Problem statement:

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:

Input: piles = [3,6,7,11], H = 8
Output: 4

Solution

public int minEatingSpeed(int[] piles, int H) {
    int max = findMax(piles);
    int lo = findMinStart(piles, H, max), hi = max + 1;
    while (lo < hi) {
        int mid = lo + (hi - lo) / 2;
        if (canEat(mid, piles, H)) {
            hi = mid;
        } else {
            lo = mid + 1;
        }
    }
    return lo;
}

//not really needed, small optimization
private int findMinStart(int[] piles, int H, int max) {
    int len = piles.length - 1;
    int hoursLeft = H - len;
    return max / hoursLeft + (max % hoursLeft == 0 ? 0 : 1);
}

private int findMax(int[] piles) {
    int max = 0;
    for (int num : piles) {
        max = Math.max(num, max);
    }
    return max;
}

public boolean canEat(int k, int[] piles, int H) {
    for (int pile : piles) {
        int hours = pile / k + (pile % k == 0 ? 0 : 1);
        H -= hours;
        if (H < 0) return false;
    }
    return true;
}

Time complexity:

O(N + NlogM)

Space complexity:

O(1)

Approach:

1. do binary search of K
2. check if koko can eat all of the bananas with specific K, then go either right or left