<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:
        heapify(buses)
        heapify(passengers)

        res = 1
        prev = 1
        cap = 0
        while len(buses) > 0:
            bus = heappop(buses)
            cap = 0

            while cap < capacity and len(passengers) > 0 and passengers[0] <= bus:
                cap += 1
                if prev != passengers[0] - 1:
                    res = passengers[0] - 1
                prev = heappop(passengers)
            else:
                if cap < capacity and prev != bus:
                    res = bus

        return res
632