875. Koko Eating Bananas

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875. Koko Eating Bananas

Description

Difficulty: Medium

Related Topics: Array, Binary Search

Koko loves to eat bananas. There are n piles of bananas, the ith 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:

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

Example 2:

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Input: piles = [30,11,23,4,20], h = 5
Output: 30

Example 3:

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Input: piles = [30,11,23,4,20], h = 6
Output: 23

Constraints:

  • 1 <= piles.length <= 104
  • piles.length <= h <= 109
  • 1 <= piles[i] <= 109

Hints/Notes

  • 2023/08/12
  • Helper function: eating hour, if eat k bananas per hour, how many hours in total would be needed?
  • Make it a binary search for left boundary
  • 0x3F’s solution(checked)

Solution

Language: C++

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class Solution {
public:
    int eatingHour(vector<int>& piles, int s) {
        int hour = 0;
        for (int pile : piles) {
            hour += pile / s;
            if (pile % s != 0) {
                hour++;
            }
        }
        return hour;
    }

    int minEatingSpeed(vector<int>& piles, int h) {
        int left = 1, right = 0;
        for (int pile : piles) {
            if (right < pile) {
                right = pile;
            }
        }
        while (left < right) {
            int mid = left + (right - left) / 2;
            int hour = eatingHour(piles, mid);
            if (hour == h) {
                right = mid;
            } else if (hour > h) {
                left = mid + 1;
            } else if (hour < h) {
                right = mid;
            }
        }
        // for the worst case left = right = max(piles)
        return left;
    }
};