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Formatted question description: https://leetcode.ca/all/2355.html
2355. Maximum Number of Books You Can Take
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 <= 1050 <= books[i] <= 105
Solutions
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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; } } -
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 (!stk.empty() && nums[stk.top()] >= nums[i]) stk.pop(); if (!stk.empty()) left[i] = stk.top(); stk.push(i); } vector<ll> dp(n); dp[0] = books[0]; ll ans = 0; for (int i = 0; i < n; ++i) { int v = books[i]; int j = left[i]; int cnt = min(v, i - j); int u = v - cnt + 1; ll s = 1ll * (u + v) * cnt / 2; dp[i] = s + (j == -1 ? 0 : dp[j]); ans = max(ans, dp[i]); } return ans; } }; -
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 -
func maximumBooks(books []int) int64 { n := len(books) nums := make([]int, n) left := make([]int, n) for i, v := range books { nums[i] = v - i left[i] = -1 } stk := []int{} for i, v := range nums { for len(stk) > 0 && nums[stk[len(stk)-1]] >= v { stk = stk[:len(stk)-1] } if len(stk) > 0 { left[i] = stk[len(stk)-1] } stk = append(stk, i) } dp := make([]int, n) dp[0] = books[0] ans := 0 for i, v := range books { j := left[i] cnt := min(v, i-j) u := v - cnt + 1 s := (u + v) * cnt / 2 dp[i] = s if j != -1 { dp[i] += dp[j] } ans = max(ans, dp[i]) } return int64(ans) } func max(a, b int) int { if a > b { return a } return b } func min(a, b int) int { if a < b { return a } return b }