feat: add solutions to lc problems: No.2351~2354

* No.2351.First Letter to Appear Twice
* No.2352.Equal Row and Column Pairs
* No.2353.Design a Food Rating System
* No.2354.Number of Excellent Pairs

Author: yanglbme

lcof2/剑指 Offer II 043. 往完全二叉树添加节点/README.md

@@ -136,7 +136,7 @@ class CBTInserter {
             }
         }
     }
-    
+
     public int insert(int v) {
         int pid = (tree.size() - 1) >> 1;
         TreeNode node = new TreeNode(v);
@@ -149,7 +149,7 @@ class CBTInserter {
         }
         return p.val;
     }
-    
+
     public TreeNode get_root() {
         return tree.get(0);
     }
@@ -192,7 +192,7 @@ public:
             if (node->right) q.push(node->right);
         }
     }
-    
+
     int insert(int v) {
         int pid = tree.size() - 1 >> 1;
         TreeNode* node = new TreeNode(v);
@@ -202,7 +202,7 @@ public:
         else p->right = node;
         return p->val;
     }
-    
+
     TreeNode* get_root() {
         return tree[0];
     }

solution/0900-0999/0919.Complete Binary Tree Inserter/README.md

@@ -141,7 +141,7 @@ class CBTInserter {
             }
         }
     }
-    
+
     public int insert(int val) {
         int pid = (tree.size() - 1) >> 1;
         TreeNode node = new TreeNode(val);
@@ -154,7 +154,7 @@ class CBTInserter {
         }
         return p.val;
     }
-    
+
     public TreeNode get_root() {
         return tree.get(0);
     }
@@ -197,7 +197,7 @@ public:
             if (node->right) q.push(node->right);
         }
     }
-    
+
     int insert(int val) {
         int pid = tree.size() - 1 >> 1;
         TreeNode* node = new TreeNode(val);
@@ -207,7 +207,7 @@ public:
         else p->right = node;
         return p->val;
     }
-    
+
     TreeNode* get_root() {
         return tree[0];
     }

solution/0900-0999/0919.Complete Binary Tree Inserter/README_EN.md

@@ -128,7 +128,7 @@ class CBTInserter {
             }
         }
     }
-    
+
     public int insert(int val) {
         int pid = (tree.size() - 1) >> 1;
         TreeNode node = new TreeNode(val);
@@ -141,7 +141,7 @@ class CBTInserter {
         }
         return p.val;
     }
-    
+
     public TreeNode get_root() {
         return tree.get(0);
     }
@@ -184,7 +184,7 @@ public:
             if (node->right) q.push(node->right);
         }
     }
-    
+
     int insert(int val) {
         int pid = tree.size() - 1 >> 1;
         TreeNode* node = new TreeNode(val);
@@ -194,7 +194,7 @@ public:
         else p->right = node;
         return p->val;
     }
-    
+
     TreeNode* get_root() {
         return tree[0];
     }

solution/1100-1199/1184.Distance Between Bus Stops/Solution.py

@@ -1,7 +1,9 @@
 class Solution:
-    def distanceBetweenBusStops(self, distance: List[int], start: int, destination: int) -> int:
+    def distanceBetweenBusStops(
+        self, distance: List[int], start: int, destination: int
+    ) -> int:
         if start > destination:
             start, destination = destination, start
-        a = sum(distance[start: destination])
+        a = sum(distance[start:destination])
         b = sum(distance[:start]) + sum(distance[destination:])
         return min(a, b)

solution/2000-2099/2062.Count Vowel Substrings of a String/Solution.py

@@ -2,4 +2,4 @@ class Solution:
     def countVowelSubstrings(self, word: str) -> int:
         n = len(word)
         s = set('aeiou')
-        return sum(set(word[i: j]) == s for i in range(n) for j in range(i + 1, n + 1))
+        return sum(set(word[i:j]) == s for i in range(n) for j in range(i + 1, n + 1))

solution/2300-2399/2348.Number of Zero-Filled Subarrays/README.md

@@ -48,7 +48,6 @@
 	<li><code>-10<sup>9</sup> &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
 </ul>
 
-
 ## 解法
 
 <!-- 这里可写通用的实现逻辑 -->

solution/2300-2399/2348.Number of Zero-Filled Subarrays/README_EN.md

@@ -47,7 +47,6 @@ There is no occurrence of a subarray with a size more than 3 filled with 0. Ther
 	<li><code>-10<sup>9</sup> &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
 </ul>
 
-
 ## Solutions
 
 <!-- tabs:start -->

solution/2300-2399/2349.Design a Number Container System/README.md

@@ -53,7 +53,6 @@ nc.find(10); // 数字 10 所在下标为 2 ，3 和 5 。最小下标为 2 ，
 	<li>调用&nbsp;<code>change</code> 和&nbsp;<code>find</code>&nbsp;的&nbsp;<strong>总次数</strong>&nbsp;不超过&nbsp;<code>10<sup>5</sup></code> 次。</li>
 </ul>
 
-
 ## 解法
 
 <!-- 这里可写通用的实现逻辑 -->
@@ -106,7 +105,7 @@ class NumberContainers {
     public NumberContainers() {
 
     }
-    
+
     public void change(int index, int number) {
         if (mp.containsKey(index)) {
             int v = mp.get(index);
@@ -118,7 +117,7 @@ class NumberContainers {
         mp.put(index, number);
         t.computeIfAbsent(number, k -> new TreeSet<>()).add(index);
     }
-    
+
     public int find(int number) {
         return t.containsKey(number) ? t.get(number).first() : -1;
     }
@@ -143,7 +142,7 @@ public:
     NumberContainers() {
 
     }
-    
+
     void change(int index, int number) {
         auto it = mp.find(index);
         if (it != mp.end())
@@ -154,7 +153,7 @@ public:
         else mp[index] = number;
         t[number].insert(index);
     }
-    
+
     int find(int number) {
         auto it = t.find(number);
         return it == t.end() || it->second.empty() ? -1 : *it->second.begin();

solution/2300-2399/2349.Design a Number Container System/README_EN.md

@@ -49,7 +49,6 @@ nc.find(10); // Number 10 is at the indices 2, 3, and 5. The smallest index that
 	<li>At most <code>10<sup>5</sup></code> calls will be made <strong>in total</strong> to <code>change</code> and <code>find</code>.</li>
 </ul>
 
-
 ## Solutions
 
 <!-- tabs:start -->
@@ -94,7 +93,7 @@ class NumberContainers {
     public NumberContainers() {
 
     }
-    
+
     public void change(int index, int number) {
         if (mp.containsKey(index)) {
             int v = mp.get(index);
@@ -106,7 +105,7 @@ class NumberContainers {
         mp.put(index, number);
         t.computeIfAbsent(number, k -> new TreeSet<>()).add(index);
     }
-    
+
     public int find(int number) {
         return t.containsKey(number) ? t.get(number).first() : -1;
     }
@@ -131,7 +130,7 @@ public:
     NumberContainers() {
 
     }
-    
+
     void change(int index, int number) {
         auto it = mp.find(index);
         if (it != mp.end())
@@ -142,7 +141,7 @@ public:
         else mp[index] = number;
         t[number].insert(index);
     }
-    
+
     int find(int number) {
         auto it = t.find(number);
         return it == t.end() || it->second.empty() ? -1 : *it->second.begin();

solution/2300-2399/2349.Design a Number Container System/Solution.py

@@ -2,7 +2,6 @@
 
 
 class NumberContainers:
-
     def __init__(self):
         self.mp = {}
         self.t = defaultdict(SortedSet)

solution/2300-2399/2350.Shortest Impossible Sequence of Rolls/README.md

@@ -55,7 +55,6 @@
 	<li><code>1 &lt;= rolls[i] &lt;= k &lt;= 10<sup>5</sup></code></li>
 </ul>
 
-
 ## 解法
 
 <!-- 这里可写通用的实现逻辑 -->

solution/2300-2399/2350.Shortest Impossible Sequence of Rolls/README_EN.md

@@ -51,7 +51,6 @@ Note that there are other sequences that cannot be taken from rolls but [4] is t
 	<li><code>1 &lt;= rolls[i] &lt;= k &lt;= 10<sup>5</sup></code></li>
 </ul>
 
-
 ## Solutions
 
 <!-- tabs:start -->

solution/2300-2399/2351.First Letter to Appear Twice/README.md

@@ -0,0 +1,128 @@
+# [2351. 第一个出现两次的字母](https://leetcode.cn/problems/first-letter-to-appear-twice)
+
+[English Version](/solution/2300-2399/2351.First%20Letter%20to%20Appear%20Twice/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>给你一个由小写英文字母组成的字符串 <code>s</code> ，请你找出并返回第一个出现 <strong>两次</strong> 的字母。</p>
+
+<p><strong>注意：</strong></p>
+
+<ul>
+	<li>如果 <code>a</code> 的 <strong>第二次</strong> 出现比 <code>b</code> 的 <strong>第二次</strong> 出现在字符串中的位置更靠前，则认为字母 <code>a</code> 在字母 <code>b</code> 之前出现两次。</li>
+	<li><code>s</code> 包含至少一个出现两次的字母。</li>
+</ul>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre><strong>输入：</strong>s = "abccbaacz"
+<strong>输出：</strong>"c"
+<strong>解释：</strong>
+字母 'a' 在下标 0 、5 和 6 处出现。
+字母 'b' 在下标 1 和 4 处出现。
+字母 'c' 在下标 2 、3 和 7 处出现。
+字母 'z' 在下标 8 处出现。
+字母 'c' 是第一个出现两次的字母，因为在所有字母中，'c' 第二次出现的下标是最小的。
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre><strong>输入：</strong>s = "abcdd"
+<strong>输出：</strong>"d"
+<strong>解释：</strong>
+只有字母 'd' 出现两次，所以返回 'd' 。
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>2 &lt;= s.length &lt;= 100</code></li>
+	<li><code>s</code> 由小写英文字母组成</li>
+	<li><code>s</code> 包含至少一个重复字母</li>
+</ul>
+
+## 解法
+
+<!-- 这里可写通用的实现逻辑 -->
+
+<!-- tabs:start -->
+
+### **Python3**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```python
+class Solution:
+    def repeatedCharacter(self, s: str) -> str:
+        cnt = Counter()
+        for v in s:
+            cnt[v] += 1
+            if cnt[v] == 2:
+                return v
+```
+
+### **Java**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```java
+class Solution {
+    public char repeatedCharacter(String s) {
+        int[] cnt = new int[26];
+        for (char c : s.toCharArray()) {
+            if (++cnt[c - 'a'] == 2) {
+                return c;
+            }
+        }
+        return '.';
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    char repeatedCharacter(string s) {
+        vector<int> cnt(26);
+        for (char c : s) if (++cnt[c - 'a'] == 2) return c;
+        return '.';
+    }
+};
+```
+
+### **Go**
+
+```go
+func repeatedCharacter(s string) byte {
+	cnt := make([]int, 26)
+	for _, c := range s {
+		cnt[c-'a']++
+		if cnt[c-'a'] == 2 {
+			return byte(c)
+		}
+	}
+	return '.'
+}
+```
+
+### **TypeScript**
+
+```ts
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->