feat: add solutions to lc problems: No.2404~2407

* No.2404.Most Frequent Even Element
* No.2405.Optimal Partition of String
* No.2406.Divide Intervals Into Minimum Number of Groups
* No.2407.Longest Increasing Subsequence II

Author: yanglbme

lcci/03.03.Stack of Plates/README.md

@@ -23,6 +23,7 @@
 </pre>
 
 ## 解法
+
 <!-- 这里可写通用的实现逻辑 -->
 
 <!-- tabs:start -->
@@ -34,6 +35,7 @@
 ```python
 
 ```
+
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->

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/0400-0499/0484.Find Permutation/Solution.py

@@ -7,6 +7,6 @@ def findPermutation(self, s: str) -> List[int]:
             j = i
             while j < n and s[j] == 'D':
                 j += 1
-            ans[i: j + 1] = ans[i: j + 1][::-1]
+            ans[i : j + 1] = ans[i : j + 1][::-1]
             i = max(i + 1, j)
         return ans

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/0500-0599/0576.Out of Boundary Paths/Solution.py

@@ -1,5 +1,7 @@
 class Solution:
-    def findPaths(self, m: int, n: int, maxMove: int, startRow: int, startColumn: int) -> int:
+    def findPaths(
+        self, m: int, n: int, maxMove: int, startRow: int, startColumn: int
+    ) -> int:
         @cache
         def dfs(i, j, k):
             if i < 0 or j < 0 or i >= m or j >= n:

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/0700-0799/0776.Split BST/Solution.py

@@ -5,7 +5,9 @@
 #         self.left = left
 #         self.right = right
 class Solution:
-    def splitBST(self, root: Optional[TreeNode], target: int) -> List[Optional[TreeNode]]:
+    def splitBST(
+        self, root: Optional[TreeNode], target: int
+    ) -> List[Optional[TreeNode]]:
         def dfs(root):
             if root is None:
                 return [None, None]

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/0800-0899/0857.Minimum Cost to Hire K Workers/Solution.py

@@ -1,5 +1,7 @@
 class Solution:
-    def mincostToHireWorkers(self, quality: List[int], wage: List[int], k: int) -> float:
+    def mincostToHireWorkers(
+        self, quality: List[int], wage: List[int], k: int
+    ) -> float:
         t = sorted(zip(quality, wage), key=lambda x: x[1] / x[0])
         ans, tot = inf, 0
         h = []

solution/1000-1099/1032.Stream of Characters/Solution.py

@@ -25,7 +25,6 @@ def search(self, w):
 
 
 class StreamChecker:
-
     def __init__(self, words: List[str]):
         self.trie = Trie()
         self.s = []
@@ -36,6 +35,7 @@ def query(self, letter: str) -> bool:
         self.s.append(letter)
         return self.trie.search(self.s[-201:])
 
+
 # Your StreamChecker object will be instantiated and called as such:
 # obj = StreamChecker(words)
 # param_1 = obj.query(letter)

solution/1500-1599/1593.Split a String Into the Max Number of Unique Substrings/Solution.py

@@ -6,10 +6,10 @@ def dfs(i, t):
                 ans = max(ans, t)
                 return
             for j in range(i + 1, len(s) + 1):
-                if s[i: j] not in vis:
-                    vis.add(s[i: j])
+                if s[i:j] not in vis:
+                    vis.add(s[i:j])
                     dfs(j, t + 1)
-                    vis.remove(s[i: j])
+                    vis.remove(s[i:j])
 
         vis = set()
         ans = 1

solution/1600-1699/1600.Throne Inheritance/Solution.py

@@ -1,5 +1,4 @@
 class ThroneInheritance:
-
     def __init__(self, kingName: str):
         self.g = defaultdict(list)
         self.dead = set()
@@ -22,6 +21,7 @@ def dfs(x):
         dfs(self.king)
         return ans
 
+
 # Your ThroneInheritance object will be instantiated and called as such:
 # obj = ThroneInheritance(kingName)
 # obj.birth(parentName,childName)

solution/1600-1699/1603.Design Parking System/Solution.py

@@ -1,5 +1,4 @@
 class ParkingSystem:
-
     def __init__(self, big: int, medium: int, small: int):
         self.cnt = [0, big, medium, small]
 

solution/2400-2499/2403.Minimum Time to Kill All Monsters/README.md

@@ -297,7 +297,6 @@ func min(a, b int64) int64 {
 }
 ```
 
-
 ### **TypeScript**
 
 ```ts

solution/2400-2499/2403.Minimum Time to Kill All Monsters/README_EN.md

@@ -177,7 +177,7 @@ public:
         n = power.size();
         f.assign(1 << n, -1);
         this->power = power;
-        return dfs(0);    
+        return dfs(0);
     }
 
     ll dfs(int mask) {

solution/2400-2499/2404.Most Frequent Even Element/README.md

@@ -0,0 +1,158 @@
+# [2404. 出现最频繁的偶数元素](https://leetcode.cn/problems/most-frequent-even-element)
+
+[English Version](/solution/2400-2499/2404.Most%20Frequent%20Even%20Element/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>给你一个整数数组 <code>nums</code> ，返回出现最频繁的偶数元素。</p>
+
+<p>如果存在多个满足条件的元素，只需要返回 <strong>最小</strong> 的一个。如果不存在这样的元素，返回 <code>-1</code> 。</p>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre><strong>输入：</strong>nums = [0,1,2,2,4,4,1]
+<strong>输出：</strong>2
+<strong>解释：</strong>
+数组中的偶数元素为 0、2 和 4 ，在这些元素中，2 和 4 出现次数最多。
+返回最小的那个，即返回 2 。</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre><strong>输入：</strong>nums = [4,4,4,9,2,4]
+<strong>输出：</strong>4
+<strong>解释：</strong>4 是出现最频繁的偶数元素。
+</pre>
+
+<p><strong>示例 3：</strong></p>
+
+<pre><strong>输入：</strong>nums = [29,47,21,41,13,37,25,7]
+<strong>输出：</strong>-1
+<strong>解释：</strong>不存在偶数元素。
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 2000</code></li>
+	<li><code>0 &lt;= nums[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：哈希表**
+
+用哈希表统计所有偶数元素出现的次数，然后找出出现次数最多且值最小的偶数元素。
+
+时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是数组的长度。
+
+<!-- tabs:start -->
+
+### **Python3**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```python
+class Solution:
+    def mostFrequentEven(self, nums: List[int]) -> int:
+        cnt = Counter(v for v in nums if v % 2 == 0)
+        ans, mx = -1, 0
+        for v, t in cnt.items():
+            if mx < t or (mx == t and ans > v):
+                mx = t
+                ans = v
+        return ans
+```
+
+### **Java**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```java
+class Solution {
+    public int mostFrequentEven(int[] nums) {
+        Map<Integer, Integer> cnt = new HashMap<>();
+        for (int v : nums) {
+            if (v % 2 == 0) {
+                cnt.put(v, cnt.getOrDefault(v, 0) + 1);
+            }
+        }
+        int ans = -1, mx = 0;
+        for (var e : cnt.entrySet()) {
+            int v = e.getKey(), t = e.getValue();
+            if (mx < t || (mx == t && ans > v)) {
+                mx = t;
+                ans = v;
+            }
+        }
+        return ans;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    int mostFrequentEven(vector<int>& nums) {
+        unordered_map<int, int> cnt;
+        for (int v : nums) {
+            if (v % 2 == 0) {
+                ++cnt[v];
+            }
+        }
+        int ans = -1, mx = 0;
+        for (auto [v, t] : cnt) {
+            if (mx < t || (mx == t && ans > v)) {
+                mx = t;
+                ans = v;
+            }
+        }
+        return ans;
+    }
+};
+```
+
+### **Go**
+
+```go
+func mostFrequentEven(nums []int) int {
+	cnt := map[int]int{}
+	for _, v := range nums {
+		if v%2 == 0 {
+			cnt[v]++
+		}
+	}
+	ans, mx := -1, 0
+	for v, t := range cnt {
+		if mx < t || (mx == t && ans > v) {
+			mx = t
+			ans = v
+		}
+	}
+	return ans
+}
+```
+
+### **TypeScript**
+
+```ts
+
+```
+
+### **...**
+
+```
+
+
+```
+
+<!-- tabs:end -->