feat: add solutions to lc problems: No.2389~2392

* No.2389.Longest Subsequence With Limited Sum
* No.2390.Removing Stars From a String
* No.2391.Minimum Amount of Time to Collect Garbage
* No.2392.Build a Matrix With Condition

Author: yanglbme

solution/0300-0399/0336.Palindrome Pairs/README.md

@@ -32,8 +32,6 @@
 <strong>输出：</strong>[[0,1],[1,0]]
 </pre>
 
-
-
 <p><strong>提示：</strong></p>
 
 <ul>

solution/0500-0599/0554.Brick Wall/README.md

@@ -28,8 +28,6 @@
 <strong>输出：</strong>3
 </pre>
 
-
-
 <p><strong>提示：</strong></p>
 
 <ul>

solution/0600-0699/0637.Average of Levels in Binary Tree/README.md

@@ -243,7 +243,7 @@ var averageOfLevels = function (root) {
  * @param {TreeNode} root
  * @return {number[]}
  */
-var averageOfLevels = function(root) {
+var averageOfLevels = function (root) {
     let s = [];
     let cnt = [];
     function dfs(root, i) {

solution/0600-0699/0637.Average of Levels in Binary Tree/README_EN.md

@@ -212,7 +212,6 @@ var averageOfLevels = function (root) {
 };
 ```
 
-
 ```js
 /**
  * Definition for a binary tree node.
@@ -226,7 +225,7 @@ var averageOfLevels = function (root) {
  * @param {TreeNode} root
  * @return {number[]}
  */
-var averageOfLevels = function(root) {
+var averageOfLevels = function (root) {
     let s = [];
     let cnt = [];
     function dfs(root, i) {

solution/0700-0799/0775.Global and Local Inversions/README.md

@@ -42,8 +42,6 @@
 <strong>解释：</strong>有 2 个全局倒置，和 1 个局部倒置。
 </pre>
 
-
-
 <p><strong>提示：</strong></p>
 
 <ul>

solution/0800-0899/0830.Positions of Large Groups/README.md

@@ -48,8 +48,6 @@
 <strong>输出：</strong>[]
 </pre>
 
-
-
 <p><strong>提示：</strong></p>
 
 <ul>

solution/1400-1499/1472.Design Browser History/README.md

@@ -111,19 +111,19 @@ class BrowserHistory {
     public BrowserHistory(String homepage) {
         visit(homepage);
     }
-    
+
     public void visit(String url) {
         stk1.push(url);
         stk2.clear();
     }
-    
+
     public String back(int steps) {
         for (; steps > 0 && stk1.size() > 1; --steps) {
             stk2.push(stk1.pop());
         }
         return stk1.peek();
     }
-    
+
     public String forward(int steps) {
         for (; steps > 0 && !stk2.isEmpty(); --steps) {
             stk1.push(stk2.pop());
@@ -152,20 +152,20 @@ public:
     BrowserHistory(string homepage) {
         visit(homepage);
     }
-    
+
     void visit(string url) {
         stk1.push(url);
         stk2 = stack<string>();
     }
-    
+
     string back(int steps) {
         for (; steps && stk1.size() > 1; --steps) {
             stk2.push(stk1.top());
             stk1.pop();
         }
         return stk1.top();
     }
-    
+
     string forward(int steps) {
         for (; steps && !stk2.empty(); --steps) {
             stk1.push(stk2.top());

solution/1400-1499/1472.Design Browser History/README_EN.md

@@ -98,19 +98,19 @@ class BrowserHistory {
     public BrowserHistory(String homepage) {
         visit(homepage);
     }
-    
+
     public void visit(String url) {
         stk1.push(url);
         stk2.clear();
     }
-    
+
     public String back(int steps) {
         for (; steps > 0 && stk1.size() > 1; --steps) {
             stk2.push(stk1.pop());
         }
         return stk1.peek();
     }
-    
+
     public String forward(int steps) {
         for (; steps > 0 && !stk2.isEmpty(); --steps) {
             stk1.push(stk2.pop());
@@ -139,20 +139,20 @@ public:
     BrowserHistory(string homepage) {
         visit(homepage);
     }
-    
+
     void visit(string url) {
         stk1.push(url);
         stk2 = stack<string>();
     }
-    
+
     string back(int steps) {
         for (; steps && stk1.size() > 1; --steps) {
             stk2.push(stk1.top());
             stk1.pop();
         }
         return stk1.top();
     }
-    
+
     string forward(int steps) {
         for (; steps && !stk2.empty(); --steps) {
             stk1.push(stk2.top());

solution/1400-1499/1472.Design Browser History/Solution.py

@@ -1,5 +1,4 @@
 class BrowserHistory:
-
     def __init__(self, homepage: str):
         self.stk1 = []
         self.stk2 = []

solution/2300-2399/2389.Longest Subsequence With Limited Sum/README.md

@@ -0,0 +1,167 @@
+# [2389. 和有限的最长子序列](https://leetcode.cn/problems/longest-subsequence-with-limited-sum)
+
+[English Version](/solution/2300-2399/2389.Longest%20Subsequence%20With%20Limited%20Sum/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>给你一个长度为 <code>n</code>&nbsp;的整数数组 <code>nums</code> ，和一个长度为 <code>m</code> 的整数数组 <code>queries</code> 。</p>
+
+<p>返回一个长度为 <code>m</code> 的数组<em> </em><code>answer</code><em> </em>，其中<em> </em><code>answer[i]</code><em> </em>是 <code>nums</code> 中<span style=""> </span>元素之和小于等于 <code>queries[i]</code> 的 <strong>子序列</strong> 的 <strong>最大</strong> 长度<span style="">&nbsp;</span><span style=""> </span>。</p>
+
+<p><strong>子序列</strong> 是由一个数组删除某些元素（也可以不删除）但不改变剩余元素顺序得到的一个数组。</p>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre>
+<strong>输入：</strong>nums = [4,5,2,1], queries = [3,10,21]
+<strong>输出：</strong>[2,3,4]
+<strong>解释：</strong>queries 对应的 answer 如下：
+- 子序列 [2,1] 的和小于或等于 3 。可以证明满足题目要求的子序列的最大长度是 2 ，所以 answer[0] = 2 。
+- 子序列 [4,5,1] 的和小于或等于 10 。可以证明满足题目要求的子序列的最大长度是 3 ，所以 answer[1] = 3 。
+- 子序列 [4,5,2,1] 的和小于或等于 21 。可以证明满足题目要求的子序列的最大长度是 4 ，所以 answer[2] = 4 。
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre>
+<strong>输入：</strong>nums = [2,3,4,5], queries = [1]
+<strong>输出：</strong>[0]
+<strong>解释：</strong>空子序列是唯一一个满足元素和小于或等于 1 的子序列，所以 answer[0] = 0 。</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>n == nums.length</code></li>
+	<li><code>m == queries.length</code></li>
+	<li><code>1 &lt;= n, m &lt;= 1000</code></li>
+	<li><code>1 &lt;= nums[i], queries[i] &lt;= 10<sup>6</sup></code></li>
+</ul>
+
+## 解法
+
+<!-- 这里可写通用的实现逻辑 -->
+
+**方法一：排序 + 前缀和 + 二分查找**
+
+将 `nums` 排序，对于每个 `queries[i]`，求出 `nums` 中所有元素之和小于等于 `queries[i]` 的子序列的最大长度。
+
+时间复杂度 $O(n\log n + m\log n)$。其中 $n$ 为 `nums` 的长度，$m$ 为 `queries` 的长度。
+
+<!-- tabs:start -->
+
+### **Python3**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```python
+class Solution:
+    def answerQueries(self, nums: List[int], queries: List[int]) -> List[int]:
+        nums.sort()
+        s = list(accumulate(nums))
+        return [bisect_right(s, v) for v in queries]
+```
+
+### **Java**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```java
+class Solution {
+    public int[] answerQueries(int[] nums, int[] queries) {
+        Arrays.sort(nums);
+        int n = nums.length;
+        int[] s = new int[n + 1];
+        for (int i = 0; i < n; ++i) {
+            s[i + 1] = s[i] + nums[i];
+        }
+        int m = queries.length;
+        int[] ans = new int[m];
+        for (int i = 0; i < m; ++i) {
+            ans[i] = search(s, queries[i]);
+        }
+        return ans;
+    }
+
+    private int search(int[] s, int v) {
+        int left = 1, right = s.length;
+        while (left < right) {
+            int mid = (left + right) >> 1;
+            if (s[mid] > v) {
+                right = mid;
+            } else {
+                left = mid + 1;
+            }
+        }
+        return left - 1;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    vector<int> answerQueries(vector<int>& nums, vector<int>& queries) {
+        sort(nums.begin(), nums.end());
+        int n = nums.size(), m = queries.size();
+        vector<int> s(n + 1);
+        for (int i = 0; i < n; ++i) {
+            s[i + 1] = s[i] + nums[i];
+        }
+        vector<int> ans(m);
+        for (int i = 0; i < m; ++i) {
+            ans[i] = upper_bound(s.begin() + 1, s.end(), queries[i]) - s.begin() - 1;
+        }
+        return ans;
+    }
+};
+```
+
+### **Go**
+
+```go
+func answerQueries(nums []int, queries []int) []int {
+	sort.Ints(nums)
+	n, m := len(nums), len(queries)
+	s := make([]int, n+1)
+	for i, v := range nums {
+		s[i+1] = s[i] + v
+	}
+	ans := make([]int, m)
+	for i, v := range queries {
+		left, right := 1, len(s)
+		for left < right {
+			mid := (left + right) >> 1
+			if s[mid] > v {
+				right = mid
+			} else {
+				left = mid + 1
+			}
+		}
+		ans[i] = left - 1
+	}
+	return ans
+}
+```
+
+### **TypeScript**
+
+```ts
+
+```
+
+### **...**
+
+```
+
+
+```
+
+<!-- tabs:end -->