docs(content): add 290 Word Pattern page and fix String badge count

apps/docs/content/(01-linear-structures)/hash-table/290-word-pattern.mdx

@@ -0,0 +1,57 @@
+---
+title: '290. Word Pattern'
+description: Given a pattern and a string s, find if s follows the same pattern
+icon: dot
+topics:
+  - { name: "Hash Table", slug: "hash-table" }
+  - { name: "String", slug: "string" }
+issue: "https://github.com/prdlk/leetcode/issues/35"
+sidebar:
+  label: 'Word Pattern'
+  badge: 'Easy'
+---
+
+### Example 1:
+- Input: `pattern = "abba", s = "dog cat cat dog"`
+- Output: `true`
+- Explanation: The bijection can be established as: • 'a' maps to "dog". • 'b' maps to "cat".
+
+### Example 2:
+- Input: `pattern = "abba", s = "dog cat cat fish"`
+- Output: `false`
+
+### Example 3:
+- Input: `pattern = "aaaa", s = "dog cat cat dog"`
+- Output: `false`
+
+### Constraints:
+
+- `1 <= pattern.length <= 300`
+- `pattern` contains only lower-case English letters.
+- `1 <= s.length <= 3000`
+- `s` contains only lowercase English letters and spaces ' '.
+- `s` does not contain any leading or trailing spaces.
+- All the words in `s` are separated by a single space.
+
+## Solution
+
+```py
+class Solution:
+    def wordPattern(self, pattern: str, s: str) -> bool:
+        if len(pattern) != len(s.split()):
+            return False
+
+        words = s.split()
+        map = {}
+
+        for i in range(len(pattern)):
+            c = pattern[i]
+            w = words[i]
+
+            if c in map and map[c] != w:
+                return False
+            else:
+                map[c] = w
+
+        return len(set(pattern)) == len(set(words))
+```

apps/docs/content/appendix/(1-data-structures)/string.mdx

@@ -3,7 +3,7 @@ title: "String"
 list: "https://leetcode.com/problem-list/string/"
 icon: quote
 sidebar:
-  badge: '17'
+  badge: '18'
 ---
 
 :::note