Koko Eating Bananas

Problem

https://leetcode.com/problems/koko-eating-bananas/

Koko loves to eat bananas. There are n piles of bananas, the i:sup:`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 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 <= 10:sup:`4`

• piles.length\ \ <= h <= 10:sup:`9`

• 1 <= piles[i] <= 10:sup:`9`

Pattern

Array, Binary Search

Approaches

Binary Search

Explanation

For any eating speed \(k\), the time it takes Koko to eat all the bananas is

\[hours = \sum_{i=1}^{P} \left\lceil \frac{piles_i}{k} \right\rceil\]

where \(P\) is the number of piles. We need to take the ceiling since Koko will not eat any more bananas if the pile has less than \(k\) bananas. If \(hours\) is greater than \(h\), then Koko needs to eat faster. If If \(hours\) is less than or equal to \(h\), then their might be a slower speed in which Koko will still finish the bananas. We can use binary search to efficiently search the space of eating speeds: \([0, \max_i(piles_i)]\). The value \(\max_i(piles_i)\) is the maximum eating speed to consider as it is not possible to eat faster than 1 pile per hour.

Code

import math


def minEatingSpeed(piles: list[int], h: int) -> int:
    """Return the minimum eating speed such that Koko can eat all bananas
    within ``h`` hours.
    """
    left = 1
    right = max(piles)

    result = 0

    while left <= right:
        k = (left + right) // 2

        # calculate how long she takes to eat
        hours = 0
        for pile in piles:
            hours += math.ceil(pile / k)

        if hours <= h:
            result = k
            right = k - 1
        else:
            left = k + 1

    return result

Test

>>> from koko_eating_bananas__binary_search import minEatingSpeed
>>> minEatingSpeed([3, 6, 7, 11], 8)
4
>>> minEatingSpeed([30, 11, 23, 4, 20], 5)
30
>>> minEatingSpeed([30, 11, 23, 4, 20], 6)
23

Complexity

\(P\) is the number of piles and \(M = \max_i(piles_i)\)

Measure	Complexity	Notes
Time	\(O(P \log M)\)	need to loop through piles to calculate time to eat for each iteration of binary search
Auxiliary Space	\(O(1)\)

koko_eating_bananas__binary_search.minEatingSpeed(piles: list[int], h: int) → int

Return the minimum eating speed such that Koko can eat all bananas within h hours.