docs(leetcode): add solution and documentation for 1365. How Many Numbers Are Smaller Than the Current Number

docs/content/docs/leetcode/1365.mdx

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+---
+title: LC-1365
+description: How Many Numbers Are Smaller Than the Current Number
+---
+
+## OverviewProblem solving guide for Technical Interviews
+
+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].
+
+### Examples
+
+| Control            | Purpose               | Default  |
+| ------------------ | --------------------- | -------- |
+| `--typeset-size`   | Base text size        | `1em`    |
+| `--typeset-leading`| Line height           | `1.75`   |
+| `--typeset-flow`   | Space between blocks  | `1.25em` |
+
+
+### Constraints

work/Baton-1/Easy/Array/1365.how-many-numbers-are-smaller-than-the-current-number.js

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+/*
+ * 1365. How Many Numbers Are Smaller Than the Current Number
+ * Difficulty: Easy
+ * https://leetcode.com/problems/how-many-numbers-are-smaller-than-the-current-number/
+ *
+ * ──────────────────────────────────────────────────
+ *
+ * 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].
+ *
+ * Return the answer in an array.
+ *
+ *
+ *
+ * Example 1:
+ *
+ * Input: nums = [8,1,2,2,3]
+ * Output: [4,0,1,1,3]
+ * Explanation:
+ * For nums[0]=8 there exist four smaller numbers than it (1, 2, 2 and
+ * 3).
+ * For nums[1]=1 does not exist any smaller number than it.
+ * For nums[2]=2 there exist one smaller number than it (1).
+ * For nums[3]=2 there exist one smaller number than it (1).
+ * For nums[4]=3 there exist three smaller numbers than it (1, 2 and 2).
+ *
+ * Example 2:
+ *
+ * Input: nums = [6,5,4,8]
+ * Output: [2,1,0,3]
+ *
+ * Example 3:
+ *
+ * Input: nums = [7,7,7,7]
+ * Output: [0,0,0,0]
+ *
+ *
+ *
+ * Constraints:
+ *
+ * 	• 2 <= nums.length <= 500
+ *
+ * 	• 0 <= nums[i] <= 100
+ */
+
+/**
+ * @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]);
+};

work/Baton-1/Medium/Hash Table/451.sort-characters-by-frequency.js

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+/*
+ * 451. Sort Characters By Frequency
+ * Difficulty: Medium
+ * https://leetcode.com/problems/sort-characters-by-frequency/
+ * 
+ * ──────────────────────────────────────────────────
+ * 
+ * Given a string s, sort it in decreasing order based on the frequency
+ * of the characters. The frequency of a character is the number of times
+ * it appears in the string.
+ * 
+ * Return the sorted string. If there are multiple answers, return any
+ * of them.
+ * 
+ *  
+ * 
+ * Example 1:
+ * 
+ * Input: s = "tree"
+ * Output: "eert"
+ * Explanation: 'e' appears twice while 'r' and 't' both appear once.
+ * So 'e' must appear before both 'r' and 't'. Therefore "eetr" is also
+ * a valid answer.
+ * 
+ * Example 2:
+ * 
+ * Input: s = "cccaaa"
+ * Output: "aaaccc"
+ * Explanation: Both 'c' and 'a' appear three times, so both "cccaaa"
+ * and "aaaccc" are valid answers.
+ * Note that "cacaca" is incorrect, as the same characters must be
+ * together.
+ * 
+ * Example 3:
+ * 
+ * Input: s = "Aabb"
+ * Output: "bbAa"
+ * Explanation: "bbaA" is also a valid answer, but "Aabb" is incorrect.
+ * Note that 'A' and 'a' are treated as two different characters.
+ * 
+ *  
+ * 
+ * Constraints:
+ * 
+ * 	• 1 <= s.length <= 5 * 10^5
+ * 
+ * 	• s consists of uppercase and lowercase English letters and digits.
+*/
+
+/**
+ * @param {string} s
+ * @return {string}
+ */
+var frequencySort = function(s) {
+    
+};