feat: add solutions to lc problem: No.2355

No.2355.Maximum Number of Books You Can Take

Author: yanglbme

solution/2300-2399/2355.Maximum Number of Books You Can Take/README.md

@@ -0,0 +1,256 @@
+# [2355. Maximum Number of Books You Can Take](https://leetcode.cn/problems/maximum-number-of-books-you-can-take)
+
+[English Version](/solution/2300-2399/2355.Maximum%20Number%20of%20Books%20You%20Can%20Take/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<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>
+
+<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>
+
+<p>Return <em>the <strong>maximum</strong> number of books you can take from the bookshelf.</em></p>
+
+<p>&nbsp;</p>
+<p><strong>Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> books = [8,5,2,7,9]
+<strong>Output:</strong> 19
+<strong>Explanation:</strong>
+- 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.
+</pre>
+
+<p><strong>Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> books = [7,0,3,4,5]
+<strong>Output:</strong> 12
+<strong>Explanation:</strong>
+- 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.
+</pre>
+
+<p><strong>Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> books = [8,2,3,7,3,4,0,1,4,3]
+<strong>Output:</strong> 13
+<strong>Explanation:</strong>
+- 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.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= books.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>0 &lt;= books[i] &lt;= 10<sup>5</sup></code></li>
+</ul>
+
+## 解法
+
+<!-- 这里可写通用的实现逻辑 -->
+
+**方法一：单调栈 + 动态规划**
+
+设 $dp[i]$ 表示以 $books[i]$ 结尾时能取走的书的最大数量。
+
+若从 $i$ 到 $0$ 可以取成一个公差为 $1$ 的等差数列，那么 $dp[i]$ 可以直接通过等差数列求和算出。
+
+若从 $i$ 到 $0$ 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（$0 \le j \lt i$），那么一定有 $books[j] \lt books[i] - (i-j)$，也即 $books[j] - j \lt books[i] - i$，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 $j$，此时 $dp[i]=dp[j] + \sum_{k=j+1}^{i} (books[k]-k)$。
+
+答案为 $max(dp[i])$。
+
+时间复杂度 $O(n)$。
+
+<!-- tabs:start -->
+
+### **Python3**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```python
+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
+```
+
+### **Java**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```java
+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;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+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;
+    }
+};
+```
+
+### **Go**
+
+```go
+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
+}
+```
+
+### **TypeScript**
+
+```ts
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->