class Solution:
    def minEatingSpeed(self, piles: List[int], h: int) -> int:
        '''
        NOTE: Do read the problem carefully!
        
        Absolute min speed (k) at which Koko can eat bananas would be 1
        and max speed would be max(piles). Of course, these might not
        be correct and might help Koko finish eating before `h` hours.
        
        So, we need to check each k from 1:max(piles) and figure out the
        min k that lets us consume <= h
        
        We can iterate over each k value and check or we can do binary
        search and figure out least k.
        '''
        
        start, end = 1, max(piles)
        k = end
        
        while start <= end:
            mid = (start + end) // 2
            
            total = 0
            for p in piles:
                # If pile has 3 bananas and k = 2, then 3 / 2 = 1.5
                # but we don't want fractional values, we will consider
                # next greatest integer which'd be ceil(p / mid)
                total += math.ceil(p / mid)
            
            # k is valid
            if total <= h:
                # Update k if mid is smaller
                if mid < k:
                    k = mid
                    end = mid - 1
                # If mid starts to go bigger than k, then
                # we can't possibly find any better results
                # so just break
                else:
                    break
            # k is too small 👉 total is large so we need
            # to move `start`
            else:
                start = mid + 1
        
        return k