<p>You are given an integer array <code>ranks</code> representing the <strong>ranks</strong> of some mechanics. <font face="monospace">ranks<sub>i</sub></font> is the rank of the <font face="monospace">i<sup>th</sup></font> mechanic<font face="monospace">.</font> A mechanic with a rank <code>r</code> can repair <font face="monospace">n</font> cars in <code>r * n<sup>2</sup></code> minutes.</p>

<p>You are also given an integer <code>cars</code> representing the total number of cars waiting in the garage to be repaired.</p>

<p>Return <em>the <strong>minimum</strong> time taken to repair all the cars.</em></p>

<p><strong>Note:</strong> All the mechanics can repair the cars simultaneously.</p>

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

<pre>
<strong>Input:</strong> ranks = [4,2,3,1], cars = 10
<strong>Output:</strong> 16
<strong>Explanation:</strong> 
- The first mechanic will repair two cars. The time required is 4 * 2 * 2 = 16 minutes.
- The second mechanic will repair two cars. The time required is 2 * 2 * 2 = 8 minutes.
- The third mechanic will repair two cars. The time required is 3 * 2 * 2 = 12 minutes.
- The fourth mechanic will repair four cars. The time required is 1 * 4 * 4 = 16 minutes.
It can be proved that the cars cannot be repaired in less than 16 minutes.​​​​​
</pre>

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

<pre>
<strong>Input:</strong> ranks = [5,1,8], cars = 6
<strong>Output:</strong> 16
<strong>Explanation:</strong> 
- The first mechanic will repair one car. The time required is 5 * 1 * 1 = 5 minutes.
- The second mechanic will repair four cars. The time required is 1 * 4 * 4 = 16 minutes.
- The third mechanic will repair one car. The time required is 8 * 1 * 1 = 8 minutes.
It can be proved that the cars cannot be repaired in less than 16 minutes.​​​​​
</pre>

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

<ul>
	<li><code>1 &lt;= ranks.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= ranks[i] &lt;= 100</code></li>
	<li><code>1 &lt;= cars &lt;= 10<sup>6</sup></code></li>
</ul>

'''
r * n**2 = T --> n = int (sqrt(T/r))

T = 20

4 --> n_0 = int (sqrt(20/4))
2 --> n_1 = int (sqrt(20/2))
3 --> n_2 = int (sqrt(20/3))
1 --> n_3 = int (sqrt(20/1))

if n_0 + n_1 + n_2 + n_3 >cars:
    then T is to big
else: 
    T is to small

'''

class Solution:
    def repairCars(self, ranks: List[int], cars: int) -> int:

        def check(T):
            '''
            return the maximum number of cars that can be checked in T mins
            '''
            return(sum([int(math.sqrt(T/ranks[i])) for i in range(len(ranks))]))
        
        T_low = 0 
        T_high = 10**15+1
        i = 0
        while T_low < T_high:
            #print(T_low,T_high)
            mid_point = int((T_low+T_high)/2)
            if check(mid_point)>=cars:
               T_high = mid_point
            else:
                T_low = mid_point+1
            i = i + 1
        #print(T_low,T_high)
        return(T_low)
    
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