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2355. Maximum Number of Books You Can Take πŸ”’

Description

You are given a 0-indexed integer array books of length n where books[i] denotes the number of books on the ith shelf of a bookshelf.

You are going to take books from a contiguous section of the bookshelf spanning from l to r where 0 <= l <= r < n. For each index i in the range l <= i < r, you must take strictly fewer books from shelf i than shelf i + 1.

Return the maximum number of books you can take from the bookshelf.

 

Example 1:

Input: books = [8,5,2,7,9]
Output: 19
Explanation:
- Take 1 book from shelf 1.
- Take 2 books from shelf 2.
- Take 7 books from shelf 3.
- Take 9 books from shelf 4.
You have taken 19 books, so return 19.
It can be proven that 19 is the maximum number of books you can take.

Example 2:

Input: books = [7,0,3,4,5]
Output: 12
Explanation:
- Take 3 books from shelf 2.
- Take 4 books from shelf 3.
- Take 5 books from shelf 4.
You have taken 12 books so return 12.
It can be proven that 12 is the maximum number of books you can take.

Example 3:

Input: books = [8,2,3,7,3,4,0,1,4,3]
Output: 13
Explanation:
- Take 1 book from shelf 0.
- Take 2 books from shelf 1.
- Take 3 books from shelf 2.
- Take 7 books from shelf 3.
You have taken 13 books so return 13.
It can be proven that 13 is the maximum number of books you can take.

 

Constraints:

  • 1 <= books.length <= 105
  • 0 <= books[i] <= 105

Solutions

Solution 1: Simulation

We directly compare each row and column of the matrix $grid$. If they are equal, then it is a pair of equal row-column pairs, and we increment the answer by one.

The time complexity is $O(n^3)$, where $n$ is the number of rows or columns in the matrix $grid$. The space complexity is $O(1)$.

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class Solution:
    def maximumBooks(self, books: List[int]) -> int:
        nums = [v - i for i, v in enumerate(books)]
        n = len(nums)
        left = [-1] * n
        stk = []
        for i, v in enumerate(nums):
            while stk and nums[stk[-1]] >= v:
                stk.pop()
            if stk:
                left[i] = stk[-1]
            stk.append(i)
        ans = 0
        dp = [0] * n
        dp[0] = books[0]
        for i, v in enumerate(books):
            j = left[i]
            cnt = min(v, i - j)
            u = v - cnt + 1
            s = (u + v) * cnt // 2
            dp[i] = s + (0 if j == -1 else dp[j])
            ans = max(ans, dp[i])
        return ans

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class Solution {
    public long maximumBooks(int[] books) {
        int n = books.length;
        int[] nums = new int[n];
        for (int i = 0; i < n; ++i) {
            nums[i] = books[i] - i;
        }
        int[] left = new int[n];
        Arrays.fill(left, -1);
        Deque<Integer> stk = new ArrayDeque<>();
        for (int i = 0; i < n; ++i) {
            while (!stk.isEmpty() && nums[stk.peek()] >= nums[i]) {
                stk.pop();
            }
            if (!stk.isEmpty()) {
                left[i] = stk.peek();
            }
            stk.push(i);
        }
        long ans = 0;
        long[] dp = new long[n];
        dp[0] = books[0];
        for (int i = 0; i < n; ++i) {
            int j = left[i];
            int v = books[i];
            int cnt = Math.min(v, i - j);
            int u = v - cnt + 1;
            long s = (long) (u + v) * cnt / 2;
            dp[i] = s + (j == -1 ? 0 : dp[j]);
            ans = Math.max(ans, dp[i]);
        }
        return ans;
    }
}

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using ll = long long;

class Solution {
public:
    long long maximumBooks(vector<int>& books) {
        int n = books.size();
        vector<int> nums(n);
        for (int i = 0; i < n; ++i) nums[i] = books[i] - i;
        vector<int> left(n, -1);
        stack<int> stk;
        for (int i = 0; i < n; ++i) {
            while (!