LeetCode 2355: Maximum Number of Books You Can Take Solution

Master LeetCode problem 2355 (Maximum Number of Books You Can Take), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2355. Maximum Number of Books You Can Take

Hard

Array Dynamic Programming Stack Monotonic Stack

Problem Explanation

Explanation

To solve this problem, we can use dynamic programming with a 2D array to keep track of the maximum number of books we can take at each position. At each position, we can either choose to take the current book or skip it. If we take the book, we need to consider the maximum number of books we can take at the previous position. If we skip the book, we can take the maximum number of books we could have taken at the previous position. We then populate the 2D array accordingly until we reach the end of the array. Finally, the maximum number of books we can take will be the value at the last position in the 2D array.

Steps:

Initialize a 2D array dp of size n x 2, where n is the number of books.
Initialize dp[0][0] to 1 (if we take the first book) and dp[0][1] to 0 (if we skip the first book).
Iterate through the books starting from the second book.
For each book at index i, update dp[i][0] to be the maximum of:
- dp[i-1][1] + 1 (if we take the current book)
- dp[i-1][0] (if we skip the current book)
Update dp[i][1] to be the maximum of:
- dp[i-1][1] (if we skip the current book)
- dp[i-1][0] (if we take the current book)
The maximum number of books we can take will be the maximum of dp[n-1][0] and dp[n-1][1].

Time Complexity

The time complexity of this solution is O(n), where n is the number of books.

Space Complexity

The space complexity of this solution is O(n) for the 2D array dp.

Solution Code

class Solution {
    public int maxBooks(int[] books) {
        int n = books.length;
        int[][] dp = new int[n][2];
        
        dp[0][0] = 1;
        dp[0][1] = 0;
        
        for (int i = 1; i < n; i++) {
            dp[i][0] = Math.max(dp[i-1][1] + 1, dp[i-1][0]);
            dp[i][1] = Math.max(dp[i-1][1], dp[i-1][0]);
        }
        
        return Math.max(dp[n-1][0], dp[n-1][1]);
    }
}

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