README_EN.md

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true	Medium	https://github.com/doocs/leetcode/edit/main/solution/0800-0899/0875.Koko%20Eating%20Bananas/README_EN.md	Array Binary Search

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

Solution 1: Binary Search

We notice that if Koko can eat all the bananas at a speed of $k$ within $h$ hours, then she can also eat all the bananas at a speed of $k' > k$ within $h$ hours. This shows monotonicity, so we can use binary search to find the smallest $k$ that satisfies the condition.

We define the left boundary of the binary search as $l = 1$, and the right boundary as $r = \max(\text{piles})$. For each binary search, we take the middle value $mid = \frac{l + r}{2}$, and then calculate the time $s$ required to eat bananas at a speed of $mid$. If $s \leq h$, it means that the speed of $mid$ can meet the condition, and we update the right boundary $r$ to $mid$; otherwise, we update the left boundary $l$ to $mid + 1$. Finally, when $l = r$, we find the smallest $k$ that satisfies the condition.

The time complexity is $O(n \times \log M)$, where $n$ and $M$ are the length and maximum value of the array piles respectively. The space complexity is $O(1)$.

Python3

class Solution:
    def minEatingSpeed(self, piles: List[int], h: int) -> int:
        def check(k: int) -> bool:
            return sum((x + k - 1) // k for x in piles) <= h

        return 1 + bisect_left(range(1, max(piles) + 1), True, key=check)

Java

class Solution {
    public int minEatingSpeed(int[] piles, int h) {
        int l = 1, r = (int) 1e9;
        while (l < r) {
            int mid = (l + r) >> 1;
            int s = 0;
            for (int x : piles) {
                s += (x + mid - 1) / mid;
            }
            if (s <= h) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    }
}

C++

class Solution {
public:
    int minEatingSpeed(vector<int>& piles, int h) {
        int l = 1, r = ranges::max(piles);
        while (l < r) {
            int mid = (l + r) >> 1;
            int s = 0;
            for (int x : piles) {
                s += (x + mid - 1) / mid;
            }
            if (s <= h) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    }
};

Go

func minEatingSpeed(piles []int, h int) int {
	return 1 + sort.Search(slices.Max(piles), func(k int) bool {
		k++
		s := 0
		for _, x := range piles {
			s += (x + k - 1) / k
		}
		return s <= h
	})
}

TypeScript

function minEatingSpeed(piles: number[], h: number): number {
    let [l, r] = [1, Math.max(...piles)];
    while (l < r) {
        const mid = (l + r) >> 1;
        const s = piles.map(x => Math.ceil(x / mid)).reduce((a, b) => a + b);
        if (s <= h) {
            r = mid;
        } else {
            l = mid + 1;
        }
    }
    return l;
}

Rust

impl Solution {
    pub fn min_eating_speed(piles: Vec<i32>, h: i32) -> i32 {
        let mut l = 1;
        let mut r = *piles.iter().max().unwrap_or(&0);
        while l < r {
            let mid = (l + r) >> 1;
            let mut s = 0;
            for x in piles.iter() {
                s += (x + mid - 1) / mid;
            }
            if s <= h {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        l
    }
}

C#

public class Solution {
    public int MinEatingSpeed(int[] piles, int h) {
        int l = 1, r = (int) 1e9;
        while (l < r) {
            int mid = (l + r) >> 1;
            int s = 0;
            foreach (int x in piles) {
                s += (x + mid - 1) / mid;
            }
            if (s <= h) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    }
}