Koko Eating Bananas Problem Solution

Here's how we can solve the Koko eating bananas problem efficiently.

1. Naive Approach (Brute Force)

A simple, but inefficient, way to solve this problem is to iterate through all possible values of k (bananas per hour) starting from 1, and check if Koko can eat all the bananas within h hours for each k. The first k that satisfies the condition is the minimum k we are looking for.

Algorithm:

1. Start with k = 1.
2. For each k, calculate the total hours needed to eat all bananas.
3. If the total hours are less than or equal to h, return k.
4. Increment k and repeat steps 2 and 3.

Code (Python):

def can_eat_all(piles, h, k):
    hours = 0
    for pile in piles:
        hours += (pile + k - 1) // k  # Ceil division
    return hours <= h

def min_eating_speed_brute_force(piles, h):
    k = 1
    while True:
        if can_eat_all(piles, h, k):
            return k
        k += 1

Big O Analysis:

• Time Complexity: O(K * N), where K is the minimum eating speed and N is the number of piles. In the worst case, K can be as large as the largest pile.
• Space Complexity: O(1)

2. Optimal Approach (Binary Search)

We can use binary search to find the optimal value of k. The minimum possible value of k is 1, and the maximum possible value of k is the largest pile size. We can perform a binary search within this range to find the minimum k that satisfies the condition.

Algorithm:

1. Initialize left = 1 and right = max(piles).
2. While left < right:
   • Calculate mid = left + (right - left) // 2 (to prevent overflow).
   • Calculate the total hours needed to eat all bananas with speed mid.
   • If the total hours are less than or equal to h, it means mid might be a valid answer, so we try to reduce k further by setting right = mid.
   • Otherwise, if the total hours are greater than h, it means mid is too small, so we increase k by setting left = mid + 1.
3. Return left (or right, since left == right at the end).

Code (Python):

def can_eat_all(piles, h, k):
    hours = 0
    for pile in piles:
        hours += (pile + k - 1) // k  # Ceil division
    return hours <= h

def min_eating_speed(piles, h):
    left, right = 1, max(piles)
    while left < right:
        mid = left + (right - left) // 2
        if can_eat_all(piles, h, mid):
            right = mid
        else:
            left = mid + 1
    return left

Big O Analysis:

• Time Complexity: O(N * log(M)), where N is the number of piles and M is the maximum value in piles. This is because we perform a binary search between 1 and the maximum pile size, and for each mid value, we iterate through all the piles to calculate the total hours.
• Space Complexity: O(1)

3. Edge Cases

• piles is empty: The problem statement says 1 <= piles.length <= 10^4, so this case is not possible.
• h is very small (close to the number of piles): This forces k to be large. The binary search handles this case efficiently.
• h is very large: This allows k to be small. The binary search also handles this case without issues.
• Overflow in calculating hours: The intermediate hours value could become large, but since we are only comparing it with h (which is at most 10^9), we do not need to worry about integer overflow issues assuming we are using a 64 bit integer.