Koko Eating Bananas

n piles of bananas, Koko eats k bananas/hour from one pile at a time, h hours available. Find the minimum k so Koko finishes everything within h hours. (LeetCode problem)

This is a binary search on the answer problem: instead of searching within an array, we binary search over a range of possible answers (here, possible eating speeds), using a feasibility check to decide which half to keep.

Brute force: try every k from 1 to max(piles), computing the hours needed for each (ceil(pile/k) summed across piles) — slow, gives TLE on large inputs.

Binary search: the answer lies in [1, max(piles)]. For each mid, compute total hours needed; if <= h, it’s a valid (possibly non-minimal) speed, so try slower (high = mid); otherwise try faster (low = mid + 1).

class Solution {
    public int minEatingSpeed(int[] piles, int h) {
        int max = 0;
        for (int p : piles) {
            max = Math.max(max, p);
        }

        int low = 1, high = max;

        while (low < high) {
            int mid = low + (high - low) / 2;
            int totalH = calculateTotalHours(piles, mid);

            if (totalH <= h) {
                high = mid; // try a slower (smaller) speed
            } else {
                low = mid + 1; // need to go faster
            }
        }

        return low;
    }

    public int calculateTotalHours(int[] piles, int hourly) {
        int totalH = 0;
        for (int i = 0; i < piles.length; i++) {
            totalH += Math.ceil((double) piles[i] / (double) hourly);
        }
        return totalH;
    }
}

Time: O(N × log(max(piles))). Space: O(1).