diff --git a/work/Default/Easy/Array/1365.how-many-numbers-are-smaller-than-the-current-number.js b/work/Default/Easy/Array/1365.how-many-numbers-are-smaller-than-the-current-number.js
deleted file mode 100644
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--- a/work/Default/Easy/Array/1365.how-many-numbers-are-smaller-than-the-current-number.js
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@@ -1,77 +0,0 @@
-/*
- * 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) {
-  // Step 1: Begin by initializing a [Frequency Map]()
-  const freq = {};
-  for (let n of nums) freq[n] = (freq[n] || 0) + 1;
-
-  // Step 2: Sort the numbers by ascending order
-  const sorted = Object.keys(freq).sort((a, b) => a - b);
-
-  // Step 3: Init a count of numbers smaller than the active number
-  let count = 0;
-
-  // Step 4: Init a map to track number of values smaller for each number
-  const smaller = {};
-
-  // Step 5: Iterate over the sorted list
-  for (let num of sorted) {
-    // Set count for active number
-    smaller[num] = count;
-
-    // Update the count by frequency
-    count += freq[num];
-  }
-
-  // Step 6: Use original list and find number of smaller values than it
-  return nums.map((n) => smaller[n]);
-};
diff --git a/work/Default/Easy/Array/1636.sort-array-by-increasing-frequency.js b/work/Default/Easy/Array/1636.sort-array-by-increasing-frequency.js
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index 2ace9f9..0000000
--- a/work/Default/Easy/Array/1636.sort-array-by-increasing-frequency.js
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-/*
- * 1636. Sort Array by Increasing Frequency
- * Difficulty: Easy
- * https://leetcode.com/problems/sort-array-by-increasing-frequency/
- *
- * ──────────────────────────────────────────────────
- *
- * 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.
- *
- *
- *
- * Example 1:
- *
- * Input: nums = [1,1,2,2,2,3]
- * Output: [3,1,1,2,2,2]
- * Explanation: '3' has a frequency of 1, '1' has a frequency of 2, and
- * '2' has a frequency of 3.
- *
- * Example 2:
- *
- * Input: nums = [2,3,1,3,2]
- * Output: [1,3,3,2,2]
- * Explanation: '2' and '3' both have a frequency of 2, so they are
- * sorted in decreasing order.
- *
- * Example 3:
- *
- * Input: nums = [-1,1,-6,4,5,-6,1,4,1]
- * Output: [5,-1,4,4,-6,-6,1,1,1]
- *
- *
- *
- * Constraints:
- *
- * 	• 1 <= nums.length <= 100
- *
- * 	• -100 <= nums[i] <= 100
- */
-
-/**
- * @param {number[]} nums
- * @return {number[]}
- */
-var frequencySort = function (nums) {
-  const freq = {};
-  for (let n of nums) freq[n] = (freq[n] || 0) + 1;
-  return nums.sort((a, b) => freq[a] - freq[b] || b - a);
-};
diff --git a/work/Default/Easy/Hash Table/387.first-unique-character-in-a-string.js b/work/Default/Easy/Hash Table/387.first-unique-character-in-a-string.js
deleted file mode 100644
index d849c6d..0000000
--- a/work/Default/Easy/Hash Table/387.first-unique-character-in-a-string.js	
+++ /dev/null
@@ -1,59 +0,0 @@
-/*
- * 387. First Unique Character in a String
- * Difficulty: Easy
- * https://leetcode.com/problems/first-unique-character-in-a-string/
- *
- * ──────────────────────────────────────────────────
- *
- * Given a string s, find the first non-repeating character in it and
- * return its index. If it does not exist, return -1.
- *
- *
- *
- * Example 1:
- *
- * Input: s = "leetcode"
- *
- * Output: 0
- *
- * Explanation:
- *
- * The character 'l' at index 0 is the first character that does not
- * occur at any other index.
- *
- * Example 2:
- *
- * Input: s = "loveleetcode"
- *
- * Output: 2
- *
- * Example 3:
- *
- * Input: s = "aabb"
- *
- * Output: -1
- *
- *
- *
- * Constraints:
- *
- * 	• 1 <= s.length <= 10^5
- *
- * 	• s consists of only lowercase English letters.
- */
-
-/**
- * @param {string} s
- * @return {number}
- */
-var firstUniqChar = function (s) {
-  const freq = {};
-  for (let c of s) freq[c] = (freq[c] || 0) + 1;
-
-  for (let i = 0; i < s.length; i++) {
-    if (freq[s[i]] === 1) {
-      return i;
-    }
-  }
-  return -1;
-};
diff --git a/work/Default/Medium/Array/347.top-k-frequent-elements.js b/work/Default/Medium/Array/347.top-k-frequent-elements.js
deleted file mode 100644
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--- a/work/Default/Medium/Array/347.top-k-frequent-elements.js
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@@ -1,62 +0,0 @@
-/*
- * 347. Top K Frequent Elements
- * Difficulty: Medium
- * https://leetcode.com/problems/top-k-frequent-elements/
- *
- * ──────────────────────────────────────────────────
- *
- * Given an integer array nums and an integer k, return the k most
- * frequent elements. You may return the answer in any order.
- *
- *
- *
- * 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.
- *
- *
- *
- * Follow up: Your algorithm's time complexity must be better than O(n
- * log n), where n is the array's size.
- */
-
-/**
- * @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);
-};
diff --git a/work/Default/Medium/Hash Table/451.sort-characters-by-frequency.js b/work/Default/Medium/Hash Table/451.sort-characters-by-frequency.js
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index 0357d5c..0000000
--- a/work/Default/Medium/Hash Table/451.sort-characters-by-frequency.js	
+++ /dev/null
@@ -1,64 +0,0 @@
-/*
- * 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) {
-  // count the frequency of each character
-  const freq = {};
-  for (let c of s) freq[c] = (freq[c] || 0) + 1;
-
-  // sort the characters by frequency
-  return s
-    .split("")
-    .sort((a, b) => freq[b] - freq[a] || a.localeCompare(b))
-    .join("");
-};