feat(content): add new problem pages, reorganize docs into (problems) subfolders, and update meta display to page

apps/docs/content/(01-linear-structures)/hash-table/(problems)/49_group-anagrams.mdx

@@ -0,0 +1,72 @@
+---
+title: '49. Group Anagrams'
+description: Given an array of strings strs, group the anagrams together. You can return the answer in any order
+icon: dot
+topics:
+  - { name: "Array", slug: "array" }
+  - { name: "Hash Table", slug: "hash-table" }
+  - { name: "String", slug: "string" }
+  - { name: "Sorting", slug: "sorting" }
+video: "vzdNOK2oB2E"
+issue: "https://github.com/prdlk/leetcode/issues/30"
+sidebar:
+  label: 'Group Anagrams'
+  badge: 'Medium'
+---
+
+### Example 1:
+- Input: `strs = ["eat","tea","tan","ate","nat","bat"]`
+- Output: `[["bat"],["nat","tan"],["ate","eat","tea"]]`
+- Explanation: There is no string in `strs` that can be rearranged to form `"bat"`. The strings `"nat"` and `"tan"` are anagrams as they can be rearranged to form each other. The strings `"ate"`, `"eat"`, and `"tea"` are anagrams as they can be rearranged to form each other.
+
+### Example 2:
+- Input: `strs = [""]`
+- Output: `[[""]]`
+
+### Example 3:
+- Input: `strs = ["a"]`
+- Output: `[["a"]]`
+
+### Constraints:
+
+- `1 <= strs.length <= 10^4`
+- `0 <= strs[i].length <= 100`
+- `strs[i]` consists of lowercase English letters.
+
+## Approach
+
+```mermaid
+flowchart TD
+  S(["groupAnagrams(strs)"]) --> K["sorted[i] = letters of strs[i], sorted — the anagram key"]
+  K --> I["anagrams = {}"]
+  I --> L{"more i in 0..n-1?"}
+  L -- no --> E(["return Object.values(anagrams)"])
+  L -- yes --> Q{"anagrams has sorted[i]?"}
+  Q -- no --> N["anagrams[sorted[i]] = [strs[i]]"]
+  Q -- yes --> P["anagrams[sorted[i]].push(strs[i])"]
+  N --> L
+  P --> L
+```
+
+## Solution
+
+```js
+/**
+ * @param {string[]} strs
+ * @return {string[][]}
+ */
+var groupAnagrams = function(strs) {
+  let sorted = strs.map(str => str.split("").sort().join(""));
+  let anagrams = {};
+
+  for (let i = 0; i < strs.length; i++){
+    if(!anagrams[sorted[i]]){
+      anagrams[sorted[i]] = [strs[i]]
+    }else{
+      anagrams[sorted[i]].push(strs[i])
+    }
+  }
+
+  return Object.values(anagrams);
+};
+```