Understanding the Problem
We are given a list of banana piles, and a total number of hours h
within which Koko must finish eating all the bananas. Each hour, Koko chooses one pile and eats up to k
bananas from it. If the pile has fewer than k
bananas, she eats all of them and doesn’t continue to another pile in the same hour.
The goal is to find the minimum integer speed k
such that Koko can finish eating all the bananas within h
hours. This is essentially an optimization problem — we want the smallest value of k
for which the total time taken does not exceed h
.
Step-by-Step Solution with Example
Step 1: Understand how time is calculated
If Koko eats at a speed k
, then for each pile, the number of hours she needs is ceil(pile / k)
. We need to compute this for all piles and add it up to find the total time.
Step 2: Choose an example
Let's say the piles are [3, 6, 7, 11]
and h = 8
. Koko must eat all bananas in at most 8 hours.
Step 3: Identify possible range for k
The minimum speed is 1
banana per hour. The maximum speed is the largest pile, 11
, because if she can finish the biggest pile in one hour, she can finish any pile in one hour.
Step 4: Use Binary Search to find the minimum k
We now perform binary search between 1 and 11:
- Mid = 6: Total time = ceil(3/6) + ceil(6/6) + ceil(7/6) + ceil(11/6) = 1 + 1 + 2 + 2 = 6 (less than 8) → try smaller speed.
- Mid = 3: Total time = ceil(3/3) + ceil(6/3) + ceil(7/3) + ceil(11/3) = 1 + 2 + 3 + 4 = 10 (too much) → try higher speed.
- Mid = 4: Total time = 1 + 2 + 2 + 3 = 8 (just fits) → try smaller speed to optimize further.
- Mid = 3: already tested, doesn’t work → so answer is 4.
Step 5: Return the result
The minimum speed k
that allows Koko to finish within h = 8
hours is 4
.
Edge Cases
- Exact Fit: If
h
equals the number of piles, then Koko must eat one pile per hour. So the speed must be at least as large as the largest pile.
- Very Large h: If
h
is very high, Koko can eat slowly. Even speed = 1 might be enough.
- Only One Pile: Then the speed is just
ceil(pile / h)
.
- No Piles: If the list is empty, return
0
since there's nothing to eat.
- All piles are size 1: The total time is equal to the number of piles, and any
k ≥ 1
would work if h
is enough.
Finally
This problem is a great example of combining simulation with optimization using binary search. The key insight is understanding how ceil(pile / k)
affects total time and narrowing the search space smartly. Always consider edge cases — especially empty input, minimum and maximum bounds — when crafting such a solution.
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