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)/347_top-k-frequent-elements.mdx

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+---
+title: '347. Top K Frequent Elements'
+description: Given an integer array nums and an integer k, return the k most frequent elements. You may return the answer in any order
+icon: dot
+topics:
+  - { name: "Array", slug: "array" }
+  - { name: "Hash Table", slug: "hash-table" }
+  - { name: "Divide and Conquer", slug: "divide-and-conquer" }
+  - { name: "Sorting", slug: "sorting" }
+  - { name: "Heap (Priority Queue)", slug: "heap-priority-queue" }
+  - { name: "Bucket Sort", slug: "bucket-sort" }
+  - { name: "Counting", slug: "counting" }
+  - { name: "Quickselect", slug: "quickselect" }
+video: "YPTqKIgVk-k"
+issue: "https://github.com/prdlk/leetcode/issues/31"
+sidebar:
+  label: 'Top K Frequent Elements'
+  badge: 'Medium'
+---
+
+::::warning
+Your algorithm's time complexity must be better than O(n log n), where n is the array's size.
+::::
+
+### Example 1:
+- Input: `nums = [1,1,1,2,2,3], k = 2`
+- Output: `[1,2]`
+
+### Example 2:
+- Input: `nums = [1], k = 1`
+- Output: `[1]`
+
+### Example 3:
+- Input: `nums = [1,2,1,2,1,2,3,1,3,2], k = 2`
+- Output: `[1,2]`
+
+### Constraints:
+
+- `1 <= nums.length <= 10^5`
+- `-10^4 <= nums[i] <= 10^4`
+- `k` is in the range `[1, the number of unique elements in the array]`.
+- It is guaranteed that the answer is unique.
+
+## Approach
+
+```mermaid
+flowchart TD
+  S(["topKFrequent(nums, k)"]) --> F["freq: count every value in nums"]
+  F --> K["keys of freq, cast to Number"]
+  K --> O["sort by freq[b] - freq[a] — most frequent first"]
+  O --> R(["return slice(0, k)"])
+```
+
+## Solution
+
+```js
+/**
+ * @param {number[]} nums
+ * @param {number} k
+ * @return {number[]}
+ */
+var topKFrequent = function (nums, k) {
+  const freq = {};
+  for (let n of nums) freq[n] = (freq[n] || 0) + 1;
+
+  return Object.keys(freq)
+    .map(Number)
+    .sort((a, b) => freq[b] - freq[a])
+    .slice(0, k);
+};
+```