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875. Koko Eating Bananas

中文文档

Description

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:

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

Solutions

Binary search.

Python3

class Solution:
    def minEatingSpeed(self, piles: List[int], h: int) -> int:
        left, right = 1, max(piles)
        while left < right:
            mid = (left + right) >> 1
            s = sum([(pile + mid - 1) // mid for pile in piles])
            if s <= h:
                right = mid
            else:
                left = mid + 1
        return left

Java

class Solution {
    public int minEatingSpeed(int[] piles, int h) {
        int mx = 0;
        for (int pile : piles) {
            mx = Math.max(mx, pile);
        }
        int left = 1, right = mx;
        while (left < right) {
            int mid = (left + right) >>> 1;
            int s = 0;
            for (int pile : piles) {
                s += (pile + mid - 1) / mid;
            }
            if (s <= h) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
}

C++

class Solution {
public:
    int minEatingSpeed(vector<int>& piles, int h) {
        int left = 1, right = *max_element(piles.begin(), piles.end());
        while (left < right)
        {
            int mid = left + right >> 1;
            int s = 0;
            for (int pile : piles) s += (pile + mid - 1) / mid;
            if (s <= h) right = mid;
            else left = mid + 1;
        }
        return left;
    }
};

Go

func minEatingSpeed(piles []int, h int) int {
	mx := 0
	for _, pile := range piles {
		mx = max(mx, pile)
	}
	left, right := 1, mx
	for left < right {
		mid := (left + right) >> 1
		s := 0
		for _, pile := range piles {
			s += (pile + mid - 1) / mid
		}
		if s <= h {
			right = mid
		} else {
			left = mid + 1
		}
	}
	return left
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

C#

public class Solution {
    public int MinEatingSpeed(int[] piles, int h) {
        int left = 1, right = piles.Max();
        while (left < right)
        {
            int mid = (left + right) >> 1;
            int s = 0;
            foreach (int pile in piles)
            {
                s += (pile + mid - 1) / mid;
            }
            if (s <= h)
            {
                right = mid;
            }
            else
            {
                left = mid + 1;
            }
        }
        return left;
    }
}

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