using previously computed values, and the two-pointer technique is not explicitly used but is conceptually present as the solution involves evaluating the sum of pairs of prefix and suffix solutions.

Example Walkthrough

Let's illustrate the solution approach with a small example. Assume we have a binary string s = "001011" where 1 indicates the presence of illegal goods in a train car.

1. Initialize pre and suf Arrays: First, we initialize pre and suf arrays to store the minimum removal times from the left and from the right, respectively.

   For s having a length of 6, pre and suf will be of length 7.

2. Fill the pre Array: Next, we iterate from left to right:

   • For s[0] = '0': pre[1] = pre[0] = 0 (no illegal goods, so no removal time needed).
   • For s[1] = '0': pre[2] = pre[1] = 0.
   • For s[2] = '1': pre[3] = min(pre[2] + 2, 3) = min(0 + 2, 3) = 2 (cheaper to remove this car only).
   • For s[3] = '0': pre[4] = pre[3] = 2.
   • For s[4] = '1': pre[5] = min(pre[4] + 2, 5) = min(2 + 2, 5) = 4 (cheaper to remove both bad cars separately).
   • For s[5] = '1': pre[6] = min(pre[5] + 2, 6) = min(4 + 2, 6) = 6 (cheaper to remove all cars from the left).

   Now, pre = [0, 0, 0, 2, 2, 4, 6].

3. Fill the suf Array: We mirror the above process, but from right to left:

   • For s[5] = '1': suf[5] = min(suf[6] + 2, 1) = min(0 + 2, 1) = 1 (cheaper to remove cars from the right).
   • For s[4] = '1': suf[4] = min(suf[5] + 2, 2) = min(1 + 2, 2) = 2.
   • For s[3] = '0': suf[3] = suf[4] = 2.
   • For s[2] = '1': suf[2] = min(suf[3] + 2, 4) = min(2 + 2, 4) = 4 (either way is the same, so minimal is 4).
   • For s[1] = '0': suf[1] = suf[2] = 4.
   • For s[0] = '0': suf[0] = suf[1] = 4.

   Now, suf = [4, 4, 4, 2, 2, 1, 0].

4. Find Minimal Combined Time: Lastly, we iterate through the combined view of pre (excluding the zeroth index) and suf (excluding the last index) and find the minimum sum.

   • For i = 1: sum is pre[1] + suf[1] = 0 + 4 = 4.
   • For i = 2: sum is pre[2] + suf[2] = 0 + 4 = 4.
   • For i = 3: sum is pre[3] + suf[3] = 2 + 2 = 4.
   • For i = 4: sum is pre[4] + suf[4] = 2 + 2 = 4.
   • For i = 5: sum is pre[5] + suf[5] = 4 + 1 = 5.
   • For i = 6: sum is pre[6] + suf[6] = 6 + 0 = 6.

The minimum time from these sums is 4 units of time, so that is the least amount of time required to remove all the cars containing illegal goods. This can be interpreted as removing the first bad car individually and the other two bad cars from the right end of the train.

Solution Implementation

Time and Space Complexity

The time complexity of the provided code is O(n), where n is the length of the input string s. This is because the code iterates over the string twice: once to fill the pre array with the minimum operations required to remove '1's from the start of the string and once to fill the suf array with the minimum operations required to remove '1's from the end of the string. Each element is computed in constant time, hence the first loop takes O(n) time and the second also takes O(n) time.

The space complexity of the code is also O(n). Two extra arrays, pre and suf, each of size n + 1, are used to store the intermediate results for the prefix and suffix operations, respectively. The space required by these auxiliary arrays is proportional to the length of the string.

Learn more about how to find time and space complexity quickly using problem constraints.