feat: add solutions to lc problems: No.2496~2503

* No.2496.Maximum Value of a String in an Array
* No.2497.Maximum Star Sum of a Graph
* No.2498.Frog Jump II
* No.2500.Delete Greatest Value in Each Row
* No.2501.Longest Square Streak in an Array
* No.2502.Design Memory Allocator
* No.2503.Maximum Number of Points From Grid Queries

Author: yanglbme

solution/0100-0199/0129.Sum Root to Leaf Numbers/README_EN.md

@@ -73,7 +73,7 @@ class Solution:
             if root.left is None and root.right is None:
                 return s
             return dfs(root.left, s) + dfs(root.right, s)
-        
+
         return dfs(root, 0)
 ```
 

solution/0300-0399/0374.Guess Number Higher or Lower/README.md

@@ -216,7 +216,7 @@ func guessNumber(n int) int {
 ### **C#**
 
 ```cs
-/** 
+/**
  * Forward declaration of guess API.
  * @param  num   your guess
  * @return 	     -1 if num is higher than the picked number

solution/0300-0399/0374.Guess Number Higher or Lower/README_EN.md

@@ -192,7 +192,7 @@ func guessNumber(n int) int {
 ### **C#**
 
 ```cs
-/** 
+/**
  * Forward declaration of guess API.
  * @param  num   your guess
  * @return 	     -1 if num is higher than the picked number

solution/2200-2299/2291.Maximum Profit From Trading Stocks/README.md

@@ -1,50 +1,44 @@
-# [2291. Maximum Profit From Trading Stocks](https://leetcode.cn/problems/maximum-profit-from-trading-stocks)
+# [2291. 最大股票收益](https://leetcode.cn/problems/maximum-profit-from-trading-stocks)
 
 [English Version](/solution/2200-2299/2291.Maximum%20Profit%20From%20Trading%20Stocks/README_EN.md)
 
 ## 题目描述
 
 <!-- 这里写题目描述 -->
 
-<p>You are given two <strong>0-indexed</strong> integer arrays of the same length <code>present</code> and <code>future</code> where <code>present[i]</code> is the current price of the <code>i<sup>th</sup></code> stock and <code>future[i]</code> is the price of the <code>i<sup>th</sup></code> stock a year in the future. You may buy each stock at most <strong>once</strong>. You are also given an integer <code>budget</code> representing the amount of money you currently have.</p>
+<p>给你两个下标从 <strong>0</strong>&nbsp;开始的数组 <code>present</code> 和 <code>future</code> ，<code>present[i]</code> 和 <code>future[i]</code> 分别代表第 <code>i</code> 支股票现在和将来的价格。每支股票你最多购买 <strong>一次</strong> ，你的预算为 <code>budget</code> 。</p>
 
-<p>Return <em>the maximum amount of profit you can make.</em></p>
+<p>求最大的收益。</p>
 
 <p>&nbsp;</p>
-<p><strong class="example">Example 1:</strong></p>
+
+<p><strong>示例 1：</strong></p>
 
 <pre>
-<strong>Input:</strong> present = [5,4,6,2,3], future = [8,5,4,3,5], budget = 10
-<strong>Output:</strong> 6
-<strong>Explanation:</strong> One possible way to maximize your profit is to:
-Buy the 0<sup>th</sup>, 3<sup>rd</sup>, and 4<sup>th</sup> stocks for a total of 5 + 2 + 3 = 10.
-Next year, sell all three stocks for a total of 8 + 3 + 5 = 16.
-The profit you made is 16 - 10 = 6.
-It can be shown that the maximum profit you can make is 6.
+<strong>输入：</strong>present = [5,4,6,2,3], future = [8,5,4,3,5], budget = 10
+<strong>输出：</strong>6
+<strong>解释：</strong>你可以选择购买第 0,3,4 支股票获得最大收益：6 。总开销为：5 + 2 + 3 = 10 , 总收益是: 8 + 3 + 5 - 10 = 6 。
 </pre>
 
-<p><strong class="example">Example 2:</strong></p>
+<p><strong>示例 2：</strong></p>
 
 <pre>
-<strong>Input:</strong> present = [2,2,5], future = [3,4,10], budget = 6
-<strong>Output:</strong> 5
-<strong>Explanation:</strong> The only possible way to maximize your profit is to:
-Buy the 2<sup>nd</sup> stock, and make a profit of 10 - 5 = 5.
-It can be shown that the maximum profit you can make is 5.
+<strong>输入：</strong>present = [2,2,5], future = [3,4,10], budget = 6
+<strong>输出：</strong>5
+<strong>解释：</strong>你可以选择购买第 2 支股票获得最大收益：5 。总开销为：5 , 总收益是: 10 - 5 = 5 。
 </pre>
 
-<p><strong class="example">Example 3:</strong></p>
+<p><strong>示例 3：</strong></p>
 
 <pre>
-<strong>Input:</strong> present = [3,3,12], future = [0,3,15], budget = 10
-<strong>Output:</strong> 0
-<strong>Explanation:</strong> One possible way to maximize your profit is to:
-Buy the 1<sup>st</sup> stock, and make a profit of 3 - 3 = 0.
-It can be shown that the maximum profit you can make is 0.
+<strong>输入：</strong>present = [3,3,12], future = [0,3,15], budget = 10
+<strong>输出：</strong>0
+<strong>解释：</strong>你无法购买唯一一支正收益股票 2 ，因此你的收益是 0 。
 </pre>
 
 <p>&nbsp;</p>
-<p><strong>Constraints:</strong></p>
+
+<p><strong>提示：</strong></p>
 
 <ul>
 	<li><code>n == present.length == future.length</code></li>

solution/2400-2499/2496.Maximum Value of a String in an Array/README.md

@@ -0,0 +1,152 @@
+# [2496. 数组中字符串的最大值](https://leetcode.cn/problems/maximum-value-of-a-string-in-an-array)
+
+[English Version](/solution/2400-2499/2496.Maximum%20Value%20of%20a%20String%20in%20an%20Array/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>一个由字母和数字组成的字符串的 <strong>值</strong>&nbsp;定义如下：</p>
+
+<ul>
+	<li>如果字符串 <strong>只</strong> 包含数字，那么值为该字符串在 <code>10</code>&nbsp;进制下的所表示的数字。</li>
+	<li>否则，值为字符串的 <strong>长度&nbsp;</strong>。</li>
+</ul>
+
+<p>给你一个字符串数组&nbsp;<code>strs</code>&nbsp;，每个字符串都只由字母和数字组成，请你返回 <code>strs</code>&nbsp;中字符串的 <strong>最大值</strong>&nbsp;。</p>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre>
+<strong>输入：</strong>strs = ["alic3","bob","3","4","00000"]
+<b>输出：</b>5
+<b>解释：</b>
+- "alic3" 包含字母和数字，所以值为长度 5 。
+- "bob" 只包含字母，所以值为长度 3 。
+- "3" 只包含数字，所以值为 3 。
+- "4" 只包含数字，所以值为 4 。
+- "00000" 只包含数字，所以值为 0 。
+所以最大的值为 5 ，是字符串 "alic3" 的值。
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre>
+<b>输入：</b>strs = ["1","01","001","0001"]
+<b>输出：</b>1
+<b>解释：</b>
+数组中所有字符串的值都是 1 ，所以我们返回 1 。</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= strs.length &lt;= 100</code></li>
+	<li><code>1 &lt;= strs[i].length &lt;= 9</code></li>
+	<li><code>strs[i]</code>&nbsp;只包含小写英文字母和数字。</li>
+</ul>
+
+## 解法
+
+<!-- 这里可写通用的实现逻辑 -->
+
+**方法一：模拟**
+
+根据题意模拟即可。
+
+时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 是数组 `strs` 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 maximumValue(self, strs: List[str]) -> int:
+        def f(s):
+            return int(s) if all(c.isdigit() for c in s) else len(s)
+
+        return max(f(s) for s in strs)
+```
+
+### **Java**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```java
+class Solution {
+    public int maximumValue(String[] strs) {
+        int ans = 0;
+        for (String s : strs) {
+            ans = Math.max(ans, f(s));
+        }
+        return ans;
+    }
+
+    private int f(String s) {
+        for (int i = 0; i < s.length(); ++i) {
+            if (s.charAt(i) >= 'a' && s.charAt(i) <= 'z') {
+                return s.length();
+            }
+        }
+        return Integer.parseInt(s);
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    int maximumValue(vector<string>& strs) {
+        auto f = [](string& s) {
+            int n = s.size(), m = 0;
+            for (char& c : s) {
+                if (!isdigit(c)) return n;
+                m = m * 10 + (c - '0');
+            }
+            return m;
+        };
+        int ans = 0;
+        for (auto& s : strs) ans = max(ans, f(s));
+        return ans;
+    }
+};
+```
+
+### **Go**
+
+```go
+func maximumValue(strs []string) (ans int) {
+	f := func(s string) int {
+		n, m := len(s), 0
+		for _, c := range s {
+			if c >= 'a' && c <= 'z' {
+				return n
+			}
+			m = m*10 + int(c-'0')
+		}
+		return m
+	}
+	for _, s := range strs {
+		if t := f(s); ans < t {
+			ans = t
+		}
+	}
+	return
+}
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/2400-2499/2496.Maximum Value of a String in an Array/README_EN.md

@@ -0,0 +1,137 @@
+# [2496. Maximum Value of a String in an Array](https://leetcode.com/problems/maximum-value-of-a-string-in-an-array)
+
+[中文文档](/solution/2400-2499/2496.Maximum%20Value%20of%20a%20String%20in%20an%20Array/README.md)
+
+## Description
+
+<p>The <strong>value</strong> of an alphanumeric string can be defined as:</p>
+
+<ul>
+	<li>The <strong>numeric</strong> representation of the string in base <code>10</code>, if it comprises of digits <strong>only</strong>.</li>
+	<li>The <strong>length</strong> of the string, otherwise.</li>
+</ul>
+
+<p>Given an array <code>strs</code> of alphanumeric strings, return <em>the <strong>maximum value</strong> of any string in </em><code>strs</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> strs = [&quot;alic3&quot;,&quot;bob&quot;,&quot;3&quot;,&quot;4&quot;,&quot;00000&quot;]
+<strong>Output:</strong> 5
+<strong>Explanation:</strong> 
+- &quot;alic3&quot; consists of both letters and digits, so its value is its length, i.e. 5.
+- &quot;bob&quot; consists only of letters, so its value is also its length, i.e. 3.
+- &quot;3&quot; consists only of digits, so its value is its numeric equivalent, i.e. 3.
+- &quot;4&quot; also consists only of digits, so its value is 4.
+- &quot;00000&quot; consists only of digits, so its value is 0.
+Hence, the maximum value is 5, of &quot;alic3&quot;.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> strs = [&quot;1&quot;,&quot;01&quot;,&quot;001&quot;,&quot;0001&quot;]
+<strong>Output:</strong> 1
+<strong>Explanation:</strong> 
+Each string in the array has value 1. Hence, we return 1.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= strs.length &lt;= 100</code></li>
+	<li><code>1 &lt;= strs[i].length &lt;= 9</code></li>
+	<li><code>strs[i]</code> consists of only lowercase English letters and digits.</li>
+</ul>
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **Python3**
+
+```python
+class Solution:
+    def maximumValue(self, strs: List[str]) -> int:
+        def f(s):
+            return int(s) if all(c.isdigit() for c in s) else len(s)
+
+        return max(f(s) for s in strs)
+```
+
+### **Java**
+
+```java
+class Solution {
+    public int maximumValue(String[] strs) {
+        int ans = 0;
+        for (String s : strs) {
+            ans = Math.max(ans, f(s));
+        }
+        return ans;
+    }
+
+    private int f(String s) {
+        for (int i = 0; i < s.length(); ++i) {
+            if (s.charAt(i) >= 'a' && s.charAt(i) <= 'z') {
+                return s.length();
+            }
+        }
+        return Integer.parseInt(s);
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    int maximumValue(vector<string>& strs) {
+        auto f = [](string& s) {
+            int n = s.size(), m = 0;
+            for (char& c : s) {
+                if (!isdigit(c)) return n;
+                m = m * 10 + (c - '0');
+            }
+            return m;
+        };
+        int ans = 0;
+        for (auto& s : strs) ans = max(ans, f(s));
+        return ans;
+    }
+};
+```
+
+### **Go**
+
+```go
+func maximumValue(strs []string) (ans int) {
+	f := func(s string) int {
+		n, m := len(s), 0
+		for _, c := range s {
+			if c >= 'a' && c <= 'z' {
+				return n
+			}
+			m = m*10 + int(c-'0')
+		}
+		return m
+	}
+	for _, s := range strs {
+		if t := f(s); ans < t {
+			ans = t
+		}
+	}
+	return
+}
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->