Koko Eating Bananas LT875

Koko loves to eat bananas.  There are N piles of bananas, the i-th 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 won't eat any more bananas during this hour.

Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.

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

Note:

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

Idea 1. Binary search, similar to Capacity To Ship Packages Within D Days LT1011, search space: (ceiling(sum(piles)/h), max(piles))

ceiling(sum(piles)/h) = (sum(piles)-1)/h + 1

也可以免去一次遍历数组，直接设置search space (1, 10^9)

Time complexity: O(nlogw), n is the size of piles, w is the maximum of piles.

Space complexity: O(1)

 class Solution {
     private int checkHours(int[] piles, int speed) {
         int hours = 0;
         for(int pile: piles) {
             int hour = (pile-1)/speed + 1;
             //int hour = ((pile%speed == 0) ? pile/speed : pile/speed + 1);
             hours += hour;
         }
         return hours;
     }
     public int minEatingSpeed(int[] piles, int H) {
         int min = 1, max = 0;
         for(int pile: piles) {
             max = Math.max(max, pile);
         }

         while(min < max) {
             int mid = min + (max - min)/2;
             int hours = checkHours(piles, mid);
             if(hours <= H) {
                 max = mid;
             }
             else {
                 min = mid + 1;
             }
         }

         return min;
     }
 }