309. Koko Eating Bananas

The set of eating speeds is monotonic: if Koko can finish at speed k, she can also finish at every larger speed. That monotonicity lets us binary-search the smallest feasible k in [1, max(piles)].

Approach

1. lo = 1, hi = max(piles).
2. mid = (lo + hi) // 2. Compute hours needed at speed mid: sum(ceil(p / mid) for p in piles).
3. If hours <= h, hi = mid (try smaller).
4. Else lo = mid + 1.
5. Return lo.

The hours formula

At speed k, pile of size p takes ⌈p / k⌉ hours, which in Python is -(-p // k) or (p + k - 1) // k.

class Solution {
    public int minEatingSpeed(int[] piles, int h) {
        int lo = 1, hi = 0;
        for (int p : piles) hi = Math.max(hi, p);
        while (lo < hi) {
            int mid = (lo + hi) / 2;
            long hours = 0;
            for (int p : piles) hours += (p + mid - 1) / mid;
            if (hours <= h) {
                hi = mid;
            } else {
                lo = mid + 1;
            }
        }
        return lo;
    }
}