/*
    Given array of banana piles, guards are gone for h hours
    Return min int k such that can eat all banans within h
    Ex. piles = [3,6,7,11] h = 8 -> 4 (1@3, 2@6, 2@7, 3@11)

    Binary search, for each k count hours needed, store min

    Time: O(n x log m) -> n = # of piles, m = max # in a pile
    Space: O(1)
*/

class Solution {
public:
    int minEatingSpeed(vector<int>& piles, int h) {
        int n = piles.size();
        
        int low = 1;
        int high = 0;
        for (int i = 0; i < n; i++) {
            high = max(high, piles[i]);
        }
        
        int result = high;
        
        while (low <= high) {
            int k = low + (high - low) / 2;
            long int hours = 0;
            for (int i = 0; i < n; i++) {
                hours += ceil((double) piles[i] / k);
            }
            if (hours <= h) {
                result = min(result, k);
                high = k - 1;
            } else {
                low = k + 1;
            }
        }
        
        return result;
    }
};