<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:
        # 首先对 buses 和 passengers 进行排序，以便按时间顺序处理
        buses.sort()
        passengers.sort()

        # j 是乘客列表的索引，初始为 0
        j = 0
        # 遍历每辆公交车的发车时间 t
        for t in buses:
            # c 表示当前公交车的剩余容量
            c = capacity
            # 当 c > 0 且 j 小于乘客列表长度
            # 且乘客到达时间 passengers[j] 小于等于公交车发车时间 t 时，乘客上车（j 增加，c 减少）
            while c and j < len(passengers) and passengers[j] <= t:
                j += 1
                c -= 1

        # j -= 1：将 j 回退一位，因为在上面的循环中 j 已经超出了最后一个上车的乘客
        j -= 1
        # ans 初始化为最后一班公交车的发车时间 buses[-1]，
        # 如果公交车还有剩余容量 c，
        # 否则为最后一个上车的乘客的时间 passengers[j]
        ans = buses[-1] if c else passengers[j]

        # 然后从 ans 开始往前找，直到找到一个时间点 ans，
        # 这个时间点没有乘客到达（即 ans 不等于 passengers[j]）
        while j >= 0 and ans == passengers[j]: 
            ans -= 1
            j -= 1
        
        # 最终返回 ans，这就是乘客可以搭乘最后一班公交车的最新时间
        return ans
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