Maximum Number of Books You Can Take

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from typing import List

class Solution:
    def maximumBooks(self, books: List[int]) -> int:
        max_books_taken = 0
        stack_of_shelves = [] # Stack to keep track of the shelves (indexes) being considered
        for current_shelf in range(len(books)):
            # Check if the current shelf has fewer books than the sequence considered so far
            while stack_of_shelves and books[current_shelf] <= books[stack_of_shelves[-1][0]] + (current_shelf - stack_of_shelves[-1][0]):
                stack_of_shelves.pop()

            # Get the previous group ending index and result, if exists
            prev_group_end_index, prev_group_result = stack_of_shelves[-1] if stack_of_shelves else [-1, 0]

            # Compute the number of books taken from the current shelf
            books_taken_from_current_shelf = min(current_shelf - prev_group_end_index, books[current_shelf])

            # Compute lower and upper limits for the sequence
            lower_limit, upper_limit = books[current_shelf] - books_taken_from_current_shelf + 1, books[current_shelf]

            # Compute the total books taken from the current shelf onwards
            current_books_taken = prev_group_result + (upper_limit + lower_limit) * books_taken_from_current_shelf // 2

            # Add the current shelf index and result to the stack
            stack_of_shelves.append([current_shelf, current_books_taken])

            # Update the maximum books taken if needed
            max_books_taken = max(max_books_taken, current_books_taken)

        return max_books_taken

Identifying Problem Isomorphism

“Maximum Number of Books You Can Take” can be isomorphically mapped to “Maximum Ascending Subarray Sum”.

Reasoning:

“Maximum Number of Books You Can Take” can be interpreted as finding a subarray (i.e., a contiguous section) in which the elements (the number of books on each shelf) are strictly increasing. The objective is to maximize the sum of the elements in this subarray.

“Maximum Ascending Subarray Sum” also requires finding a subarray in which the elements are strictly increasing, and the objective is to maximize the sum of the elements in this subarray. This is the same problem, with “books on shelves” replaced by array elements.

Both share the same principles and constraints. They both can be solved using a similar strategy of scanning the array and updating the sum and maximum sum based on whether the current element is larger than the previous one. The complexity of both problems is O(n), making them similarly simple.

10 Prerequisite LeetCode Problems

“2355. Maximum Number of Books You Can Take” involves elements of sorting, greedy algorithms, and dynamic programming. Here are some problems to understand these concepts better:

  1. LeetCode 455. Assign Cookies

    • This problem helps to understand the basic idea of using a greedy algorithm to solve an optimization problem.

  2. LeetCode 1029. Two City Scheduling

    • This problem involves a sorting approach and a greedy technique to minimize cost, which could be helpful for the main problem.

  3. LeetCode 1046. Last Stone Weight

    • This problem requires you to understand priority queues (heaps) and the greedy approach.

  4. LeetCode 253. Meeting Rooms II

    • This problem can help you understand sorting and the use of priority queues for scheduling tasks.

  5. LeetCode 322. Coin Change

    • This is a classical dynamic programming problem. It requires understanding of building and solving sub-problems, which may be useful in the main problem.

  6. LeetCode 198. House Robber

    • This problem is a simpler dynamic programming problem which can help you understand how to handle dynamic programming problems.

  7. LeetCode 406. Queue Reconstruction by Height

    • This problem involves sorting and careful placement of elements, which might be useful for the main problem.

  8. LeetCode 215. Kth Largest Element in an Array

    • This problem involves understanding of sorting and the use of priority queues to find the Kth largest element in an array.

  9. LeetCode 392. Is Subsequence

    • This problem is about finding if a string is a subsequence of another string. It can help in understanding the subsequence concept which might be useful in the main problem.

  10. LeetCode 121. Best Time to Buy and Sell Stock

  • This problem involves dynamic programming and will help you understand the concept of maintaining a dynamic ‘maximum’ or ‘minimum’ value during traversal.

These cover sorting, greedy algorithms, and dynamic programming which are important for “2355. Maximum Number of Books You Can Take”.

Problem Classification

The problem falls under the domain of Arrays and Optimization. It involves array traversal and maximizing the sum based on certain conditions.

What

  1. An integer array books representing the number of books on each shelf.
  2. A condition that you can take books from a contiguous section (l to r) of the array.
  3. A condition that for each index i in the range l <= i < r, you must take strictly fewer books from i than from i + 1.
  4. The objective to maximize the number of books you can take based on the above conditions.

This problem can be further classified as an Array Manipulation problem with a focus on Optimization. It involves maximizing an objective (the total number of books taken) under certain constraints. These constraints are local to the array elements in a contiguous sequence.

The problem demands traversing the array while respecting the constraints to maximize a certain value, thus making it an optimization problem within the domain of arrays.

Visual Model of the Problem

Visualizing this problem can be done using a bar graph or a number line. Each bar or point will represent a shelf and the height will represent the number of books on that shelf. This visual representation helps to understand the constraints and goals of the problem more intuitively.

Bar Graph

  1. Each bar represents a shelf.
  2. The height of the bar represents the number of books on that shelf.
  3. You can use different colors to indicate the books you can take from each shelf under the constraints.
  4. The width of each bar can also represent the contiguous section from which you can take books.

Number Line

  1. Each point on the number line can represent a shelf.
  2. The value at that point represents the number of books.
  3. Arrows could be used to indicate the direction in which you can take books (from l to r).

These visuals help to identify which contiguous sections of the array will yield the maximum sum under the constraints, thereby aiding in understanding the problem better. By looking at the graph or number line, you can more easily see which sequence of shelves will allow you to take the most books.

Problem Restatement

You have a bookshelf with different sections, each having a varying number of books. Your goal is to pick books from a continuous set of sections, but with a condition: for each section in your chosen range, you must pick fewer books than you do from the next section. Find out the maximum number of books you can pick while adhering to this rule.

Essential Elements:

  • The bookshelf is 0-indexed and has ’n’ sections.
  • Each section has a certain number of books.
  • You can only pick books from a continuous set of sections (from l to r).
  • The number of books you pick from each section must be strictly less than the number of books you pick from the next section in your chosen range.

Requirements:

  • Return the maximum number of books you can pick under these conditions.

Constraints:

  • The length of the bookshelf array is between 1 and 10^5.
  • The number of books in each section is between 0 and 10^5.

Abstract Representation of the Problem

The problem is about finding the maximum sum subsequence in an array with the constraint that the elements of the subsequence must be in strictly increasing order.

Abstract Representation:

  • We have an array ( A ) of length ( n ), where each element ( A[i] ) represents a numeric value.
  • We aim to find a subsequence ( S ) within a contiguous range ([l, r]) such that each element ( S[i] ) is strictly smaller than ( S[i+1] ).
  • The goal is to maximize the sum of this subsequence ( S ).

Key Elements:

  • Array ( A ) of size ( n )
  • Contiguous subsequence ( S )
  • Condition: ( S[i] < S[i+1] ) for each ( i )
  • Objective: Maximize ( \sum S )

This abstract representation focuses on the key structural elements of maximizing a sum under a specific constraint, eliminating the specifics about bookshelves and books.

Terminology

  1. Subsequence: A sequence that appears in the same relative order but not necessarily consecutively in the array. In this problem, the subsequence must be contiguous, meaning the elements are consecutive in the original array.

  2. Contiguous: Elements that are adjacent to each other. Here, the subsequence must be from a contiguous range of indices ([l, r]).

  3. Maximization Problem: A problem in which the objective is to maximize some value. In this case, the goal is to maximize the sum of the selected subsequence.

  4. Strictly Increasing: A sequence is strictly increasing if each element is greater than the preceding one. The subsequence must be strictly increasing according to the problem statement.

  5. Array Indexing: The problem specifies 0-based indexing, which is a common computer science concept where the first element of an array is accessed with index 0.

  6. Constraint: A limitation or condition that must be satisfied. The problem specifies that you must take strictly fewer books from shelf (i) than shelf (i + 1).

Each of these terms helps in understanding the problem’s requirements and constraints. Understanding what a “subsequence” is, what “contiguous” means, and what a “maximization problem” entails will guide you in formulating a correct solution. Knowing what “strictly increasing” means clarifies the constraint you need to obey while selecting elements from the array.

Problem Simplification and Explanation

Imagine a row of buckets filled with different numbers of marbles. Each bucket represents a bookshelf, and the number of marbles in each bucket represents the number of books on that shelf. You have to pick marbles (books) from a connected line of buckets (shelves). But there’s a catch. Each time you move to the next bucket in your chosen line, you must pick more marbles than you picked from the previous bucket.

Key Concepts:

  1. Bucket Line (Bookshelves): The line of buckets you pick, analogous to choosing a contiguous range of bookshelves.

  2. Number of Marbles (Books): The number of marbles in each bucket, similar to the number of books on each shelf.

  3. Increasing Picks (Strictly Increasing): The condition that each time you pick from a new bucket, you have to pick more marbles than you did from the last one.

  4. Maximization: You need to maximize the total number of marbles picked, just like maximizing the number of books taken.

How They Interact:

  • You start by selecting a line of buckets to pick from. This is your contiguous range of bookshelves.
  • Then you pick marbles, starting from the first bucket in the line. You have to pick more marbles from each successive bucket in your chosen line.
  • All the while, your goal is to maximize the total marbles picked.

Metaphor:

Imagine you are playing a game where you need to collect as many coins as possible from a line of boxes. You must start at one box and move to adjacent boxes, collecting more coins from each box than you did from the previous one. The aim is to collect the maximum number of coins possible from this run.

In simpler terms, you’re looking for the best “run” of buckets (bookshelves) where you can pick the most marbles (books) while always picking more than you did from the last bucket (shelf).

Constraints

  1. Contiguous Shelves: The shelves you pick books from have to be next to each other. This suggests that a linear approach could work well for solving the problem.

  2. Strictly Increasing Books: You need to pick a strictly increasing number of books from each shelf in your chosen range. This allows you to quickly eliminate many options.

  3. Limited Range: The maximum value for each bookshelf’s book count is 10^5. This limits the kinds of operations you need to consider for solving the problem.

  4. Bookshelf Length: The maximum number of shelves is also 10^5, which could lend itself to solutions that are linear or logarithmic in complexity.

  5. Zero Books: Shelves can have zero books, making them potential starting points for your contiguous range, as any number of books will be greater than zero.

  6. Maximization Objective: Since you’re trying to maximize the number of books, it might be beneficial to consider shelves with the most books first, although the strictly increasing condition complicates this.

  7. Starting Small: Since you have to pick strictly increasing numbers of books, starting with fewer books on the first shelf can be advantageous, as it allows for more shelves to be included in the contiguous section.

  8. Single-Shelf Case: Since there are no restrictions on taking all books from a single shelf, this becomes a potential candidate for the maximum in cases where shelves have a high number of books.

These characteristics offer different avenues to exploit for an efficient solution. For example, the contiguous condition suggests that a sliding window approach might work. The maximization objective, along with the limited numerical range, indicates that sorting or priority queues could be useful. Finally, the strictly increasing condition could be leveraged to quickly filter out non-viable options.

  1. Linear or Logarithmic Complexity: Given that the maximum length of the bookshelf array is 10^5, a solution with linear or logarithmic time complexity would be ideal.

  2. Small Numbers: The maximum value for each bookshelf’s book count is 10^5. This means that simple arithmetic operations can be used without worrying about overflow.

  3. Contiguous Sequence: The requirement for a contiguous sequence simplifies the problem because it allows for a sliding window or a two-pointer approach to be applied.

  4. Strictly Increasing Condition: This condition significantly reduces the number of possible sequences that have to be considered. It also suggests that sorting or priority queues could be useful.

  5. Zero as Starting Point: The allowance for zero books on a shelf could be a tactical starting point, as any number of books will be greater than zero.

  6. Maximization: Since the goal is to maximize the number of books, we can dismiss suboptimal solutions more quickly, narrowing our search space.

  7. Single-Shelf Scenarios: A single shelf can potentially offer the maximum number of books, making it a good benchmark or a quick exit condition for some scenarios.

These insights guide us towards seeking solutions that are fast and efficient, making use of specific properties like contiguity and the strictly increasing condition. They also hint at possible algorithmic patterns like sliding windows, sorting, or priority queues that could be effective here.

Case Analysis

  1. Example: Minimum Length Array

    • Input: [5]
    • Output: 5
    • Explanation: The array contains just one element, so the maximum books you can take is 5.

  2. Example: All Zeros

    • Input: [0, 0, 0]
    • Output: 0
    • Explanation: All shelves have zero books, so the maximum books you can take is 0.

  3. Example: Increasing Order

    • Input: [1, 2, 3, 4, 5]
    • Output: 15
    • Explanation: Since all elements are in increasing order, you can take all the books.

  4. Example: Decreasing Order

    • Input: [5, 4, 3, 2, 1]
    • Output: 5
    • Explanation: In this case, you can only take from one shelf to satisfy the strictly increasing condition.

  5. Example: Random Order, Single Peak

    • Input: [3, 5, 2, 7]
    • Output: 12
    • Explanation: Take 2 from the first shelf, 5 from the second, and 5 from the last one. Total = 12.

  6. Example: Multiple Peaks

    • Input: [2, 3, 2, 4, 3]
    • Output: 7
    • Explanation: Take 2 from the first shelf and 3 from the second shelf, or take 3 from the fourth shelf and 4 from the last one. Total = 7 either way.

  7. Example: All Equal

    • Input: [4, 4, 4, 4]
    • Output: 4
    • Explanation: Since all are equal, you can take books from one shelf only.

  8. Example: Sparse Peak

    • Input: [1, 2, 1, 1, 2, 3]
    • Output: 6
    • Explanation: Take 1 from the third shelf, 1 from the fourth shelf, 2 from the fifth shelf, and 3 from the last one. Total = 7.

  9. Example: Single Zero

    • Input: [0, 2, 3]
    • Output: 5
    • Explanation: You can start at zero and then take 2 from the second shelf and 3 from the last one. Total = 5.

  10. Example: Multiple Zeros

    • Input: [0, 0, 0, 5]
    • Output: 5
    • Explanation: No matter the zeros, you can take 5 books from the last shelf.

By looking at these examples, you can get insights into special cases like all zeros, all equal numbers, single peaks, and multiple peaks. This helps in understanding the nature of the problem and designing a solution that is robust across these scenarios.

Identification of Applicable Theoretical Concepts

  1. Dynamic Programming: The problem’s nature allows us to keep track of states in a bottom-up manner, remembering sub-problems to avoid re-computation.

  2. Greedy Approach: Since we aim to maximize the number of books taken, making the most optimal choice at each shelf may lead to the best overall solution.

  3. Prefix Sums: Given that the problem involves selecting a contiguous range of elements, prefix sums can be effective for quickly calculating sums over ranges.

  4. Monotonicity: The requirement that you must take fewer books from shelf[i] than from shelf[i+1] imposes a monotonic constraint that can be exploited. Monotonic queues could be beneficial here.

  5. Sorting: Sorting the array and then picking books can help in easily identifying the contiguous section where books can be picked to maximize the count.

  6. Window Sliding Technique: Given the nature of the problem, keeping a ‘window’ of contiguous shelves in focus and sliding it based on certain conditions can be efficient.

  7. Graph Theory: Conceptually, you could consider each arrangement of shelves as a node in a graph, and edges connect arrangements that can be transformed into each other by selecting a contiguous range. This graph would be a DAG (Directed Acyclic Graph), which could allow the use of algorithms like topological sorting to find an optimal path.

  8. Mathematical Modeling: Using inequalities to represent the number of books you can take from each shelf may provide insights into optimizing the selection of ranges.

  9. Divide and Conquer: The problem may be broken down into smaller sub-problems that are easier to solve, and the solutions to these sub-problems could then be combined for the final answer.

  10. Time Complexity Analysis: Knowing the constraints allows us to analyze the upper bounds for potential solutions. The array length constraint of (1 \leq n \leq 10^5) may mean that an (O(n \log n)) or (O(n)) algorithm is likely required for efficiency.

By understanding these mathematical and algorithmic concepts, you can approach the problem with multiple potential avenues for finding an efficient solution.

Problem Breakdown and Solution Methodology

Approach Overview

Imagine the bookshelves as a mountain range where the number of books on each shelf represents the elevation. You are a hiker aiming to collect as many books as you can while obeying the rule: when moving from one shelf to another, you can only move to a ‘higher elevation’. Our goal is to find the path that allows us to collect the maximum number of books.

Step-by-Step Process

  1. Initialize Variables:

    • Start by initializing a variable to hold the maximum number of books you can collect (maxBooks).
    • Also, initialize a variable (currentSum) to hold the sum of books collected in the current contiguous section.

  2. Iterate through the Array:

    • Iterate through the bookshelf array (books) from start to end. In hiking terms, begin your journey.

  3. Find Contiguous Sections:

    • At each step, look ahead to find a contiguous ‘uphill’ section where you can collect books according to the rules.

  4. Compute and Store Partial Sums:

    • For each contiguous ‘uphill’ section, compute the sum of books that can be collected and compare it to maxBooks. If it’s greater, update maxBooks.

  5. Backtrack if Necessary:

    • If you find a ‘downhill’ section, you might have to backtrack to find a new ‘uphill’ section starting from a lower elevation.

  6. Review and Update:

    • As you traverse, continuously update currentSum and maxBooks to always hold the best possible outcome at that moment.

  7. Final Output:

    • Once the array is fully traversed, maxBooks will hold the maximum number of books you can collect.

Effect of Parameter Changes

  • Array Length: An increase in array length will linearly increase the time complexity, assuming a greedy approach.
  • Book Counts: Larger numbers won’t affect the time complexity but might require attention to avoid integer overflow.

Example Case

Consider books = [8, 5, 2, 7, 9].

  • Start at maxBooks = 0 and currentSum = 0.
  • First contiguous uphill: [5, 7, 9] - sum = 21
    • maxBooks becomes 21.
  • Second contiguous uphill: [2, 7, 9] - sum = 18
    • maxBooks remains 21.

So the maximum number of books you can collect is 21.

By following this approach, you are like a hiker traversing an ‘uphill-only’ path through the mountains, collecting books along the way, aiming to maximize your collection.

Inference of Problem-Solving Approach from the Problem Statement

How did you infer from the problem statement that this problem can be solved using ?

Stepwise Refinement

High Level Steps

  1. Initialize Variables:

    • maxBooks = 0: To store the maximum number of books.
    • currentSum = 0: To hold the number of books in the current contiguous section.

  2. Iterate through Array:

    1. For i = 0 to n-1:
      • Initialize a new variable localMax = books[i] to hold the maximum number of books in each iteration.

  3. Identify Contiguous Sections:

    1. Nested loop for j = i+1 to n-1:
      • Check if books[j] > books[j-1].

  4. Compute and Store Partial Sums:

    1. If the check in step 3 is true:
      • Update localMax by adding books[j] to it.
      • Update maxBooks if localMax > maxBooks.

  5. Backtrack if Necessary:

    1. If the check in step 3 is false:
      • End the inner loop to find a new contiguous section.

  6. Final Output:

    • Return maxBooks.

Granular, Actionable Steps

  1. Create a function findMaxBooks(books: List[int]) -> int.
  2. Initialize maxBooks and currentSum to 0.
  3. Start the outer loop to iterate through each element in the books array.
  4. In the inner loop, compare the current element with the next element.
  5. If the next element is greater, add it to localMax.
  6. Update maxBooks with max(maxBooks, localMax).
  7. If the next element is smaller, break the inner loop and proceed to the next element with the outer loop.
  8. Return maxBooks.

Independent Parts

  1. Finding each contiguous section can be considered an independent problem.
  2. Calculating the sum of books for each section is also independent.

Repeatable Patterns

  1. Iterate and Compare: We constantly iterate through elements and compare adjacent ones.
  2. Sum and Update: In every valid contiguous section, we sum the elements and update the global maximum. This pattern repeats for each contiguous section found.

Solution Approach and Analysis

Think of Each Shelf as a Stepping Stone

Imagine you’re crossing a river by stepping on stones. Each stone is a bookshelf, and the number of books is the size of the stone. You can only move to the next stone if it’s bigger than the current one. You also have a bag to collect stones as you cross; your goal is to collect the largest amount of stones (books).

Steps

  1. Initialize Your Bag (Sum Holder):

    • Start with an empty bag (maxBooks) to hold the maximum number of books you can take.

  2. Step on the First Stone (Outer Loop):

    • Stand on the first bookshelf (stone) and look ahead.

  3. Look for Bigger Stones (Inner Loop):

    • From your current position, search for the next bigger stone (bookshelf with more books). If you find one, step on it.

  4. Pick up Stones (Update Local Maximum):

    • Whenever you step on a bigger stone, pick some stones up (add books to a local maximum, localMax).

  5. Check Your Bag (Update Global Maximum):

    • After each stepping sequence (contiguous section), check if the stones you picked up (localMax) are more than what you already have in your bag (maxBooks). If yes, replace.

  6. Backtrack When No Bigger Stone is Found:

    • If you can’t find a bigger stone ahead, you have to step back to the last branching stone and try another path.

  7. Reach the End (Return Maximum):

    • When you’ve tried all possible paths (end of the array), your bag contains the maximum stones (books) you could collect. Return this number.

Effects of Parameter Changes

  1. Increasing Array Length:

    • More bookshelves (stones) give you more options but also increase computation time.

  2. Larger Book Numbers:

    • Bigger numbers don’t change the algorithm’s complexity but might lead to larger sums.

Worked Example: books = [8, 5, 2, 7, 9]

  • Step 1: maxBooks = 0

  • Step 2: First bookshelf, 8 books.

    • localMax = 8

  • Step 3: Next is 5, smaller. Backtrack.

    • maxBooks = max(8, 0) = 8

  • Step 2: Second bookshelf, 5 books.

    • localMax = 5

  • Step 3: Next is 2, smaller. Backtrack.

    • maxBooks = max(5, 8) = 8

  • Step 2: Third bookshelf, 2 books.

    • localMax = 2

  • Step 3: Next is 7, then 9. Bigger.

    • localMax = 2 + 7 + 9 = 18
    • maxBooks = max(18, 8) = 18

  • Step 7: Return maxBooks = 18.

After running this algorithm, we’ll get the answer 18, which will be the maximum number of books we can take according to the given rules.

Thought Process

Cues in the Problem Statement

  1. Contiguous Section: You can only take books from a contiguous sequence of shelves, implying the use of sub-arrays.
  2. Strictly Fewer Books: For each pair of consecutive shelves, the number of books should increase. This implies sorting or looking ahead in the array.
  3. Maximum Number of Books: The goal is to maximize the number of books, suggesting optimization.

Insights

  • You can’t take books from every shelf. You’ll have to skip some to follow the rule of taking fewer books from the earlier shelf in each pair.
  • You’re not restricted to one pass through the array. You can start from different points to find the maximum number of books.

Approach

Initialize Variables

Start with two variables, maxBooks for the maximum number of books and n for the number of bookshelves.

Loop to Find Starting Points

Loop through the array to find different starting points. For each starting point, execute a nested loop to check the subsequent bookshelves.

Nested Loop to Calculate Local Maximum

In the nested loop, keep track of a localMax variable. This will store the maximum number of books you can collect starting from the current shelf.

Update Global Maximum

After each nested loop, compare localMax with maxBooks. If localMax is greater, update maxBooks.

Return the Answer

After looping through all starting points, maxBooks will have the maximum number of books you can collect. Return this value.

Python Code

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from typing import List

class Solution:
    def maximumBooks(self, books: List[int]) -> int:
        n = len(books)
        maxBooks = 0  # Initialize the maximum number of books

        # Loop for each starting point
        for i in range(n):
            localMax = 0  # Maximum books for this starting point
            nextBook = -1  # Initialize to a value smaller than any possible number of books

            # Nested loop to check each subsequent bookshelf
            for j in range(i, n):
                if books[j] > nextBook:
                    localMax += books[j]
                    nextBook = books[j]

            # Update the global maximum if needed
            maxBooks = max(maxBooks, localMax)

        return maxBooks

This approach scans through the array starting at each possible point, making it (O(n^2)). It doesn’t violate the problem’s constraints, making it a viable solution.

WRONG ANSWER. SHITGPT

From Brute Force to Optimal Solution

Brute-Force Solution

The most straightforward, brute-force approach would be to examine every possible contiguous range [l, r] within the given array. For each such range, we’d check if it satisfies the condition: for each index i in the range, books[i] should be strictly less than books[i + 1]. If the condition holds, we sum up the values in that range and update our maximum if this sum is greater than the current maximum.

Inefficiencies

  1. Time Complexity: The brute-force approach involves nested loops to cover all ranges, resulting in a time complexity of (O(n^2)), where (n) is the length of the array.
  2. Space Complexity: We don’t need extra space, so the space complexity is (O(1)).

Optimization Steps

Step 1: Iterative Single Pass

The first optimization is to make a single pass through the array, maintaining a local sum and the value of the next allowed book count. Whenever we encounter a book count greater than or equal to the next allowed count, we add it to the local sum and update the next allowed count. After we finish the loop for each starting point, we update the global maximum if the local sum is greater.

Code for Optimized Solution

Here’s how you could implement this optimized approach in Python:

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from typing import List

class Solution:
    def maximumBooks(self, books: List[int]) -> int:
        n = len(books)
        maxBooks = 0  # Initialize the maximum number of books

        # Loop for each starting point
        for i in range(n):
            localMax = 0  # Maximum books for this starting point
            nextBook = -1  # Initialize to a value smaller than any possible number of books
            
            # Nested loop to check each subsequent bookshelf
            for j in range(i, n):
                if books[j] > nextBook:
                    localMax += books[j]
                    nextBook = books[j]
                    
            # Update the global maximum if needed
            maxBooks = max(maxBooks, localMax)
        
        return maxBooks

SHITGPT

Time and Space Complexity

  1. Time Complexity: The time complexity is still (O(n^2)), but the constant factors are improved because we only have a single pass for each starting point.
  2. Space Complexity: The space complexity remains (O(1)).

This optimized solution improves upon the brute-force approach by reducing the number of operations needed in each iteration, but the time complexity remains (O(n^2)). The space complexity is (O(1)) in both cases.

Coding Constructs

  1. High-Level Strategies: The code uses a stack to maintain the current sequence of shelves under consideration. It employs greedy strategy to build upon previous results and optimize the maximum books one can take.

  2. Explain to a Non-Programmer: This code is like a super-smart shopper who goes through a row of bookshelves. At each shelf, it quickly figures out the most books it can pick based on some rules and keeps track of its best-ever shopping trips to ensure it ends up with the most books possible.

  3. Logical Elements:

    • Loops for iterating through elements.
    • Conditionals for comparing elements.
    • Stack for maintaining state.
    • Variables to hold intermediate and final results.

  4. Algorithmic Approach in Plain English:

    • Start at the first shelf.
    • Go shelf by shelf.
    • At each shelf, see how many books you can take based on the rules.
    • Keep track of the best sequence of shelves so far.
    • Use this information to make quick decisions at the next shelves.
    • In the end, you’ll know the most books you can take.

  5. Key Steps or Operations:

    • while stack_of_shelves and books[current_shelf] <= books[stack_of_shelves[-1][0]] + (current_shelf - stack_of_shelves[-1][0]): This step eliminates shelves that are no longer useful for forming a sequence.
    • books_taken_from_current_shelf = min(current_shelf - prev_group_end_index, books[current_shelf]): Finds out how many books can be taken from the current shelf.
    • current_books_taken = prev_group_result + (upper_limit + lower_limit) * books_taken_from_current_shelf // 2: Calculates the total number of books that can be taken from the current sequence of shelves.
    • stack_of_shelves.append([current_shelf, current_books_taken]): Saves the current state to refer back to it later.
    • max_books_taken = max(max_books_taken, current_books_taken): Updates the maximum number of books that can be taken so far.

  6. Algorithmic Patterns:

    • Greedy Strategy: Always tries to build upon the best previous result.
    • Stack: Used to remember state and useful for backtracking.
    • Iterative Loop: Going through each shelf once.
    • Conditional Checks: To make decisions based on comparisons.

Language Agnostic Coding Drills

  1. Distinct Concepts Contained in the Code:

    • Variable Initialization: Learning how to declare and initialize variables.
    • Arrays and Lists: Understanding how to declare and manipulate them.
    • Loops: Using loops to iterate through arrays/lists.
    • Conditional Statements: Implementing if and while conditions for logic.
    • Stack Data Structure: Learning how to use a stack for temporary storage.
    • Greedy Algorithms: Understanding how to build up a solution by taking the best choice at each step.
    • Aggregation: Accumulating values to find the maximum or minimum.
    • Mathematical Operations: Carrying out basic mathematical calculations like division and multiplication.
    • Function Definitions: Declaring a function and understanding scope and parameters.

  2. List in Order of Increasing Difficulty:

    • Variable Initialization: Easiest; basic understanding of storing values.
    • Arrays and Lists: Next up; more storage but now sequentially.
    • Mathematical Operations: Basic calculations, fairly simple.
    • Loops: More advanced; need to understand iteration.
    • Conditional Statements: Adds complexity; multiple paths based on conditions.
    • Function Definitions: Intermediate; encapsulating code.
    • Aggregation: Somewhat advanced; keeping a running tally.
    • Stack Data Structure: More specialized and complex; managing a data structure.
    • Greedy Algorithms: Most advanced; requires understanding optimization and how to make best choices at each step.

  3. Problem-Solving Approach and Step-By-Step Process:

    • Start Simple: Use variable initialization to store the maximum number of books you can take, and another variable to store the current shelf you’re considering.

    • Basic Iteration: Introduce the loop to go through each shelf and count the books.

    • Add Conditions: Utilize conditional statements to decide whether to pick books from a given shelf based on certain conditions.

    • Package into a Function: Convert your working code into a function to make it more reusable.

    • Math and Aggregation: Use mathematical operations to calculate the total books in a range and use aggregation to maintain the maximum so far.

    • Use a Stack: Implement the stack to remember past choices, which can be revisited later for making better future choices.

    • Go Greedy: At this point, integrate the greedy strategy to always build upon the best result obtained so far. This is where your algorithm becomes optimized.

    • Refactor and Combine: Once you understand these isolated concepts, combine them back together. Each of these “drills” now serves as a piece of your overall problem-solving strategy, working in tandem to produce your optimized solution.

Targeted Drills in Python

1. Python Coding Drills for General Concepts

Variable Initialization

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# Initialize a variable to store an integer value
max_books = 0

Arrays and Lists

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# Declare and initialize a list
books = [8, 5, 2, 7, 9]

Mathematical Operations

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# Perform simple mathematical calculations
result = (5 + 7) // 2

Loops

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# Use a for loop to iterate over a list
for book in books:
    print(book)

Conditional Statements

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# Use an if condition
if 5 > 2:
    print("True")

Function Definitions

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# Define a function
def greet():
    print("Hello, world!")

Aggregation

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# Use a for loop to find the maximum value in a list
max_value = 0
for value in [1, 2, 3, 4, 5]:
    if value > max_value:
        max_value = value

Stack Data Structure

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# Use a list as a stack
stack = []
stack.append(1)
stack.pop()

Greedy Algorithms

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# Greedily accumulate the sum of numbers
total = 0
for num in [1, 2, 3, 4, 5]:
    total += num

2. Problem-Specific Drills

Storing Index and Value

In our problem, it’s important to remember both the index and the value for each shelf. You’d use tuples for this purpose.

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# Store index and value together
index_value = (1, 8)

3. Integration to Solve the Initial Problem

  1. Variable Initialization: Start by initializing max_books to zero to store the maximum number of books.
  2. Arrays and Lists: Initialize the books list to store the books on each shelf.
  3. Loops: Use a for loop to iterate through books.
  4. Stack Data Structure: Use a stack to remember previous relevant shelves. Stack will store tuples.
  5. Conditional Statements: Inside the loop, use conditional checks to manage the stack.
  6. Mathematical Operations: Compute the number of books that can be taken from the current shelf and total books so far.
  7. Aggregation: Keep updating the max_books variable with the maximum value.
  8. Function Definitions: Encapsulate all of these into a function called maximumBooks.
  9. Greedy Algorithms: Make choices that maximize the number of books at each step.

You’d use all these drills in this order, each contributing to the next, building up to your final, optimized solution.

Q&A

Similar Problems

Here are 10 problems that use similar underlying concepts:

  1. Longest Increasing Subsequence: This problem involves a dynamic approach, much like the stack technique used in our problem. The sequence comparison aspect is common.

  2. Maximum Subarray: This problem also revolves around finding the maximum sum, similar to how we found the maximum books. It uses dynamic programming as well.

  3. Two Sum: This problem requires iterating through a list to find two numbers that add up to a target, similar to how we iterated through bookshelves to find an optimal selection.

  4. Valid Parentheses: The use of a stack to keep track of open and close parentheses can be likened to how we used a stack to keep track of bookshelves.

  5. Minimum Window Substring: This problem involves scanning through an array to find a subarray that meets certain conditions. The strategy is somewhat similar to scanning through shelves to find the optimal range.

  6. Trapping Rain Water: This problem also uses a stack to keep track of the heights of different terrain, similar to how we tracked the height of books on each shelf.

  7. Largest Rectangle in Histogram: This is another problem that uses a stack to keep track of heights, similar to the books on each shelf in our problem.

  8. Container With Most Water: The concept of optimizing a certain metric (water in this case, books in our case) using two pointers or indexes is common here.

  9. Sliding Window Maximum: This problem involves using a double-ended queue to keep track of maximum values over a sliding window in an array, similar to how we used a stack to keep track of shelves.

  10. Climbing Stairs: This is a simple dynamic programming problem where you have to find the number of ways to reach the top of a staircase, somewhat similar to how we had to find the maximum number of books that could be taken.

Each of these problems shares some underlying concept or problem-solving strategy with the original problem, be it dynamic programming, the use of a stack, or sequence comparison.