We use cookies (including Google cookies) to personalize ads and analyze traffic. By continuing to use our site, you accept our Privacy Policy.

SimplyLeet
TagsDisclaimersPrivacyTermsContactAbout Us

Minimum Time to Remove All Cars Containing Illegal Goods

Difficulty: Hard

Problem Description

You are given a 0-indexed binary string s which represents a sequence of train cars. s[i] = '0' denotes that the i-th car does not contain illegal goods and s[i] = '1' denotes that the i-th 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:

  1. Remove a train car from the left end (i.e., remove s[0]) which takes 1 unit of time.
  2. Remove a train car from the right end (i.e., remove s[s.length - 1]) which takes 1 unit of time.
  3. 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 that an empty sequence of cars is considered to have no cars containing illegal goods.

Key Insights

  • Cars containing illegal goods are represented by '1's in the string.
  • Removing cars from the ends (left or right) is cheaper than removing them from the middle.
  • The minimum time can be achieved by removing cars from the left and right ends until all illegal goods are removed.
  • The solution requires determining the leftmost and rightmost positions of '1's in the string.

Space and Time Complexity

Time Complexity: O(n)
Space Complexity: O(1)

Solution

To solve the problem, the algorithm follows these steps:

  1. Traverse the string to find the first and last occurrence of '1'.
  2. Calculate the number of operations needed to remove all '1's by considering three options:
    • Remove all '1's by removing cars from the left end.
    • Remove all '1's by removing cars from the right end.
    • Remove '1's from both ends and in the middle.
  3. The minimum time will be the smallest of these calculated times.

This approach ensures a linear scan of the string, making it efficient given the constraints.

Code Solutions

def minimumTime(s: str) -> int:
    left = s.find('1')  # Find the first '1'
    right = s.rfind('1')  # Find the last '1'
    
    if left == -1:  # No '1' found
        return 0

    # Calculate the number of operations needed
    remove_from_left = right + 1  # Removing cars from the left to the last '1'
    remove_from_right = len(s) - left  # Removing cars from the right to the first '1'
    remove_from_middle = (right - left + 1) * 2  # Removing '1's from the middle

    return min(remove_from_left, remove_from_right, remove_from_middle)
← Back to All Questions