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875. Koko Eating Bananas

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.

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

  • class Solution {
public:
    int minEatingSpeed(vector<int>& piles, int h) {
        int left = 1, right = 1e9;
        while (left < right) {
            int mid = (left + right) >> 1;
            int s = 0;
            for (int& x : piles) s += (x + mid - 1) / mid;
            if (s <= h)
                right = mid;
            else
                left = mid + 1;
        }
        return left;
    }
};

  • class Solution:
    def minEatingSpeed(self, piles: List[int], h: int) -> int:
        left, right = 1, int(1e9)
        while left < right:
            mid = (left + right) >> 1
            s = sum((x + mid - 1) // mid for x in piles)
            if s <= h:
                right = mid
            else:
                left = mid + 1
        return left


  • func minEatingSpeed(piles []int, h int) int {
	return sort.Search(1e9, func(i int) bool {
		if i == 0 {
			return false
		}
		s := 0
		for _, x := range piles {
			s += (x + i - 1) / i
		}
		return s <= h
	})
}

  • function minEatingSpeed(piles: number[], h: number): number {
    let left = 1;
    let right = Math.max(...piles);
    while (left < right) {
        const mid = (left + right) >> 1;
        let s = 0;
        for (const x of piles) {
            s += Math.ceil(x / mid);
        }
        if (s <= h) {
            right = mid;
        } else {
            left = mid + 1;
        }
    }
    return left;
}


  • 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;
    }
}

  • 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
    }
}

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