Koko loves to eat bananas. There are n
piles of bananas, the ith
pile has piles[i]
bananas. The guards have gone and will come back in h
hours.
Koko can decide her bananasperhour eating speed of k
. Each hour, she chooses some pile of bananas and eats k
bananas from that pile. If the pile has less than k
bananas, she eats all of them instead and will not eat any more bananas during this hour.
Koko likes to eat slowly but still wants to finish eating all the bananas before the guards return.
Return the minimum integer k
such that she can eat all the bananas within h
hours.
Example 1:
Input: piles = [3,6,7,11], h = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], h = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], h = 6
Output: 23
Constraints:
1 <= piles.length <= 104
piles.length <= h <= 109
1 <= piles[i] <= 109
Algorithm
We can use Binary Search to find the minimum k

 if
hours > h
, that indicatesk
is too small, thenlow = k + 1
 if
hours < h
, that indicatesk
is too large, thenhigh = k  1
 if
hours == h
, we can try a smallerk
, then alsohigh = k  1
 intuitively, we can initialize
low = 1, high = 1000000000
orlow = 1, high = max(piles)
 Note that we are searching
k
via Binary Search, we need not sort any array or list, the condition of Binary Search is  The search space is limited
 Every time after checking for the current
mid
, we know exactly where to search next (greater thanmid
or lower thanmid
)
 Space complexity: O(1)
 Time complexity: O(n*log(max(piles)))