refactor(work): restructure solution directories with bucket level and add .gitkeep placeholders

work/1/Easy/Two Pointers/125.valid-palindrome.js

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+/*
+ * 125. Valid Palindrome
+ * Difficulty: Easy
+ * https://leetcode.com/problems/valid-palindrome/
+ * 
+ * ──────────────────────────────────────────────────
+ * 
+ * A phrase is a palindrome if, after converting all uppercase letters
+ * into lowercase letters and removing all non-alphanumeric characters,
+ * it reads the same forward and backward. Alphanumeric characters
+ * include letters and numbers.
+ * 
+ * Given a string s, return true if it is a palindrome, or false
+ * otherwise.
+ * 
+ *  
+ * 
+ * Example 1:
+ * 
+ * Input: s = "A man, a plan, a canal: Panama"
+ * Output: true
+ * Explanation: "amanaplanacanalpanama" is a palindrome.
+ * 
+ * Example 2:
+ * 
+ * Input: s = "race a car"
+ * Output: false
+ * Explanation: "raceacar" is not a palindrome.
+ * 
+ * Example 3:
+ * 
+ * Input: s = " "
+ * Output: true
+ * Explanation: s is an empty string "" after removing non-alphanumeric
+ * characters.
+ * Since an empty string reads the same forward and backward, it is a
+ * palindrome.
+ * 
+ *  
+ * 
+ * Constraints:
+ * 
+ * 	• 1 <= s.length <= 2 * 10^5
+ * 
+ * 	• s consists only of printable ASCII characters.
+*/
+
+/**
+ * @param {string} s
+ * @return {boolean}
+ */
+var isPalindrome = function(s) {
+  let normal = s.replace(/[^a-zA-Z0-9]/g, "").toLowerCase()    
+  let i = 0, j = normal.length - 1;
+
+  while (i < j) {
+    if(normal[i] !== normal[j]) {
+      return false;
+    }
+    i++;
+    j--;
+  }
+  return true;
+};

work/1/Easy/Two Pointers/344.reverse-string.js

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+/*
+ * 344. Reverse String
+ * Difficulty: Easy
+ * https://leetcode.com/problems/reverse-string/
+ * 
+ * ──────────────────────────────────────────────────
+ * 
+ * Write a function that reverses a string. The input string is given as
+ * an array of characters s.
+ * 
+ * You must do this by modifying the input array in-place with O(1)
+ * extra memory.
+ * 
+ *  
+ * 
+ * Example 1:
+ * 
+ * Input: s = ["h","e","l","l","o"]
+ * Output: ["o","l","l","e","h"]
+ * 
+ * Example 2:
+ * 
+ * Input: s = ["H","a","n","n","a","h"]
+ * Output: ["h","a","n","n","a","H"]
+ * 
+ *  
+ * 
+ * Constraints:
+ * 
+ * 	• 1 <= s.length <= 10^5
+ * 
+ * 	• s[i] is a printable ascii character.
+*/
+
+/**
+ * @param {character[]} s
+ * @return {void} Do not return anything, modify s in-place instead.
+ */
+var reverseString = function(s) {
+    let i = 0;
+    let j = s.length - 1;
+    
+    while (i < j) {
+        [s[i], s[j]] = [s[j], s[i]];
+        j--;
+        i++;
+    }   
+};

work/1/Easy/Two Pointers/392.is-subsequence.py

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+"""
+392. Is Subsequence
+Difficulty: Easy
+https://leetcode.com/problems/is-subsequence/
+
+──────────────────────────────────────────────────
+
+Given two strings s and t, return true if s is a subsequence of t, or
+false otherwise.
+
+A subsequence of a string is a new string that is formed from the
+original string by deleting some (can be none) of the characters
+without disturbing the relative positions of the remaining characters.
+(i.e., "ace" is a subsequence of "abcde" while "aec" is not).
+
+
+
+Example 1:
+
+Input: s = "abc", t = "ahbgdc"
+Output: true
+
+Example 2:
+
+Input: s = "axc", t = "ahbgdc"
+Output: false
+
+
+
+Constraints:
+
+        • 0 <= s.length <= 100
+
+        • 0 <= t.length <= 10^4
+
+        • s and t consist only of lowercase English letters.
+
+
+
+Follow up: Suppose there are lots of incoming s, say s1, s2, ..., sk
+where k >= 10^9, and you want to check one by one to see if t has its
+subsequence. In this scenario, how would you change your code?
+"""
+
+
+class Solution:
+    def isSubsequence(self, s: str, t: str) -> bool:
+        if len(s) > len(t):
+            return False
+
+        i, j = 0, 0
+        while i < len(s) and j < len(t):
+            if s[i] == t[j]:
+                i += 1
+            j += 1
+
+        return i == len(s)