Problem Description

You are given an integer array piles where piles[i] is the number of bananas in the i^{th} pile. You are also given an integer h, which represents the number of hours you have to eat all the bananas.

You may decide your bananas-per-hour eating rate of k. Each hour, you may choose a pile of bananas and eats k bananas from that pile. If the pile has less than k bananas, you may finish eating the pile but you can not eat from another pile in the same hour.

Return the minimum integer k such that you can eat all the bananas within h hours.

Example 1:

Input: piles = [1,4,3,2], h = 9

Output: 2 Explanation: With an eating rate of 2, you can eat the bananas in 6 hours. With an eating rate of 1, you would need 10 hours to eat all the bananas (which exceeds h=9), thus the minimum eating rate is 2.

Example 2:

Input: piles = [25,10,23,4], h = 4

Output: 25

Constraints:

• 1 <= piles.length <= 1,000
• piles.length <= h <= 1,000,000
• 1 <= piles[i] <= 1,000,000,000

Approach

• Brute Force Approach: Iterate through potential eating speeds k starting from 1 up to the maximum pile size. For each k, compute the total hours required to finish all piles by taking the ceiling of division for each pile. Return the first k that meets the condition hours <= h.

Solution

1. Brute Force (Current By User)

class Solution {
    public int minEatingSpeed(int[] piles, int h) {

        int n = piles.length;
        Arrays.sort(piles);// 1 2 3 4
        int k = 0;
        for(k=1;k<=piles[n-1];k++){ //2
            int index = 0;
            int hour = 0;
            while(index<n){
                if(piles[index]>k){
                    hour += Math.ceil((double)piles[index]/k);
                    index++;
                }
                else{
                    index++;
                    hour++;
                } //1 1 1 1 1 1

            }
            if(hour<=h){
                return k;
            }
        }
        return k;
    }
}