<p>You are given a <strong>0-indexed</strong> integer array <code>buses</code> of length <code>n</code>, where <code>buses[i]</code> represents the departure time of the <code>i<sup>th</sup></code> bus. You are also given a <strong>0-indexed</strong> integer array <code>passengers</code> of length <code>m</code>, where <code>passengers[j]</code> represents the arrival time of the <code>j<sup>th</sup></code> passenger. All bus departure times are unique. All passenger arrival times are unique.</p>

<p>You are given an integer <code>capacity</code>, which represents the <strong>maximum</strong> number of passengers that can get on each bus.</p>

<p>When a passenger arrives, they will wait in line for the next available bus. You can get on a bus that departs at <code>x</code> minutes if you arrive at <code>y</code> minutes where <code>y &lt;= x</code>, and the bus is not full. Passengers with the <strong>earliest</strong> arrival times get on the bus first.</p>

<p>More formally when a bus arrives, either:</p>

<ul>
	<li>If <code>capacity</code> or fewer passengers are waiting for a bus, they will <strong>all</strong> get on the bus, or</li>
	<li>The <code>capacity</code> passengers with the <strong>earliest</strong> arrival times will get on the bus.</li>
</ul>

<p>Return <em>the latest time you may arrive at the bus station to catch a bus</em>. You <strong>cannot</strong> arrive at the same time as another passenger.</p>

<p><strong>Note: </strong>The arrays <code>buses</code> and <code>passengers</code> are not necessarily sorted.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> buses = [10,20], passengers = [2,17,18,19], capacity = 2
<strong>Output:</strong> 16
<strong>Explanation:</strong> Suppose you arrive at time 16.
At time 10, the first bus departs with the 0<sup>th</sup> passenger. 
At time 20, the second bus departs with you and the 1<sup>st</sup> passenger.
Note that you may not arrive at the same time as another passenger, which is why you must arrive before the 1<sup>st</sup> passenger to catch the bus.</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> buses = [20,30,10], passengers = [19,13,26,4,25,11,21], capacity = 2
<strong>Output:</strong> 20
<strong>Explanation:</strong> Suppose you arrive at time 20.
At time 10, the first bus departs with the 3<sup>rd</sup> passenger. 
At time 20, the second bus departs with the 5<sup>th</sup> and 1<sup>st</sup> passengers.
At time 30, the third bus departs with the 0<sup>th</sup> passenger and you.
Notice if you had arrived any later, then the 6<sup>th</sup> passenger would have taken your seat on the third bus.</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == buses.length</code></li>
	<li><code>m == passengers.length</code></li>
	<li><code>1 &lt;= n, m, capacity &lt;= 10<sup>5</sup></code></li>
	<li><code>2 &lt;= buses[i], passengers[i] &lt;= 10<sup>9</sup></code></li>
	<li>Each element in <code>buses</code> is <strong>unique</strong>.</li>
	<li>Each element in <code>passengers</code> is <strong>unique</strong>.</li>
</ul>

class Solution:
    def latestTimeCatchTheBus(self, buses: List[int], passengers: List[int], capacity: int) -> int:
        # passengers.sort()
        # buses.sort()
        # p_bus_map = defaultdict(list)
        # for p in passengers:
        #     for b in buses:
        #         if p <= b and len(p_bus_map[b]) < capacity:
        #             p_bus_map[b].append(p)
        #             break
        # if not p_bus_map:
        #     return buses[-1]
        # if buses[-1] not in p_bus_map:
        #     return buses[-1]
        # latest_bus = max(p_bus_map)
        # latest_bus_passengers = p_bus_map[latest_bus]
        # i = len(latest_bus_passengers) - 1
        # if capacity > i+1 and latest_bus not in latest_bus_passengers:
        #     return latest_bus
        # while i > 0 and latest_bus_passengers[i] - 1 == latest_bus_passengers[i-1]:
        #     i -= 1
        # j = 1
        # while latest_bus_passengers[i] - j in passengers:
        #     j += 1
        # return latest_bus_passengers[i] - j
        passengers.sort()
        buses.sort()
        
        # Use a set to quickly check if a time is taken by a passenger
        passengers_set = set(passengers)
        
        # Initialize a pointer for passengers
        p_index = 0
        n_passengers = len(passengers)
        
        # Iterate over each bus to determine the latest possible time
        for bus in buses:
            # Count the number of passengers this bus can take
            count = 0
            while p_index < n_passengers and passengers[p_index] <= bus and count < capacity:
                count += 1
                p_index += 1

        # After the last bus, either it has full capacity or less than full
        # Start checking from the bus time backwards to find the latest time slot
        last_time = buses[-1]  # The last bus time
        
        # If the last bus is not fully occupied, return the last bus time directly
        if count < capacity:
            while last_time in passengers_set:
                last_time -= 1
            return last_time
        
        # If the bus is fully occupied, check the last passenger it took
        last_time = passengers[p_index - 1] - 1
        
        # Find the last available time that isn't occupied by another passenger
        while last_time in passengers_set:
            last_time -= 1
            
        return last_time
        
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