Koko Eating Bananas
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
Koko can decide her bananas-per-hour 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
piles = [3,6,7,11], h = 8
piles = [30,11,23,4,20], h = 5
piles = [30,11,23,4,20], h = 6
1 <= piles.length <= 104
piles.length <= h <= 109
1 <= piles[i] <= 109
We want to apply the binary search template, and implement the
Here, the feasible function is whether Koko
can_finish_eating all piles within
h hours while eating at speed
k per hour.
Since Koko eats at only one pile during each hour,
ceil(float(p)/k) is the time Koko takes to finish pile
p/k does not work here because we want a whole number of hours so we needed to round up
Therefore, the feasiblity is determined by whether Koko's
hours_used is within
h hours, where
hours_used is the total hours to finish all piles.
Recall example 1 with Input:
piles = [3,6,7,11], h = 8, and Output:
can_finish_eating function returns the following results.
1def can_finish_eating(piles, h, k): 2 hours_used = 0 3 for p in piles: 4 hours_used += ceil(float(p)/k) 5 return hours_used <= h 6 7def minEatingSpeed(piles, h): 8 left, right = 1, 1000000000 # 10^9 max length of the piles 9 ans = -1 10 while left <= right: 11 mid = (left + right) // 2 12 if can_finish_eating(piles, h, mid): 13 ans = mid 14 right = mid - 1 15 else: 16 left = mid + 1 17 return ans
Got a question? Ask the Teaching Assistant anything you don't understand.