<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, b: List[int], p: List[int], c: int) -> int:
        b.sort()
        p.sort()
        i,j=0,0
        cp=0
        p2b={}
        nf_last={}
        while i<len(b) and j<len(p):
            if cp<c:
                if b[i]>= p[j]:
                    p2b[j]=i
                    j+=1
                    cp+=1
                    last_index=j
                elif b[i]< p[j]:
                    nf_last[i]=j-1
                    i+=1
                    cp=0
            elif cp==c:
                i+=1
                cp=0

        
        if cp<c and i<len(b) and i not in nf_last:
            nf_last[i]= len(p)-1
        empty=[]
        for index in range(i+1, len(b)):
            empty.append(index)
            
        #print('b ',b)
        #print('p', p )
        #print('p2b ',p2b)
        #print('nf_last ',nf_last)
        #print('empty ',empty)
        res=float('-inf')
        if empty:
            max_b_index=max(empty)
            res= b[max_b_index]
            
        else:
            for p_index,b_index in p2b.items():
                if p_index>=1:
                    if (p[p_index]-1) != p[p_index-1]:
                        res=max(res, p[p_index]-1)

                elif p_index==0:
                    res=max(res, p[0]-1)
            for b_index in nf_last:
                last_p_index= nf_last[b_index]
                if b[b_index]!= p[last_p_index]: 
                    res=max(res, b[b_index])
        return res
    
632