Koko Eating Bananas LeetCode Solution 875 [Python]

Koko Eating Bananas LeetCode solution to the problem of determining the minimum eating speed to consume all the bananas within a time frame

Maybe Koko loves bananas, and in this problem, we are presented with the task of helping Koko determine the minimum integer eating speed, denoted as 'k,' to consume all the bananas within a given time frame, 'h' hours.

Koko Eating Bananas Problem Overview

We are provided with an array 'piles' containing ‘n’ piles of bananas, each represented by an integer value indicating the number of bananas in that pile.

The guards have left and will return in ‘h’ hours.

Koko has the freedom to choose her eating speed (‘k’) per hour. During each hour, she can select a pile of bananas and eat ‘k’ bananas from it.

If the pile has fewer than ‘k’ bananas, she will consume them all during that hour and won’t consume any more in that hour.

Koko’s goal is to eat all the bananas before the guards return.

Our task is to find the minimum ‘k’ that allows Koko to achieve this.

Understanding Constraints

Before diving into the solution, it’s important to note the constraints:

1 <= piles.length <= 10^4
piles.length <= h <= 10^9
1 <= piles[i] <= 10^9

These constraints give us some important insights:

The number of piles will always be less than or equal to the number of hours available. This means it’s possible to eat all the bananas within the time frame.
The minimum possible value for ‘h’ is the length of the ‘piles’ array since each pile can be consumed in one hour, and the maximum ‘h’ can be as large as 10^9.

Bruteforce Approach to Solve Koko Eating Bananas

Let’s start with a straightforward brute-force approach to tackle this problem.

We will try different values of ‘k’ starting from 1, and for each ‘k,’ we will calculate the time required to consume all the bananas.

If the total time is less than or equal to ‘h,’ we will update the result with the minimum ‘k’ found.

In this approach, we iterate through possible ‘k’ values using a binary search approach. We calculate the total time needed for each ‘k’ and check if it is less than or equal to ‘h’. If it is, we update our result and continue searching for a smaller ‘k’. If the total time exceeds ‘h,’ we increase ‘k’ to find a rate that allows Koko to eat within the time limit.

Efficient Solution

The brute force approach helps us arrive at the solution. However, it’s clear that binary search is a more efficient way to find the minimum ‘k.’ By applying binary search, we significantly reduce the number of iterations.

Python Code Solution

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def minEatingSpeed(self, piles: List[int], h: int) -> int:
    left, right = 1, max(piles)
    result = right

    while left &lt;&equals; right:
        k = (left + right) // 2
        total_time = 0
        
        for p in piles:
            total_time += math.ceil(float(p) / k)
        
        if total_time &lt;&equals; h:
            result = k
            right = k - 1
        else:
            left = k + 1
    
    return result

How The Code Works

left and right are initially set to 1 and the maximum number of bananas in the piles, respectively. These values define the search space for the eating speed. result is initialized with the maximum possible eating speed.
The code enters a binary search loop with while left <= right. In each iteration, it calculates a mid-point value k as (left + right) // 2. This k represents the current guess for Koko’s eating speed.
A loop iterates through all the piles, calculating the time it would take Koko to eat each pile at speed k. math.ceil(float(p) / k) calculates the time required to eat a pile of p bananas at the current speed k, ensuring Koko finishes completely.
The total_time accumulates the time required to eat all piles at the current speed k.
After calculating total_time, the code checks whether Koko can eat all the bananas within the given time limit h. If total_time is less than or equal to h, it means Koko can eat the bananas within the time limit, so the result is updated to k.
If total_time is greater than h, it means Koko cannot eat all the bananas in time, so the right bound of the search space is updated to k - 1 to narrow the search space to the left.
If total_time is less than or equal to h, the search space is narrowed to the left by updating left to k + 1, trying to find a smaller eating speed that still satisfies the time limit.
The binary search continues until left exceeds right, at which point the minimum eating speed that allows Koko to eat all the bananas within the time limit is found, and it is stored in result.
The function returns result, which is the minimum eating speed for Koko to finish eating the bananas within h hours.

The use of math.ceil and converting p to float before division serves specific purposes in this code:

The math.ceil function is used to ensure that the time taken to eat each pile of bananas is rounded up to the nearest integer. This is important because Koko can’t eat a fraction of a banana. If the division of p by k results in a non-integer value (meaning there are leftover bananas that cannot be consumed in a fractional amount of time), math.ceil rounds it up to the next whole unit of time, ensuring that Koko eats the entire pile. This is a necessary step to calculate the total time accurately.
Converting p to float, in Python, integer division (division between two integers) results in an integer quotient without the fractional part. By converting p to float before dividing by k, it ensures that the division operation results in a floating-point number, allowing for accurate time calculation. If p and k were both integers, integer division could round down the result and underestimate the time required to eat a pile, potentially leading to an incorrect total time calculation.

Time and Space Complexity

The time complexity of this solution is O(n * log(max(p))), where ‘n’ is the number of piles and ‘max(p)’ is the maximum value in the ‘piles’ array.

Binary search reduces the number of iterations, making it efficient.

The space complexity is O(1) since we only use a few variables to store results.

Edge Cases a Valid Solution Must Consider

Minimum and maximum values of ‘piles’: Ensure your solution works for small and large pile values.
Minimum and maximum values of ‘h’: Test the lower and upper limits of ‘h.’
Piles with the same number of bananas: Check if the code handles equal pile sizes correctly.
Piles with varying sizes: Verify that your code correctly processes piles with varying banana quantities.

Conclusion

In summary, the efficient binary search approach provides a fast and accurate solution to the Koko Eating Bananas problem.

It allows Koko to determine the minimum eating speed required to consume all the bananas within the given time frame. The solution considers the constraints and edge cases, making it a reliable choice for this problem.

For more details and to try the code, you can refer to the LeetCode problem.

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