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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0801_0900.s0875_koko_eating_bananas;

// #Medium #Array #Binary_Search #Binary_Search_II_Day_4
// #2022_03_28_Time_15_ms_(91.32%)_Space_55_MB_(6.01%)

/**
 * 875 - Koko Eating Bananas\.
 *
 * Medium
 *
 * 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 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 <= 10⁴
 * *   piles.length <= h <= 10⁹
 * *   1 <= piles[i] <= 10⁹
**/
public class Solution {
    public int minEatingSpeed(int[] piles, int h) {
        int maxP = piles[0];
        long sumP = 0L;
        for (int pile : piles) {
            maxP = Math.max(maxP, pile);
            sumP += pile;
        }
        // binary search
        int low = (int) ((sumP - 1) / h + 1);
        int high = maxP;
        while (low < high) {
            int mid = low + (high - low) / 2;
            if (isPossible(piles, mid, h)) {
                high = mid;
            } else {
                low = mid + 1;
            }
        }
        return low;
    }

    private boolean isPossible(int[] piles, int k, int h) {
        int sum = 0;
        for (int pile : piles) {
            sum += (pile - 1) / k + 1;
        }
        return sum <= h;
    }
}