The key insight is that you're not searching through the piles, you're binary searching on the speed itself.
This is a classic optimization problem where we need to find the minimum valid eating speed. The key observation is that the feasibility of a speed k is monotonic: if k works, any larger speed also works. Therefore, we can use binary search on the answer, checking feasibility in O(n) time per candidate speed.
Pro tip: Always clarify edge cases and constraints upfront, such as whether h is at least the number of piles (otherwise it's impossible) and the range of pile sizes. This shows attention to detail and prevents incorrect assumptions.
Restate the problem in your own words. Identify that each hour you choose one pile and eat up to k bananas, and you need to finish all piles within h hours. Note that h must be at least n (the number of piles), otherwise it's impossible.
For a given speed k, compute the total hours needed as sum(ceil(pile / k)) for all piles. If this sum is <= h, then k is feasible.
Observe that if speed k is feasible, any speed > k is also feasible. Thus, we can binary search for the minimum k in the range [1, max(piles)].
Set low = 1, high = max(piles). While low < high, compute mid = (low + high) // 2. If feasible(mid), set high = mid; else set low = mid + 1. Return low as the minimum speed.
Time complexity: O(n log m) where m is the maximum pile size. Space complexity: O(1). Handle edge cases: h < n (impossible), single pile, large h (speed 1 works).
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