Solution Explanation: The Latest Time to Catch a Bus

This problem asks to find the latest time a passenger can arrive at the bus station and still catch a bus. The solution involves sorting, simulating bus departures, and a bit of clever post-processing.

Approach:

1. Sort: We start by sorting both the buses and passengers arrays. This allows us to efficiently process passengers in arrival order and check for bus availability.

2. Simulation: We iterate through the sorted buses array. For each bus, we simulate passengers boarding. We maintain a capacity counter representing the remaining seats on the bus. We iterate through the passengers array using a pointer j. If a passenger arrives before the bus departs (passengers[j] <= t, where t is the current bus departure time) and there's space on the bus (c > 0), the passenger boards, capacity decreases, and we move to the next passenger.

3. Determining the Latest Arrival Time: After simulating all bus departures, we need to determine the latest possible arrival time.

   • If there are remaining seats on the last bus (c > 0): The latest arrival time is the departure time of the last bus (buses[-1]). However, we need to adjust this if another passenger arrives at the same time. We decrement the time until we find a time where no passenger arrives.

   • If the last bus is full (c == 0): The latest arrival time is the arrival time of the last passenger who boarded the last bus (passengers[j]). Again, we need to adjust this to ensure we don't arrive at the same time as another passenger. We decrement the time until we find a time where no passenger arrives.

Time Complexity:

• Sorting the buses and passengers arrays takes O(n log n) and O(m log m) time respectively, where n is the number of buses and m is the number of passengers.
• The simulation loop iterates through the buses once, which takes O(n) time. Within the loop, we might iterate through all passengers in the worst case, which takes O(m) time.
• The post-processing to adjust the arrival time takes at most O(m) time in the worst case.

Therefore, the overall time complexity is dominated by sorting and is O(n log n + m log m).

Space Complexity:

The space complexity is determined by the space used for sorting (which can vary depending on the sorting algorithm used, but is often O(log n) and O(log m) in-place algorithms), and any auxiliary variables used during the simulation. This results in a space complexity of O(log n + log m) (or O(1) if using in-place sorting).

Code Examples (Python):

class Solution:
    def latestTimeCatchTheBus(self, buses: List[int], passengers: List[int], capacity: int) -> int:
        buses.sort()
        passengers.sort()
        n = len(buses)
        m = len(passengers)
        j = 0
        for i in range(n):
            count = capacity
            while j < m and count > 0 and passengers[j] <= buses[i]:
                count -= 1
                j += 1
 
        j -= 1  # Adjust j to point to the last passenger on the last bus
 
        latest_time = buses[-1] if count > 0 else passengers[j]
 
        while j >= 0 and passengers[j] == latest_time:
            latest_time -= 1
            j -= 1
 
        return latest_time

This Python code directly implements the algorithm described above. The other language examples follow the same logic. Note that minor variations might exist across implementations due to differences in how sorting and array handling is performed in each language.