Koko Eating Bananas#
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#
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.
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.
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.
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:
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.
(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:
Constant Eating Rate: Koko can only eat a constant number of bananas per hour (
k
), from a single pile.One Pile at a Time: Koko will start a new pile only after finishing the current one.
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.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 inh
hours or fewer.
Constraints#
These assumptions lead to certain constraints (in no particular order):
Minimum Speed: Koko can’t have a speed of zero; she must eat at least one banana per hour.
Time Constraint: Koko has only
h
hours, a hard deadline to finish eating all bananas.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.
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:
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.Problem Scope: By setting minimum and maximum values, they define the problem’s scope more precisely, thereby allowing for standardized assessment of solutions.
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.
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#
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#
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.
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.
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,
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:
For convenience, we also define a shorthand notation for the maximum number of bananas in a pile:
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:
where \(\lceil \cdot \rceil\) is the ceiling function, which rounds a real number \(x\) up to the nearest integer. For example, \(\lceil 2.3 \rceil = 3\).
(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:
The optimization problem can thus be formally stated as:
or equivalently:
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.
\[ k^* = \min \{ k \in \mathcal{K} \,|\, \mathcal{H}\left(k, \mathcal{P}\right) \leq h \} \]
Solution: Binary Search#
In solving this problem, the objective is to efficiently identify the minimum eating speed \(k\) that allows Koko to consume all the bananas in \(h\) hours. A straightforward method is to iterate over all potential speeds \(k\) in the range \(1 \leq k \leq M\)[1] and evaluate if \(\mathcal{H}\left(k, \mathcal{P}\right) \leq h\), where \(\mathcal{H}\left(k, \mathcal{P}\right)\) represents the hours needed to eat all bananas at speed \(k\). This naive approach results in a time complexity of \(\mathcal{O}(N \times M)\), which could be computationally prohibitive when \(N\), the number of piles, and/or \(M\), the maximum size of a pile, is large. However, we can improve this to \(\mathcal{O}(N \log M)\) by employing a binary search technique.
Solution Intuition#
The first question that arises is, how can we be sure that binary search is an appropriate technique for solving this problem? The answer often depends on one’s level of experience. However, beyond experience, there is an underlying intuition that experienced programmers have developed. This intuition allows them to instantly recognize problems that are well-suited for binary search. So, what constitutes this intuition, and how can it be developed? In general, if a specific type of algorithm adheres to a recognizable pattern or template, then that pattern serves as the intuition behind choosing that algorithm.
Main intuition following the binary search framework.
Minimize \(k\) such that koko can eat all the bananas within \(h\) hours.
The Precondition for Binary Search#
The main precondition for applying binary search to this problem is the property
of monotonicity in the feasibility
function[2]. If a feasibility function
\(\mathcal{F}(\cdot)\) is monotonic, then the problem lends itself well to a
binary search solution. Here, the feasible function \(\mathcal{F}(\cdot)\) is a
boolean function that returns either True
or False
, or 1
or 0
since we
want to talk about monotonicity, which is better understood if the output is a
“number”.
The second precondition is implicit. The search space \(\mathcal{K}\) must be inherently ordered in which you can sort it (sortable). This is because binary search requires the search space to be sorted.
Consequently, the domain of the feasibility function \(\mathcal{F}(\cdot)\) is the search space \(\mathcal{K}\), and the range is \(\{0, 1\}\).
But why is having this monotonicity property important? Why is it that we need this feasibility function to be monotonic? The reason is that we want to be able to discard half of the search space at each step. This is the key to binary search. The connection is that monotonicity implies there is a tipping point. There exists a \(k\) such that if \(\mathcal{F}(k) = 1\), then \(\mathcal{F}(k') = 1\) for all \(k' > k\).
Understanding the problem through the lens of feasibility and monotonicity lends itself to the concept of a sorted boolean array[3] where we can easily apply binary search!
The only difficulty lies in defining the feasibility function \(\mathcal{F}(\cdot)\).
Framing the Problem as a Sorted Boolean Array#
Feasibility Function#
First, we define the feasibility function formally.
(Feasibility Function)
The feasibility function \(\mathcal{F}\) is a mapping from a decision variable \(k \in \mathcal{K}\) and a set of parameters \(\Theta\) to a binary outcome, subject to a set of constraints \(C\).
Here, \(\mathcal{K}\) is the domain of possible values for the decision variable, \(\Theta\) represents other relevant parameters, and \(C\) represents the set of constraints.
The function is defined as:
Here, \(C(k, \Theta)\) is a function that returns True
(\(1\)) or False
(\(0\))
based on whether all constraints are satisfied for the given \(k\) and \(\Theta\).
For our problem, the decision variable is \(k\), the set of parameters is \(\Theta = \{ \mathcal{P}, h \}\), and the constraints is just whether Koko can finish eating all the bananas in \(h\) hours or fewer.
(Feasibility Function for Koko Eating Bananas)
The feasibility function \(\mathcal{F}\) for the Koko Eating Bananas problem is defined as:
where \(\mathcal{H}\left(k, \mathcal{P}\right)\) is the total number of hours it takes Koko to eat all the bananas at speed \(k\).
This feasibility function looks reasonable, we remain to verify if this function is monotonic on the search space \(\mathcal{K}\).
Monotonicity#
Here we first define the concept of monotonicity formally.
(Monotonicity)
In the context of a binary search problem, if the feasibility function \(\mathcal{F}: \mathcal{K} \to \{0, 1\}\) is monotonic over the search space \(\mathcal{K}\), it would satisfy one of the following conditions:
For all \(k_1, k_2 \in \mathcal{K}\) such that \(k_1 \leq k_2\), it holds that \(\mathcal{F}(k_1) \leq \mathcal{F}(k_2)\) (Monotonically Increasing).
For all \(k_1, k_2 \in \mathcal{K}\) such that \(k_1 \leq k_2\), it holds that \(\mathcal{F}(k_1) \geq \mathcal{F}(k_2)\) (Monotonically Decreasing).
We claim that the feasibility function \(\mathcal{F}(k, \mathcal{P}, h)\) is monotonically increasing. If Koko can eat all bananas in \(h\) hours at a speed \(k_1\), then she can surely do so at any greater speed \(k_2 > k_1\). Note that the total number of hours to finish, \(\mathcal{H}\left(k, \mathcal{P}\right)\), however, is monotonically decreasing with respect to \(k\) because if Koko eats faster, she will take fewer hours to finish.
More concretely:
The logic
Proof
Proof. We start off by a claim:
Claim: \(\mathcal{F}(k_1, \mathcal{P}, h) \leq \mathcal{F}(k_2, \mathcal{P}, h)\) for all \(k_1 < k_2\).
In other words, if \(\mathcal{F}(k_1, \mathcal{P}, h) = 1\), then \(\mathcal{F}(k_2, \mathcal{P}, h) = 1\) for all \(k_2 > k_1\).
Proof:
Given that \(\mathcal{F}(k_1, \mathcal{P}, h) = 1\), we have:
We aim to prove:
Since \(k_1 < k_2\), \(\frac{p_n}{k_1} \geq \frac{p_n}{k_2}\) for all \(n\).
This implies that \(\left \lceil \frac{p_n}{k_1} \right \rceil \geq \left \lceil \frac{p_n}{k_2} \right \rceil\) for all \(n\).
Hence, if \(\sum_{n=1}^{N} \left \lceil \frac{p_n}{k_1} \right \rceil \leq h\), then \(\sum_{n=1}^{N} \left \lceil \frac{p_n}{k_2} \right \rceil \leq h\).
This concludes the proof that \(\mathcal{F}(k, \mathcal{P}, h)\) is monotonically increasing with respect to \(k\).
Translating to a First True in a Sorted Boolean Array Problem#
Given the monotonic property of \(\mathcal{F}\), we can effectively sort the
boolean outputs of \(\mathcal{F}\) across the search space \(\mathcal{K}\). That is,
as we traverse \(\mathcal{K}\) from the slowest to the fastest eating speed, the
function \(\mathcal{F}\) will transition from returning False
to returning
True
. Our objective is to find the smallest \(k\) for which \(\mathcal{F}\)
returns True
. Essentially, we have transformed the problem into a first
true in a sorted boolean array problem.
To rigorously connect these ideas, consider the feasibility function
\(\mathcal{F}\) as a sequence
\(\mathcal{F} = \{ \mathcal{F}(\mathcal{P}, k_1, h), \mathcal{F}(\mathcal{P}, k_2, h), \ldots \}\)
where \(k_1 < k_2 < \ldots\) are the sorted eating speeds in \(\mathcal{K}\). The
sequence \(\mathcal{F}\) is essentially a “Sorted Boolean Array”. Given the
monotonic property of \(\mathcal{F}\), \(\mathcal{F}\) is a sorted boolean sequence
with a threshold point where it transitions from False
to True
. Finding this
point is equivalent to solving the original problem, and binary search is the
efficient way to do it.
Thus, we can see that the Koko Eating Bananas problem is fundamentally a “First True in a Sorted Boolean Array” problem, revealed through the lens of feasibility and its monotonic properties:
Ordered Search Space: The space of all possible eating speeds \(k\) is well-defined and ordered (from 1 to \(M\)).
Monotonicity of Feasibility: The problem exhibits a monotonic property with respect to feasibility; if Koko can finish all the bananas in \(h\) hours at speed \(k\), she can also finish them at any speed greater than \( k\).
First True in Sorted Boolean Array: With the above properties, one can treat the search space \(\mathcal{K}\) as a sorted boolean array defined by the feasibility function \(\mathcal{F}\). The problem reduces to finding the “first true” in this sorted boolean array.
To this end, we can reframe the earlier minimization statement:
to
Visualization#
Visual representation of the problem and solution (if applicable).
I do not have any visualization, here’s an image of a monkey eating bananas generated by OpenAI’s DALL·E 3 model xd.
Algorithm#
Whiteboarding#
The domain of our search space is defined as \(k \in [1, \max(\text{{piles}})]\):
The lower bound \(k = 1\) is set because Koko cannot eat 0 bananas per hour.
The upper bound \(k = \max(\text{{piles}})\) is chosen based on the constraint that eating at a faster rate than the largest pile is unproductive since Koko can only eat from one pile at a time.
The problem lends itself to a binary search approach due to its inherent
monotonicity. To illustrate, consider an example with piles = [3, 6, 7, 11]
and \(h = 8\). The initial search space spans from \(k = 1\) to \(k = 11\), with a
mid-point at \(k = 6\).
We then evaluate whether Koko can consume all the bananas within \(h = 8\) hours
at a speed of \(k = 6\) bananas per hour. The answer is affirmative, requiring
[1, 1, 2, 2]
hours for the respective piles. Consequently, all speeds
\(k' > k = 6\) must also satisfy this condition. This enables us to discard the
right half of the search space, focusing our search on \(k' \leq k = 6\).
Conversely, if we arrive at an insufficient speed, say \(k = 3\), then all speeds \(k' < k = 3\) will also be insufficient. Therefore, we can discard the left half of the search space, concentrating our search on \(k' \geq k = 3\).
The monotonic property guarantees that if a speed \(k\) enables Koko to consume all bananas within \(h\) hours, all speeds \(k' > k\) will also satisfy this condition. Hence, the binary search can be efficiently employed to halve the search space iteratively until the minimum \(k\) that meets the requirement is found.
Pseudocode#
(Binary Search Pseudocode)
Algorithm: min_eating_speed(P, h, M)
Input: P = [p_1, p_2, ..., p_N] (list of piles),
h (total hours available),
M (maximum number of bananas in a pile)
Output: k* (optimal eating speed)
1: Initialize l = 1
2: Initialize r = M
3: while l < r do
4: m = floor((l + r) / 2)
5: if feasible(P, m, h) == 1 then
6: r = m
7: else
8: l = m + 1
9: end if
10: end while
11: return l
Mathematical Representation#
Given the search space \(\mathcal{K}\) and the feasibility function \(\mathcal{F}(k, \mathcal{P}, h)\), we seek to find the optimal eating speed \(k^*\) by exploring \(\mathcal{K}\) via binary search.
We employ binary search in \(\mathcal{K}\), the search space defined as:
The algorithm initiates with the lower bound as \(l = 1\) and the upper bound as \(r = M\), and iteratively updates these bounds until \(l \geq r\). Let \(m_t\) be the candidate speed at iteration \(t\) calculated as \(m_t = \left\lfloor \frac{l_t + r_t}{2} \right\rfloor\).
The algorithm uses these bounds to iteratively tighten the search space.
Initialization:
\(l = 1\)
\(r = M\)
Iterative Procedure: For \(k = m_t\):
Compute \(\mathcal{F}(\mathcal{P}, k, h)\).
If \(\mathcal{F}(\mathcal{P}, k, h) = 1\), then \(r_{t+1} = m_t\).
Otherwise, \(l_{t+1} = m_t + 1\).
Termination: The algorithm terminates when \(l_{t+1} \geq r_{t+1}\). The optimal eating speed \(k^* = l\).
Correctness#
Proving the correctness of a binary search algorithm typically involves demonstrating two properties: invariance and termination.
Proof. We prove both properties here.
Invariant: At each iteration of the binary search, we maintain the invariant that the optimal eating speed \(k^*\) must lie within the interval \([l, r]\). This is because we only adjust \(l\) or \(r\) based on the feasibility function, which correctly tells us whether a certain speed is too slow (and thus, the optimal speed must be higher) or feasible (and thus, the optimal speed could be lower or equal).
Initialization: At the start, \(l = 1\) and \(r = M\). Clearly, \(k^*\) must lie within this range because it’s defined to be a positive integer that’s no greater than \(M\).
Maintenance: At each step, we calculate \(m = \left\lfloor \frac{l + r}{2} \right\rfloor\). Then, based on \(\mathcal{F}(\mathcal{P}, m, h)\), we either set \(r = m\) or \(l = m + 1\).
If \(\mathcal{F}(\mathcal{P}, m, h) = 1\), meaning that eating speed \(m\) is feasible, we set \(r = m\). This is correct because if \(m\) is feasible, then any speed greater than \(m\) might also be feasible, but the optimal speed (the smallest feasible speed) must be less than or equal to \(m\). Hence, we adjust the interval to \([l, m]\).
If \(\mathcal{F}(\mathcal{P}, m, h) = 0\), meaning that eating speed \(m\) is not feasible, we set \(l = m + 1\). This is correct because if \(m\) is not feasible, then any speed less than \(m\) is also not feasible. Hence, the optimal speed must be greater than \(m\), and we adjust the interval to \([m + 1, r]\).
By updating \(l\) and \(r\) in this manner, we ensure that the interval \([l, r]\) always contains \(k^*\).
Termination: The algorithm terminates when \(l \geq r\). Due to the properties of the floor function and the update rules for \(l\) and \(r\), this condition will eventually be met. At this point, the interval has been narrowed down to a single point, \(l\), which must be the optimal eating speed \(k^*\). This follows from the invariant; since \(k^*\) was always within the interval, and the interval has been reduced to a single point, that point must be \(k^*\).
This completes the proof of correctness for the binary search algorithm in the context of the Koko eating bananas problem. The invariance ensures that the optimal solution is never excluded from the search space, and the termination condition guarantees that the algorithm will converge to the optimal solution.
Implementation#
1class Solution:
2 def feasible(self, piles: Sequence[int], speed: int, h: int) -> bool:
3 return self.total_hours_to_finish_eating(piles, speed) <= h
4
5 def total_hours_to_finish_eating(self, piles: Sequence[int], speed: int) -> int:
6 total_hours = 0
7 for pile in piles:
8 hours = pile / speed # num_bananas / speed -> hours needed to finish eating
9 hours = math.ceil(hours) # if hours = 2.5, it counts at 3
10 total_hours += hours
11 return total_hours
12
13 def minEatingSpeed(self, piles: Sequence[int], h: int) -> int:
14 l, r = 1, max(piles)
15
16 while l < r:
17 m = l + (r - l) // 2
18 if self.feasible(piles, speed=m, h=h):
19 r = m
20 else:
21 l = m + 1
22
23 return l
It is worth noting to readers that this approach corresponds more closely to the approach 3: find lower bound in Leetcode’s Binary Search editorial.
In binary search and other algorithms, the “lower bound” refers to the smallest
value in a range that meets a specified criterion. This criterion is not
necessarily about being the minimum value in a numerical sense, but rather the
first value in a sorted sequence that satisfies a certain condition. So if we
again look at the problem via the first true in a sorted boolean array lens, the
lower bound is the first True
value in the sorted boolean array.
Tests#
Set of tests for validating the algorithm.
1# Initialize the TestFramework class
2tf = TestFramework()
3
4minimum_speed = Solution().minEatingSpeed
5
6@tf.describe("Testing minimum_speed function for Koko Eating Bananas")
7def test_minimum_speed():
8 @tf.individual_test("Multiple Piles, Limited Time")
9 def _():
10 tf.assert_equals(
11 minimum_speed([30, 11, 23, 4, 20], 6),
12 23,
13 "Should return 23",
14 )
15
16 @tf.individual_test("Single Pile and Minimum Possible Input")
17 def _():
18 tf.assert_equals(minimum_speed([1], 1), 1, "Should return 1")
19
20 @tf.individual_test("Highly Skewed Piles")
21 def _():
22 tf.assert_equals(minimum_speed([100, 1, 1, 1], 4), 100, "Should return 100")
23
24 @tf.individual_test("Large Data Set")
25 def _():
26 tf.assert_equals(minimum_speed([8] * 10**6, 10**6), 8, "Should return 8.")
Time Complexity#
(Time Complexity)
The Big O notation \(\mathcal{O}(g(n))\) represents a set of functions. Specifically, a function \(f(n)\) is in \(\mathcal{O}(g(n))\) if there exist positive constants \(C\) and \(n_0\) such that for all \(n \geq n_0\), the inequality \(f(n) \leq C \cdot g(n)\) holds.
In set notation, \(\mathcal{O}(g(n))\) can be rigorously defined as:
In this definition:
\(f(n)\) is any function that represents the actual running time of an algorithm for an input of size \(n\).
\(g(n)\) is a function that represents the upper bound on the growth rate of \(f(n)\).
\(C\) is a constant multiplier that shows \(f(n)\)’s growth rate is at most as fast as \(C \cdot g(n)\) for sufficiently large \(n\).
\(n_0\) is a threshold beyond which the inequality \(f(n) \leq C \cdot g(n)\) consistently holds.
If an actual time complexity function \(f(n)\) falls within the set defined by \(\mathcal{O}(g(n))\), we say that \(f(n)\) is in \(\mathcal{O}(g(n))\), indicating that the growth rate of \(f(n)\) is asymptotically upper bounded by \(g(n)\) times some constant factor.
To analyze the time complexity of the given code, we will break down each
function and its operations. We will denote \(N\) as the length of the piles
list.
In our context, \(\mathcal{H}\) represents the function that computes the total time taken for Koko to finish eating the bananas given a particular eating speed, while \(\mathcal{J}\) represents the function that determines the minimum eating speed Koko must have to finish the bananas within \(h\) hours.
Let’s redefine \(\mathcal{T}_{\mathcal{H}}(N)\) and \(\mathcal{T}_{\mathcal{J}}(N)\) to represent the time complexities of the functions \(\mathcal{H}\) and \(\mathcal{J}\), respectively.
Time Complexity Function \(\mathcal{T}(N)\): This notation represents the exact time complexity or actual running time of an algorithm as a function of the input size \(N\). It quantifies the precise number of steps or operations an algorithm takes in relation to the size of its input. For example, if an algorithm requires \(3N + 5\) operations for an input of size \(N\), then its time complexity function is expressed as \(\mathcal{T}(N) = 3N + 5\).
Big O Notation \(\mathcal{O}(g(N))\): In contrast, Big O notation is used in asymptotic analysis to describe an upper bound on the growth rate of a function. It is defined as a set of functions that grow no faster than \(g(N)\) times a constant factor for sufficiently large \(N\). It provides a simplified description of the algorithm’s behavior for large inputs, focusing on the dominant term and ignoring lower-order terms and constant factors. For example, if \(\mathcal{T}(N) = 3N + 5\), we would say \(\mathcal{T}(N) \in \mathcal{O}(N)\) because for large \(N\), the term \(3N\) dominates, and the running time grows linearly with \(N\).
To formalize, a function \(f(N)\) is in \(\mathcal{O}(g(N))\) if there exist positive constants \(C\) (the constant factor) and \(N_0\) (the threshold beyond which the bound holds) such that for all \(N \geq N_0\), the inequality \(f(N) \leq C \cdot g(N)\) is satisfied.
Applying this to the example given, if the time complexity function of an algorithm is \(\mathcal{T}(N) = 3N + 5\), we identify \(f(N)\) as \(\mathcal{T}(N)\) and \(g(N)\) as \(N\) in the Big O definition. We can then say that \(\mathcal{T}(N)\) is in \(\mathcal{O}(N)\) (or \(\mathcal{T}(N) \in \mathcal{O}(N)\)) because there exist constants \(C\) (in this case, \(C\) could be \(3\) or any larger number) and \(N_0\) (which could be \(1\) or any positive integer) such that for all \(N \geq N_0\), the running time \(\mathcal{T}(N)\) is bounded above by \(C \cdot N\). This indicates that the algorithm’s running time grows linearly or slower as the input size \(N\) increases.
In essence, \(\mathcal{T}(N)\) gives a precise, often more detailed description of an algorithm’s running time, while \(\mathcal{O}(N)\) provides an asymptotic upper bound that captures the dominant growth behavior for large input sizes.
Function feasible
#
This function simply calls another function total_hours_to_finish_eating
and
compares its output to \(h\). Thus, the time complexity of feasible
is the same
as that of total_hours_to_finish_eating
.
Function total_hours_to_finish_eating
#
Let \(\mathcal{H}\) be the function representing the computation performed by
total_hours_to_finish_eating
. This function iterates over each element of
piles
once and performs a constant number of operations for each pile.
For each pile, there are essentially three operations: division, ceiling, and addition. Let’s assume the division and ceiling together take a constant time of 2 units (since they are constant-time operations), and the addition takes 1 unit of time. This leads to a total of 3 operations per pile.
Therefore, the time complexity \(\mathcal{T}_{\mathcal{H}}(N)\) can be expressed as:
Therefore, the actual time complexity of \(\mathcal{H}\), denoted as \(\mathcal{T}_{\mathcal{H}}(N) \), is linear in terms of the number of piles \(N\), or in an abuse of notation, \(\mathcal{T}_{\mathcal{H}}(N) = \mathcal{O}(N)\).
We can further state formally that \(\mathcal{T}_{\mathcal{H}}(N) \in \mathcal{O}(g(N))\) where
and \(g(N) = N\) is the growth rate of the input size \(N\).
This is also equivalent to saying that
for some suitable choice of \(C\) and \(N_0\).
We can further prove this by finding a suitable \(C\) and \(N_0\) that satisfies the inequality.
Proof that \(\mathcal{T}_{\mathcal{H}}(N) = 3N\) is in \(\mathcal{O}(N)\):
Proof. We note that \(g(N) = N\). We choose \(C = 3\) and \(N_0 = 1\) (since the relation holds for all positive integers \(N\)). For all \(N \geq N_0\):
Hence, we have demonstrated that \(\mathcal{T}_{\mathcal{H}}(N) = 3N\) is bounded
above by \(C \cdot g(N)\), confirming that \(\mathcal{T}_{\mathcal{H}}(N) = O(N)\).
This shows that the time complexity of the total_hours_to_finish_eating
function grows linearly with the number of piles \(N\) and is bounded by a factor
of 3.
Function minEatingSpeed
#
This function uses a binary search to find the minimum feasible speed. The binary search will operate by repeatedly halving the search interval. The maximum possible number of bananas in a pile, denoted as \(M\), defines the initial search space, ranging from \(1\) to \(M\). Each iteration of the while loop cuts the search space in half, and the total number of iterations needed will be \(\log_2 M\).
Within each iteration of the while loop, the function feasible
is called,
which in turn calls total_hours_to_finish_eating
. As we established, the time
complexity for total_hours_to_finish_eating
is \(\mathcal{O}(N)\). Therefore,
the total time complexity for minEatingSpeed
(we now denote this function as
\(\mathcal{T}_{\mathcal{J}}(M, N)\)) is the product of the number of binary search
iterations and the time complexity of the function called within each iteration:
Best, Worst, and Average Case Analysis#
In a typical binary search problem, the best, worst, and average case time complexities are presented below. Please note the \(N\) below is the length of the search space, in contrast to our \(M\) (so don’t get confused).
Case |
Worst Case |
Average Case |
Best Case |
---|---|---|---|
Element is in the list |
\(\mathcal{O}(\log_2 N)\) |
\(\mathcal{O}(\log_2 N)\) |
\(\mathcal{O}(1)\) |
Element is not in the list |
\(\mathcal{O}(\log_2 N)\) |
\(\mathcal{O}(\log_2 N)\) |
\(\mathcal{O}(\log_2 N)\) |
In our Koko problem, the element is always in the list. Therefore, we can ignore the “element is not in the list” case. Furthermore, in each iteration of the binary search, we need to spend \(\mathcal{O}(N)\) time to check if the speed is feasible. Therefore, the best, worst, and average case time complexities for Koko eating bananas are as follows:
Case |
Worst Case |
Average Case |
Best Case |
---|---|---|---|
Element is in the list |
\(\mathcal{O}(\log_2 M \cdot N)\) |
\(\mathcal{O}(\log_2 M \cdot N)\) |
\(\mathcal{O}(N)\) |
In a typical binary search, the best case occurs when the element is at the midpoint of the search space, resulting in an \(\mathcal{O}(1)\) complexity since no further searching is required. The worst and average cases involve more searching and have a complexity of \(\mathcal{O}(\log_2 N)\), as you’ve noted.
For the Koko eating bananas problem, however, each iteration of the binary search involves checking the entire list of piles to see if Koko can eat all the bananas at a certain rate within \(h\) hours. This checking process has a time complexity of \(\mathcal{O}(N)\). Even in the best case, where the ideal rate is found in the first try, you still need to perform this check once. Therefore, the best case complexity would be \(\mathcal{O}(N)\), as you need to check every pile at least once.
Space Complexity#
The space complexity can be broken down into two components: input space complexity and auxiliary space complexity.
Input Space Complexity#
The input for the Koko eating bananas problem is the list of banana piles. This input occupies \(\mathcal{O}(N)\) space, where \(N\) is the number of piles. The speed \(k\) and the hours \(h\) are just integers, so they occupy constant space.
Auxiliary Space Complexity#
Auxiliary space refers to the extra space used by an algorithm, not including
the space taken up by the inputs. For the Koko problem, if we are implementing
the binary search iteratively, the auxiliary space complexity would be constant,
\(\mathcal{O}(1)\), because we only need a fixed number of variables: one for the
low end of our search space (l
), one for the high end (r
), and occasionally
one for the midpoint (m
) during each iteration of the binary search.
The algorithm does not use any dynamic data structures, like lists or arrays, that grow with the input size, and there is no recursive stack space to consider since it’s an iterative approach.
Total Space Complexity#
Total space complexity is the sum of input and auxiliary space complexities. Since the input space complexity is \(\mathcal{O}(N)\) and the auxiliary space complexity is \(\mathcal{O}(1)\), the total space complexity for the Koko eating bananas problem is \(\mathcal{O}(N)\) when considering the input space. If we choose not to consider the input space, the space complexity remains \(\mathcal{O}(1)\).
In summary, the algorithm for solving the Koko eating bananas problem is space-efficient, requiring only constant auxiliary space. The total space complexity is primarily determined by the input size.