Update 0242-valid-anagram.js

changing space complexity to O(1).
It's constant since we keep hashmap of the letters and their frequency, so the size of the hashmao will be the size of the alphabet -> size of 26.

Author: eliork

javascript/0242-valid-anagram.js

@@ -20,7 +20,7 @@ const reorder = (str) => str
 
 /**
  * Hash Map - Frequency Counter
- * Time O(N) | Space O(N)
+ * Time O(N) | Space O(1)
  * https://leetcode.com/problems/valid-anagram/
  * @param {string} s
  * @param {string} t
@@ -30,8 +30,8 @@ var isAnagram = (s, t, map = new Map()) => {
     const isEqual = s.length === t.length;
     if (!isEqual) return false;
 
-    addFrequency(s, map);      /* Time O(N) | Space O(N) */
-    subtractFrequency(t, map); /* Time O(N) | Space O(N) */
+    addFrequency(s, map);      /* Time O(N) | Space O(1) */
+    subtractFrequency(t, map); /* Time O(N) | Space O(1) */
 
     return checkFrequency(map);/* Time O(N) */
 };
@@ -40,7 +40,7 @@ const addFrequency = (str, map) => {
     for (const char of str) {/* Time O(N) */
         const count = (map.get(char) || 0) + 1;
 
-        map.set(char, count);   /* Space O(N) */
+        map.set(char, count);   /* Space O(1) */
     }
 }
 
@@ -50,7 +50,7 @@ const subtractFrequency = (str, map) => {
 
         const count = map.get(char) - 1;
 
-        map.set(char, count);   /* Space O(N) */
+        map.set(char, count);   /* Space O(1) */
     }
 };
 
@@ -61,4 +61,4 @@ const checkFrequency = (map) => {
     }
 
     return true;
-}
+}