3. Koko Eating Bananas

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.

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

Constraints:

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

Approach

• Determine the min and max speed and use binary search on it
• Minimum is 1 and max is whatever the highest pile is
• Use ceil to round up to next if fraction
• Use double as java would round up on its own for integer division
• Time: O(n*logm) Space: O(1) - m = max pile

class Solution {
    public int minEatingSpeed(int[] piles, int h) {
        int l = 1;
        int r = Arrays.stream(piles).max().getAsInt();
        int speed = r;
 
        while (l <= r) {
            int m = l + (r - l) / 2;
            int totalHours = 0;
            for (int pile: piles)
                totalHours += Math.ceil((double) pile / m);
            if (totalHours <= h) {
                speed = m;
                r = m - 1;
            } else {
                l = m + 1;
            }
        }
 
        return speed;
    }
}