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Koko Eating Bananas

Koko Eating Bananas#

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Learning Objectives#

  • Understand the problem statement and its significance.

  • Develop an intuition for the problem.

  • Identify the assumptions and constraints.

  • Formulate the problem mathematically.

  • Identify the appropriate algorithmic approach.

  • Implement the solution and test cases.

  • Analyze the time and space complexity.

  • Identify the tradeoffs between different approaches (if any).

Introduction#

Problem Statement#

Koko loves to eat bananas. There are N piles of bananas, the n-th pile has piles piles[n] 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.

Theoretical Foundations and Practical Implications#

The Koko Eating Bananas problem serves as a captivating example of optimization problems in computer science. At a high level, it tackles the challenge of resource allocation under time constraints. The narrative may involve a monkey and her bananas, but the core issue is universally relatable: how to perform a set of tasks in the most efficient manner possible given limited time?

If we peel back the layers of this seemingly whimsical problem, we find a scenario not unlike those encountered in areas like network bandwidth optimization, job scheduling in multi-core processors, or even time-management strategies in day-to-day activities. The bananas can be thought of as data packets needing to be processed, and Koko’s eating speed is akin to a processor’s clock speed.

For readers interested in broader contexts, this problem can be viewed as a specific case within combinatorial optimization and decision theory. Both are fields that provide general frameworks for making optimal choices from a finite set of possibilities.

Solving this problem doesn’t just offer a method for optimizing a peculiar banana-eating scenario. It provides a framework for tackling a wide range of real-world optimization challenges. Thus, understanding its solution has both theoretical and practical significance.

Analogy: Job Scheduling in Data Centers#

Imagine a data center where multiple tasks (akin to the piles of bananas piles) need to be processed by a single server (akin to Koko). The task of the data center manager (akin to the algorithm minEatingSpeed you’re trying to design) is to determine the minimum processing power (speed k) each server must have to complete all tasks within a prescribed time frame (hours h).

Breaking it Down#

  1. Tasks in Data Center: These are equivalent to the piles of bananas. Each task could require different amounts of processing, just as piles can have different numbers of bananas.

  2. Processing Power of the Server: This is akin to the speed \(k\) at which Koko eats bananas. The faster the processing power, the quicker the tasks get done.

  3. Time Frame: Just as Koko has a deadline \(h\), the data center has a service level agreement (SLA) specifying the maximum time in which all tasks must be completed.

  4. Optimization: The challenge lies in finding the minimum server speed that will allow all tasks to be completed within the time stipulated by the SLA, much like finding the minimum \(k\) in our problem.

This is a real-world issue with significant implications. Inefficient job scheduling can lead to increased electricity costs, lower throughput, and ultimately, less satisfied clients. Thus, solving this problem efficiently has both theoretical and practical significance (as mentioned earlier).

By understanding the Koko Eating Bananas problem, you gain insights into how you might approach job scheduling optimization in a data center, a concept that has far-reaching applications in the world of cloud computing and distributed systems.

Problem Intuition#

In any computational or mathematical endeavor, intuition serves as the compass that guides the journey from problem formulation to solution. While assumptions lay the groundwork and constraints define the limits, intuition enlightens us on the why behind a problem. It acts as the mental model that helps us navigate the complexities, making the abstract concrete and the intractable approachable.

Understanding the intuition behind a problem is akin to grasping the motivations behind a story—it provides context, illuminates significance, and reveals underlying patterns. It shapes our strategic approach, helping us decide which algorithms to employ or what data structures would be most effective.

In essence, intuition corresponds to the why that breathes life into the how and what of problem-solving. Just as a seasoned chess player instinctively knows which moves lead to victory, a nuanced understanding of problem intuition equips us with the foresight to make informed decisions and optimize solutions.

A Naive Approach#

The straightforward method to solve this problem is to initialize \(k = 1\) and incrementally test each value of \(k\) to see if Koko can finish all the bananas within \(h\) hours.

Why?

This makes sense because we are interested in finding the minimum \(k\) that allows Koko to meet the time constraint. The algorithm proceeds by iteratively increasing \(k\) and stops at the first \(k\) that allows Koko to consume all the bananas in \(h\) or fewer hours. This “first \(k\)” will be the minimum \(k\) satisfying the given condition.

This naive approach has a time complexity of \(\mathcal{O}(k_{\text{min}} \times N)\), where \(k_{\text{min}}\) is the smallest speed at which Koko can consume all the bananas within \(h\) hours, and \(N\) is the number of piles. Since \(k_{\text{min}}\) is not known in advance and could be large, this approach can be computationally expensive for large datasets or tight time constraints. Therefore, the crux of the problem is to find this optimal \(k\) without resorting to such a linear search through all potential speeds.

Reducing the Search Space#

The observant reader should notice that actually the \(k_{\text{min}}\) is upper bounded by the maximum number of bananas in a pile:

\[ k_{\text{min}} \leq M := \max(\text{piles}) \]

Why?

Because we know from the constraints that piles.length <= h. This is an important observation because it allows us to reduce the search space from \([1, \infty)\) to \([1, M]\). We will always yield a solution if Koko is allowed to eat at a speed of \(M\) bananas per hour. She will always be able to finish all the bananas in \(h\) hours or fewer. Therefore, we can discard all speeds greater than \(M\). So our time complexity for the naive approach becomes \(\mathcal{O}(M \times N)\) since we will at most iterate through \(M\) speeds.

We will show later that we can further optimize this time complexity to \(\mathcal{O}(M \log N)\) via a binary search approach.

Example#

In this section, we go through some small example(s) to illustrate the problem.

Example 1: Iterative Approach#

Input :
   piles = [3,6,7,11],
   h     = 8
Output:    4

The question at hand is: What is the minimum speed at which Koko can eat all the bananas in the piles within h hours?

Let’s look at each pile individually and calculate the hours it would take for Koko to finish each pile at different speeds:

  • At speed 1: Koko would take 3, 6, 7, and 11 hours respectively for each pile, which sums up to 27 hours in total. This is more than 8 hours, so the speed is too slow.

  • At speed 2: Koko would take 2, 3, 4, and 6 hours respectively for each pile, which sums up to 15 hours in total. This is still more than 8 hours, so the speed is still too slow.

  • At speed 3: Koko would take 1, 2, 3, and 4 hours respectively for each pile, which sums up to 10 hours in total. This is still more than 8 hours, so the speed is still too slow.

  • At speed 4: Koko would take 1, 2, 2, and 3 hours respectively for each pile, which sums up to 8 hours in total. This is exactly the number of hours available, so the speed is just right.

Hence, the minimum speed at which Koko can eat all the bananas in the piles within h = 8 hours is 4.

Example 2: Pigeonhole Principle#

Input :
   piles = [30,11,23,4,20],
   h     = 5
Output:    30

Again, we are looking for the minimum speed at which Koko can eat all the bananas in the piles within h hours.

Given that there are only 5 hours available and one of the piles itself has 30 bananas, the minimum speed at which Koko must eat to finish just that pile in time is 30 bananas per hour. Since no pile has more than 30 bananas, a speed of 30 bananas per hour will also suffice for all other piles.

Why?

In the scenario described, Koko has 5 hours to eat all the bananas from 5 piles, and the largest pile contains 30 bananas. The pigeonhole principle in this context dictates that each hour must be allocated to a single pile, as there are exactly as many piles as there are hours.

Since Koko can only eat from one pile each hour and one of those piles has 30 bananas, she must eat at a rate of at least 30 bananas per hour to finish just that pile in time. If her eating speed were any slower, she would not be able to finish the largest pile within the allotted hour.

Moreover, because no pile has more than 30 bananas, a speed of 30 bananas per hour is sufficient for Koko to eat any of the piles within an hour. Therefore, 30 bananas per hour is both a necessary and sufficient speed for Koko to eat all the bananas from all the piles within the 5 hours.

Assumptions and Constraints#

The relationship between assumptions and constraints in a research or problem-solving context is akin to the interplay between axioms and theorems in a mathematical framework. Assumptions serve as the foundational elements upon which the rest of the work is constructed. They provide the initial conditions or premises that guide the problem-solving approach. On the other hand, constraints are often the derived limitations or boundary conditions that naturally arise from these assumptions.

Remark 13 (Externally Derived Constraints)

It’s important to understand that constraints can also be externally imposed or arise from specific practical considerations, rather than being directly derived from the problem’s assumptions. In such cases, these constraints serve as additional rules or limitations that must be adhered to when seeking a solution.

Assumptions#

Assumptions set the stage for problem-solving by defining the initial conditions, parameters, or rules that are accepted without direct evidence. They simplify complex situations, making them more manageable and tractable for analysis or computational modeling.

In the Koko Eating Bananas problem, we make the following assumptions:

  1. Constant Eating Rate: Koko can only eat a constant number of bananas per hour (k), from a single pile.

  2. One Pile at a Time: Koko will start a new pile only after finishing the current one.

  3. Quantized Eating Periods: If a pile has fewer bananas than k, Koko will take less than an hour to finish that pile but won’t start another pile within the same hour. This implicitly implies that Koko will not eat any more bananas from other piles in the same hour even if she finishes eating the current pile in less than an hour.

  4. Existence of Solution: We assume there exists a solution to the problem. In other words, we assume that there exists a speed k such that Koko can eat all the bananas in h hours or fewer.

Constraints#

These assumptions lead to certain constraints (in no particular order):

  1. Minimum Speed: Koko can’t have a speed of zero; she must eat at least one banana per hour.

  2. Time Constraint: Koko has only h hours, a hard deadline to finish eating all bananas.

  3. Integer Hours: Time is quantized in hours. Even if Koko takes less than an hour to finish a pile, she can’t start another one within the same hour.

  4. Existence of Solution: The number of piles is less than or equal to the number of hours. This is because if the number of piles is greater than the number of hours, then it is impossible for Koko to eat all the bananas in h hours or fewer.

In addition, leetcode provides the following constraints:

  • \(1 \leq \text{piles.length} \leq 10^4\)

  • \(\text{piles.length} \leq h \leq 10^9\)

  • \(1 \leq \text{piles[i]} \leq 10^9\)

The additional constraints from LeetCode serve the following purposes:

  1. Computational Feasibility: Limiting the array length and the value of h ensures that the problem can be solved within reasonable computational time, given the algorithmic techniques that candidates are expected to use.

  2. Problem Scope: By setting minimum and maximum values, they define the problem’s scope more precisely, thereby allowing for standardized assessment of solutions.

  3. Avoiding Edge Cases: Constraints like \(1 \leq \text{piles[i]} \leq 10^9\) help in removing trivial or degenerate cases, focusing the problem on meaningful scenarios.

  4. Algorithmic Complexity: The constraints provide bounds that can guide the choice of algorithm, helping one discern whether an \(\mathcal{O}(N \log N)\) or \(\mathcal{O}(N)\) algorithm is appropriate, for example.

Test Cases#

Standard Cases#

  1. Multiple Piles, Limited Time

    • Input: piles = [30, 11, 23, 4, 20], h = 6

    • Output: 23

    • Explanation: Koko needs a minimum speed of 23 bananas/hour to finish all bananas in 6 hours or fewer.

Edge Cases#

  1. Single Pile and Minimum Possible Input

    • Input: piles = [1], h = 1

    • Output: 1

    • Explanation: This is the simplest case where the pile has the minimum number of bananas and Koko has the least amount of time. The speed is also 1 banana/hour.

  2. Highly Skewed Piles

    • Input: piles = [100, 1, 1, 1], h = 4

    • Output: 100

    • Explanation: The large pile dictates Koko’s minimum eating speed. Koko needs to eat at 100 bananas/hour to finish all bananas in 4 hours.

  3. Large Data Set

    • Input: piles = [8]*10^6, h = 10^6

    • Output: 8

    • Explanation: This tests the algorithm’s ability to handle large-scale scenarios efficiently. Koko would need to eat at least 8 bananas/hour.

Theoretical Best Time/Space Complexity and Space-Time Tradeoff#

This section presents a conundrum. Before looking at the algorithmic method, it’s often recommended to have a rough intuition on the optimal time and space complexities and their inherent tradeoffs. However, having this foundational knowledge upfront can guide us away from futile attempts at chasing an elusive optimal strategy. It essentially sets a lower boundary for our optimization efforts.

Theoretical Best Time Complexity#

In the “Koko Eating Bananas” problem, the goal is to minimize the eating speed \(k\) such that all bananas are eaten within \(h\) hours. One common algorithmic approach to solve this problem is using binary search on \(k\).

The binary search would operate on the speed range \([1, M]\), and for each candidate speed, we need to traverse all the piles to check if Koko can finish eating in \(h\) hours. This traversal takes \(\mathcal{O}(N)\) time where \(N\) is the length of the piles array. Thus, the best theoretical time complexity for solving this problem would be \(\mathcal{O}(N \log M)\), where \(M\) is the maximum number of bananas in a pile.

Theoretical Best Space Complexity#

For the binary search algorithm, we only need a constant amount of extra space to store variables such as the low, high, and mid points of the search, as well as a counter for the total hours Koko would take for a given \(k\). Therefore, the space complexity is \(\mathcal{O}(1)\), which is the best you can achieve for this problem assuming that the input size is not counted towards the space complexity.

Space-Time Tradeoff#

In this specific problem, there’s limited scope for a space-time tradeoff. The time complexity is primarily determined by the need to iterate over all the piles for each candidate \(k\), and this is not something that can be pre-computed or stored to save time later. Similarly, the space complexity is already at its theoretical minimum \(\mathcal{O}(1)\), so there isn’t room for optimization by using more space.

To sum up, this problem doesn’t offer much room for a space-time tradeoff, given its constraints and the nature of its optimal solution.

Mathematical Formulation#

First, we will introduce some mathematical notations and definitions to help us formulate the problem and solution in a more concise manner.

Given a sequence of piles,

\[ \mathcal{P} = \left\{ p_1, p_2, \ldots, p_N \,|\, 1 \leq n \leq N \right\}, \]

each containing a non-negative integer number of bananas, and a positive integer \(h\) representing the total number of hours available, our goal is to find the minimum constant integer eating speed \(k\) such that Koko can finish all the bananas in \(\mathcal{P}\) within \(h\) hours.

For reasons meantioned earlier and in the constraints, we can specify the search space for \(k\) as:

\[ \mathcal{K} = \left\{ k \in \mathbb{Z}^+ \,|\, 1 \leq k \leq \max_{n \in [1,N]} p_n \right\}. \]

For convenience, we also define a shorthand notation for the maximum number of bananas in a pile:

\[ M \stackrel{\text{def}}{=} \max_{n \in [1,N]} p_n \]

In this setting, we define the time \(\mathcal{T}(k, p_n)\) it takes Koko to eat a pile \(p_n\) at speed \(k\) as:

\[ \mathcal{T}(k, p_n) = \left \lceil \frac{p_n}{k} \right \rceil \]

where \(\lceil \cdot \rceil\) is the ceiling function, which rounds a real number \(x\) up to the nearest integer. For example, \(\lceil 2.3 \rceil = 3\).

Remark 14 (Why ceiling?)

Because Koko can only eat a constant number of bananas per hour, so she will always take at least \(\lceil \frac{p_n}{k} \rceil\) hours to finish a pile \(p_n\). Consider \(k=3\) and the pile \(p_n=5\). Then Koko will take \(2\) hours to finish this pile. This is because \(\frac{5}{3} = 1.6667\), which rounds up to \(2\) (of course you cannot round down).

Consequently, the total time \(\mathcal{H}(k, \mathcal{P})\) required to eat all the bananas in \(\mathcal{P}\) at speed \(k\) can be expressed as:

\[ \mathcal{H}\left(k, \mathcal{P}\right) = \sum_{n=1}^{N} \mathcal{T}(k, p_n) \]

The optimization problem can thus be formally stated as:

\[\begin{split} \begin{aligned} & \text{minimize} && k \in \mathcal{K} \\ & \text{s.t.} && \mathcal{H}\left(k, \mathcal{P}\right) \leq h \\ & \text{where} && k, h \in \mathbb{Z}^+, \; \mathcal{K} \subset \mathbb{Z}^+, \; 1 \leq k \leq M \end{aligned} \end{split}\]

or equivalently:

\[\begin{split} \begin{aligned} & k^* = \arg\min_{k \in \mathcal{K}} k \\ & \text{s.t.} \quad & \mathcal{H}\left(k, \mathcal{P}\right) \leq h \\ & \text{where} \quad & k, h \in \mathbb{Z}^+, \; \mathcal{K} \subset \mathbb{Z}^+ \end{aligned} \end{split}\]

Definitions#

The definitions of this problem are fairly straightforward and self-contained. The previous section already introduced most of the necessary definitions.

  • \(p_n\) is the number of bananas in the \(n\)th pile

  • \(k\) is the speed at which Koko eats bananas

  • \(h\) is the number of hours Koko has to eat all the bananas

  • \(N\) is the number of piles

  • \(\mathcal{T}(k, p_n) = \left\lceil \frac{p_n}{k} \right\rceil\) is the number of hours it takes Koko to eat the \(n\)th pile

  • \(\mathcal{H}(k, \mathcal{P}) = \sum_{n=1}^{N} \left\lceil \frac{p_n}{k} \right\rceil\) is the total number of hours it takes Koko to eat all the bananas.

  • \(M\) is the maximum number of bananas in a pile \(\max_{n \in [1,N]} p_n\). This is the upper bound of the search space for \(k\).

We include two more definitions here for completeness since we will use it later.

  • Feasibility Function \(\mathcal{F}(\mathcal{P}, k, h)\): A binary function indicating whether it is possible to eat all bananas within \(h\) hours at speed \(k\).

    \[\begin{split} \mathcal{F}(k, \mathcal{P}, h) = \begin{cases} 1, & \text{if } \mathcal{H}\left(k, \mathcal{P}\right) \leq h \\ 0, & \text{otherwise} \end{cases} \end{split}\]

  • Optimal Eating Speed \(k^*\): The minimum constant integer eating speed that satisfies the problem’s constraints.

    \[ k^* = \min \{ k \in \mathcal{K} \,|\, \mathcal{H}\left(k, \mathcal{P}\right) \leq h \} \]

References and Further Readings#

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