875. Koko Eating Bananas

1. Description

Koko loves to eat bananas. There are n piles of bananas, the ith pile has piles[i] bananas. The guards have gone and will come back in h hours.
Koko can decide her bananas-per-hour eating speed of k. Each hour, she chooses some pile of bananas and eats k bananas from that pile. If the pile has less than k bananas, she eats all of them instead and will not eat any more bananas during this hour.
Koko likes to eat slowly but still wants to finish eating all the bananas before the guards return.
Return the minimum integer k such that she can eat all the bananas within h hours.

2. Example

Example 1:
Input: piles = [3,6,7,11], h = 8
Output: 4

Example 2:
Input: piles = [30,11,23,4,20], h = 5
Output: 30

Example 3:
Input: piles = [30,11,23,4,20], h = 6
Output: 23

3. Constraints

• 1 <= piles.length <= $10^4$
• piles.length <= h <= $10^9$
• 1 <= piles[i] <= $10^9$

4. Solutions

My Accepted Solution

n = piles, m = max(piles)
Time complexity: O(nlogm)
Space complexity: O(1)

class Solution {
private:
    int eat_all_piles_time(const vector<int> &piles, int speed) {
        int result = 0;
        for (auto pile : piles) {
            result += (pile + speed - 1) / speed;
            // == ceil(double(pile) / speed);
            // == pile / speed + (pild % speed > 0);
        }

        return result;
    }

public:
    // int minEatingSpeed(vector<int> &piles, int h)
    int minEatingSpeed(const vector<int> &piles, int hours) {
        int result = 0;
        for (int left = 1, right = *max_element(piles.begin(), piles.end()) + 1; left < right;) {
            int mid = (left + right) / 2;
            int time_to_eat_piles = eat_all_piles_time(piles, mid);

            time_to_eat_piles <= hours ? result = mid, right = mid : left = mid + 1;
        }

        return result;
    }
};