Maximum Number Of Books You Can Take - Leetcode Solution

The "Maximum Number of Books You Can Take" problem on Leetcode is a dynamic programming problem.

Problem Statement:

You are given an integer array representing the number of pages of each book that Nathalie wants to read. Nathalie can read at most "k" pages per day.

She wants to finish her books as soon as possible. Write a function to return the maximum number of books that Nathalie can read.

Solution:

Step 1: Define the Subproblems

The first step in solving a dynamic programming problem is to define the subproblems.

In this problem, we can define the subproblem as the maximum number of books that Nathalie can read by reading x books from the array.

Step 2: Define the Base Cases

The next step is to define the base cases. In this problem, the base case is when Nathalie can read only 1 book.

Step 3: Define the Recurrence Relation

The recurrence relation for this problem is as follows:

f(x) = the maximum number of books that Nathalie can read if she reads x books from the array.

f(x) = max(f(x-1), f(x-2)+a[x-1])

where, a[x-1] is the number of pages in the xth book.

Step 4: Implement the Algorithm

To implement the algorithm, we need to create an array "dp" of size n+1, where n is the number of books Nathalie wants to read.

The value dp[i] represents the maximum number of books Nathalie can read if she reads i books from the array.

The base case is dp[0] = 0, as Nathalie cannot read any books if she has not started.

The recurrence relation can be implemented using a loop:

for i from 1 to n: dp[i] = max(dp[i-1], dp[i-2]+a[i-1])

The final answer is dp[n].

Step 5: Analyze Time Complexity

The time complexity of this algorithm is O(n), as we are looping through the array once.

Step 6: Write Code

Here is the code for the solution:

def max_books(a, k): n = len(a) dp = [0]*(n+1) dp[0] = 0 # Base case

for i in range(1, n+1):
    if a[i-1] > k:  # If the book is too big, Nathalie cannot read it
        dp[i] = dp[i-1]
    else:
        dp[i] = max(dp[i-1], dp[i-2]+a[i-1])

return dp[n]

a = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] k = 3 print(max_books(a, k))

Output:

6

1