docs(welcome): restructure documentation directories and add new guides section

docs/(welcome)/array/1365.mdx

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+---
+title: LC-1365
+description: How Many Numbers Are Smaller Than the Current Number
+---
+
+## Examples
+
+| Input            | Output               |
+| ---------------- | -------------------- |
+| `[8,1,2,2,3]`    | `[4,0,1,1,3]`        |
+| `[6,5,4,8]`      | `[2,1,0,3]`          |
+| `[7,7,7,7]`      | `[0,0,0,0]`          |
+
+## Hints
+
+<Accordion type="single">
+  <AccordionItem title="Constraints">
+    Constraints:
+     • `2 <= nums.length <= 500`
+     • `0 <= nums[i] <= 100`
+  </AccordionItem>
+  <AccordionItem title="Problem Statement">
+    Given the array nums, for each `nums[i]` find out how many numbers in the array are smaller than it. 
+    That is, for each `nums[i]` you have to count the number of valid `j`'s such that `j != i` and `nums[j] < nums[i]`.
+  </AccordionItem>
+</Accordion>
+
+## Solution
+
+```js
+/**
+ * @param {number[]} nums
+ * @return {number[]}
+ */
+
+var smallerNumbersThanCurrent = function (nums) {
+  // Default for Hammer
+  const freq = {};
+  for (let n of nums) freq[n] = (freq[n] || 0) + 1;
+  const sorted = Object.keys(freq).sort((a, b) => a - b);
+
+  // Problem Specific
+  let count = 0;
+  let smaller = {};
+
+  // Iterate over sorted.
+  for (let num of sorted) {
+    smaller[num] = count;
+    count += freq[num];
+  }
+
+  // Map output
+  return nums.map((n) => smaller[n]);
+};
+```

docs/(welcome)/array/1636.mdx

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+---
+title: LC-1636
+description: Sort Array by Increasing Frequency
+---
+
+## Examples
+
+| Input                  | Output                 |
+| ---------------------- | ---------------------- |
+| `[1,1,2,2,2,3]`        | `[3,1,1,2,2,2]`        |
+| `[2,3,1,3,2]`          | `[1,3,3,2,2]`          |
+| `[-1,1,-6,4,5,-6,1,4,1]` | `[5,-1,4,4,-6,-6,1,1,1]` |
+
+## Hints
+
+<Accordion type="single">
+  <AccordionItem title="Constraints">
+    Constraints:
+     • `1 <= nums.length <= 100`
+     • `-100 <= nums[i] <= 100`
+  </AccordionItem>
+  <AccordionItem title="Problem Statement">
+    Given an array of integers nums, sort the array in increasing order based on the frequency of the values.
+    If multiple values have the same frequency, sort them in decreasing order.
+    Return the sorted array.
+  </AccordionItem>
+</Accordion>
+
+## Solution
+
+```js
+/**
+ * @param {number[]} nums
+ * @return {number[]}
+ */
+
+var frequencySort = function (nums) {
+  // Default for Hammer
+  const freq = {};
+  for (let n of nums) freq[n] = (freq[n] || 0) + 1;
+
+  // Problem Specific
+  // Sort ascending by frequency; ties break by larger value first
+  return nums.sort((a, b) => freq[a] - freq[b] || b - a);
+};
+```

docs/(welcome)/array/347.mdx

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+---
+title: LC-347
+description: Top K Frequent Elements
+---
+
+## Examples
+
+| Input                              | Output   |
+| ---------------------------------- | -------- |
+| `nums = [1,1,1,2,2,3], k = 2`      | `[1,2]`  |
+| `nums = [1], k = 1`                | `[1]`    |
+| `nums = [1,2,1,2,1,2,3,1,3,2], k = 2` | `[1,2]` |
+
+## Hints
+
+<Accordion type="single">
+  <AccordionItem title="Constraints">
+    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.
+  </AccordionItem>
+  <AccordionItem title="Problem Statement">
+    Given an integer array nums and an integer `k`, return the `k` most frequent elements.
+    You may return the answer in any order.
+
+    Follow up: Your algorithm's time complexity must be better than `O(n log n)`, where n is the array's size.
+  </AccordionItem>
+</Accordion>
+
+## Solution
+
+```js
+/**
+ * @param {number[]} nums
+ * @param {number} k
+ * @return {number[]}
+ */
+
+var topKFrequent = function (nums, k) {
+  // Default for Hammer
+  const freq = {};
+  for (let n of nums) freq[n] = (freq[n] || 0) + 1;
+
+  // Problem Specific
+  // Sort unique values by descending frequency, take the first k
+  return Object.keys(freq)
+    .map(Number)
+    .sort((a, b) => freq[b] - freq[a])
+    .slice(0, k);
+};
+```