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

work/1/Easy/Hash Table/1832.check-if-the-sentence-is-pangram.js

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+/*
+ * 1832. Check if the Sentence Is Pangram
+ * Difficulty: Easy
+ * https://leetcode.com/problems/check-if-the-sentence-is-pangram/
+ * 
+ * ──────────────────────────────────────────────────
+ * 
+ * A pangram is a sentence where every letter of the English alphabet
+ * appears at least once.
+ * 
+ * Given a string sentence containing only lowercase English letters,
+ * return true if sentence is a pangram, or false otherwise.
+ * 
+ *  
+ * 
+ * Example 1:
+ * 
+ * Input: sentence = "thequickbrownfoxjumpsoverthelazydog"
+ * Output: true
+ * Explanation: sentence contains at least one of every letter of the
+ * English alphabet.
+ * 
+ * Example 2:
+ * 
+ * Input: sentence = "leetcode"
+ * Output: false
+ * 
+ *  
+ * 
+ * Constraints:
+ * 
+ * 	• 1 <= sentence.length <= 1000
+ * 
+ * 	• sentence consists of lowercase English letters.
+*/
+
+/**
+ * @param {string} sentence
+ * @return {boolean}
+ */
+var checkIfPangram = function(sentence) {
+    const freq = {};
+    for (let c of sentence) freq[c] = (freq[c] || 0) + 1;
+    
+    if (Object.keys(freq).length === 26){
+        return true;
+    }
+    return false;
+};

work/1/Easy/Hash Table/242.valid-anagram.js

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+/*
+ * 242. Valid Anagram
+ * Difficulty: Easy
+ * https://leetcode.com/problems/valid-anagram/
+ * 
+ * ──────────────────────────────────────────────────
+ * 
+ * Given two strings s and t, return true if t is an anagram of s, and
+ * false otherwise.
+ * 
+ *  
+ * 
+ * Example 1:
+ * 
+ * Input: s = "anagram", t = "nagaram"
+ * 
+ * Output: true
+ * 
+ * Example 2:
+ * 
+ * Input: s = "rat", t = "car"
+ * 
+ * Output: false
+ * 
+ *  
+ * 
+ * Constraints:
+ * 
+ * 	• 1 <= s.length, t.length <= 5 * 10^4
+ * 
+ * 	• s and t consist of lowercase English letters.
+ * 
+ *  
+ * 
+ * Follow up: What if the inputs contain Unicode characters? How would
+ * you adapt your solution to such a case?
+*/
+
+/**
+ * @param {string} s
+ * @param {string} t
+ * @return {boolean}
+ */
+var isAnagram = function(s, t) {
+  let freq = {};
+  for (let c of s) freq[c] = (freq[c] || 0) + 1;
+  
+  for (let c of t) {
+    if(!freq[c] || freq[c] === 0){
+      return false;
+    }
+    freq[c] = freq[c] - 1;
+  }
+  return true;
+};

work/1/Easy/Hash Table/387.first-unique-character-in-a-string.js

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+/*
+ * 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;
+};