Koko Eating Baacnanas

Koko loves to eat bananas. There are n piles of bananas, the i<sup>th</sup> 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.

LeetCode Problem - 875: Link | Click Here

import java.util.IntSummaryStatistics;
import java.util.stream.IntStream;

class Solution {
    // Method to find the minimum eating speed
    public int minEatingSpeed(int[] piles, int h) {

        // Get summary statistics of the array using Java streams
        IntSummaryStatistics sb = IntStream.of(piles).summaryStatistics();

        // Set the initial search range for the eating speed
        int high = sb.getMax();
        int low = 1;

        // Binary search loop to find the optimal eating speed
        while(low <= high){
            // Calculate the middle value of the current search range
            int mid = (low + high) / 2;

            // Calculate the total hours required for the current eating speed
            int speed = minEat(piles, mid);

            // Adjust the search range based on the calculated total hours
            if(speed < h){
                high = mid - 1;  // Reduce the eating speed
            }
            else{
                low = mid + 1;   // Increase the eating speed
            }
        }

        // Return the lowest eating speed that satisfies the given hours
        return low;
    }

    // Method to calculate the total hours required for a given eating speed
    public int minEat(int[] arr, int h){
        int totalHour = 0;
        // Calculate total hours using the formula: totalHour += ceil(arr[i] / h)
        for(int i = 0; i < arr.length; i++){
            totalHour += Math.ceil((double) arr[i] / h);
        }
        return totalHour;
    }
}