Added improved JS solution for problem 49

Author: imaginate

javascript/49-Group-Anagrams.js

@@ -1,3 +1,58 @@
+//////////////////////////////////////////////////////////////////////////////
+// Hash Each Word
+// Time: O(n*max(w))
+// Space: O(n*max(w))
+// This solution is faster than sorting each word.
+//////////////////////////////////////////////////////////////////////////////
+
+/** @const {!Object<string, number>} */
+const CODES = {
+    a: 0, b: 1, c: 2, d: 3, e: 4, f: 5, g: 6, h: 7, i: 8, j: 9,
+    k: 10, l: 11, m: 12, n: 13, o: 14, p: 15, q: 16, r: 17,
+    s: 18, t: 19, u: 20, v: 21, w: 22, x: 23, y: 24, z: 25
+};
+
+/**
+ * @param {string[]} words
+ * @return {string[][]}
+ */
+function groupAnagrams(words) {
+    
+    const map = Object.create(null);
+    for (const word of words) {
+        const hash = hashWord(word);
+        if (!(hash in map)) {
+            map[hash] = [];
+        }
+        map[hash].push(word);
+    }
+    
+    const groups = [];
+    for (const key in map) {
+        groups.push(map[key]);
+    }
+    return groups;
+}
+
+/**
+ * @param {string} word
+ * @return {string}
+ */
+function hashWord(word) {
+    const hash = new Array(26).fill(0);
+    for (const ch of word) {
+        ++hash[CODES[ch]];
+    }
+    return hash.toString();
+}
+
+//////////////////////////////////////////////////////////////////////////////
+// Sort Each Word
+// Time: O(n*max(w)*log(max(w)))
+// Space: O(n*max(w))
+// This solution is slower than hashing each word.
+//////////////////////////////////////////////////////////////////////////////
+
 /**
  * @param {string[]} strs
  * @return {string[][]}
@@ -8,7 +63,6 @@ const groupAnagrams = (strs) => {
   for (let i = 0; i < strs.length; i++) {
     const sorted = strs[i].split("").sort().join("");
     //! we are just splitting the string and sorting it and joining it back
-    console.log(sorted);
     if (map.has(sorted)) {
       map.get(sorted).push(strs[i]); //! if the map has the sorted string, we push the string into the array
     } else {