LeetCode 2332: The Latest Time to Catch a Bus Solution

Master LeetCode problem 2332 (The Latest Time to Catch a Bus), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2332. The Latest Time to Catch a Bus

Medium

Array Two Pointers Binary Search Sorting

Problem Explanation

Explanation:

To solve this problem, we need to find the latest time a passenger can arrive at the bus station to catch a bus. We can simulate the process of passengers arriving and buses departing to determine this time.

We first sort both the arrays buses and passengers in ascending order.
Initialize a variable time to 0 to represent the current time.
Iterate through the buses and passengers simultaneously.
For each bus, while there are passengers waiting and the bus is not full, we check if the next passenger in line can catch the bus at the current time.
If the passenger can catch the bus, we update the time to the maximum of the current time and the passenger's arrival time.
Return the latest time as the result.

Time Complexity:

The time complexity of this approach is O(n + m), where n is the number of buses and m is the number of passengers.

Space Complexity:

The space complexity is O(1) as we are not using any extra space apart from a few variables.

Solution Code

import java.util.Arrays;

class Solution {
    public int latestTimeToCatchBus(int[] buses, int[] passengers, int capacity) {
        Arrays.sort(buses);
        Arrays.sort(passengers);
        
        int time = 0;
        int passengerIdx = 0;
        
        for (int bus : buses) {
            while (passengerIdx < passengers.length && passengers[passengerIdx] <= bus && passengerIdx < capacity) {
                time = Math.max(time, passengers[passengerIdx]);
                passengerIdx++;
            }
        }
        
        return time;
    }
}

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