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feat: add solutions to lc problem: No.2355
No.2355.Maximum Number of Books You Can Take
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# [2355. Maximum Number of Books You Can Take](https://leetcode.cn/problems/maximum-number-of-books-you-can-take)
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[English Version](/solution/2300-2399/2355.Maximum%20Number%20of%20Books%20You%20Can%20Take/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<p>You are given a <strong>0-indexed</strong> integer array <code>books</code> of length <code>n</code> where <code>books[i]</code> denotes the number of books on the <code>i<sup>th</sup></code> shelf of a bookshelf.</p>
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<p>You are going to take books from a <strong>contiguous</strong> section of the bookshelf spanning from <code>l</code> to <code>r</code> where <code>0 &lt;= l &lt;= r &lt; n</code>. For each index <code>i</code> in the range <code>l &lt;= i &lt; r</code>, you must take <strong>strictly fewer</strong> books from shelf <code>i</code> than shelf <code>i + 1</code>.</p>
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<p>Return <em>the <strong>maximum</strong> number of books you can take from the bookshelf.</em></p>
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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> books = [8,5,2,7,9]
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<strong>Output:</strong> 19
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<strong>Explanation:</strong>
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- Take 1 book from shelf 1.
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- Take 2 books from shelf 2.
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- Take 7 books from shelf 3.
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- Take 9 books from shelf 4.
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You have taken 19 books, so return 19.
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It can be proven that 19 is the maximum number of books you can take.
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</pre>
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<p><strong>Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> books = [7,0,3,4,5]
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<strong>Output:</strong> 12
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<strong>Explanation:</strong>
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- Take 3 books from shelf 2.
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- Take 4 books from shelf 3.
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- Take 5 books from shelf 4.
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You have taken 12 books so return 12.
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It can be proven that 12 is the maximum number of books you can take.
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</pre>
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<p><strong>Example 3:</strong></p>
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<pre>
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<strong>Input:</strong> books = [8,2,3,7,3,4,0,1,4,3]
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<strong>Output:</strong> 13
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<strong>Explanation:</strong>
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- Take 1 book from shelf 0.
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- Take 2 books from shelf 1.
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- Take 3 books from shelf 2.
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- Take 7 books from shelf 3.
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You have taken 13 books so return 13.
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It can be proven that 13 is the maximum number of books you can take.
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt;= books.length &lt;= 10<sup>5</sup></code></li>
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<li><code>0 &lt;= books[i] &lt;= 10<sup>5</sup></code></li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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**方法一:单调栈 + 动态规划**
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设 $dp[i]$ 表示以 $books[i]$ 结尾时能取走的书的最大数量。
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若从 $i$ 到 0ドル$ 可以取成一个公差为 1ドル$ 的等差数列,那么 $dp[i]$ 可以直接通过等差数列求和算出。
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若从 $i$ 到 0ドル$ 不能取成一个公差为 $-1$ 的等差数列,即这个规律在某个 $j$ 处断掉了(0ドル \le j \lt i$),那么一定有 $books[j] \lt books[i] - (i-j),ドル也即 $books[j] - j \lt books[i] - i,ドル利用单调栈在新数组 $books[i] - i$ 的每个位置,找到左边第一个比它严格小的数的位置,可以求出符合题意的 $j,ドル此时 $dp[i]=dp[j] + \sum_{k=j+1}^{i} (books[k]-k)$。
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答案为 $max(dp[i])$。
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时间复杂度 $O(n)$。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def maximumBooks(self, books: List[int]) -> int:
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nums = [v - i for i, v in enumerate(books)]
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n = len(nums)
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left = [-1] * n
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stk = []
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for i, v in enumerate(nums):
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while stk and nums[stk[-1]] >= v:
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stk.pop()
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if stk:
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left[i] = stk[-1]
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stk.append(i)
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ans = 0
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dp = [0] * n
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dp[0] = books[0]
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for i, v in enumerate(books):
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j = left[i]
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cnt = min(v, i - j)
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u = v - cnt + 1
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s = (u + v) * cnt // 2
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dp[i] = s + (0 if j == -1 else dp[j])
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ans = max(ans, dp[i])
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return ans
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public long maximumBooks(int[] books) {
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int n = books.length;
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int[] nums = new int[n];
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for (int i = 0; i < n; ++i) {
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nums[i] = books[i] - i;
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}
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int[] left = new int[n];
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Arrays.fill(left, -1);
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Deque<Integer> stk = new ArrayDeque<>();
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for (int i = 0; i < n; ++i) {
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while (!stk.isEmpty() && nums[stk.peek()] >= nums[i]) {
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stk.pop();
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}
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if (!stk.isEmpty()) {
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left[i] = stk.peek();
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}
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stk.push(i);
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}
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long ans = 0;
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long[] dp = new long[n];
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dp[0] = books[0];
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for (int i = 0; i < n; ++i) {
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int j = left[i];
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int v = books[i];
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int cnt = Math.min(v, i - j);
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int u = v - cnt + 1;
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long s = (long) (u + v) * cnt / 2;
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dp[i] = s + (j == -1 ? 0 : dp[j]);
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ans = Math.max(ans, dp[i]);
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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using ll = long long;
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class Solution {
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public:
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long long maximumBooks(vector<int>& books) {
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int n = books.size();
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vector<int> nums(n);
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for (int i = 0; i < n; ++i) nums[i] = books[i] - i;
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vector<int> left(n, -1);
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stack<int> stk;
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for (int i = 0; i < n; ++i)
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{
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while (!stk.empty() && nums[stk.top()] >= nums[i]) stk.pop();
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if (!stk.empty()) left[i] = stk.top();
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stk.push(i);
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}
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vector<ll> dp(n);
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dp[0] = books[0];
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ll ans = 0;
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for (int i = 0; i < n; ++i)
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{
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int v = books[i];
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int j = left[i];
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int cnt = min(v, i - j);
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int u = v - cnt + 1;
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ll s = 1ll * (u + v) * cnt / 2;
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dp[i] = s + (j == -1 ? 0 : dp[j]);
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ans = max(ans, dp[i]);
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func maximumBooks(books []int) int64 {
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n := len(books)
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nums := make([]int, n)
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left := make([]int, n)
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for i, v := range books {
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nums[i] = v - i
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left[i] = -1
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}
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stk := []int{}
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for i, v := range nums {
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for len(stk) > 0 && nums[stk[len(stk)-1]] >= v {
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stk = stk[:len(stk)-1]
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}
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if len(stk) > 0 {
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left[i] = stk[len(stk)-1]
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}
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stk = append(stk, i)
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}
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dp := make([]int, n)
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dp[0] = books[0]
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ans := 0
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for i, v := range books {
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j := left[i]
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cnt := min(v, i-j)
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u := v - cnt + 1
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s := (u + v) * cnt / 2
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dp[i] = s
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if j != -1 {
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dp[i] += dp[j]
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}
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ans = max(ans, dp[i])
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}
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return int64(ans)
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}
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```
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### **TypeScript**
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```ts
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```
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### **...**
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```
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```
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<!-- tabs:end -->

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