Added inline comments to 242-Valid-Anagram

java/242-Valid-Anagram.java

@@ -1,20 +1,38 @@
 class Solution {
     public boolean isAnagram(String s, String t) {
 
+        // If the lengths of the inputs are not equal,
+        // they cannot form an anagram
         if(s.length() != t.length()) return false;
+
+        // If both the strings are same, then they
+        // form an anagram
         if(s.equals(t)) return true;
 
+        // Create 2 new HashMaps to store the characters in
+        // each string as key and their respective number of 
+        // occurences as values
         Map<Character, Integer> sMap = new HashMap<>();
         Map<Character, Integer> tMap = new HashMap<>();
 
+        // Use the merge() function of HashMap to increment
+        // the value for every occurence of a key in the string
         for(int i = 0; i < s.length(); i++) {
             sMap.merge(s.charAt(i), 1, Integer::sum);
             tMap.merge(t.charAt(i), 1, Integer::sum);
         }
 
+        // If the number of occurences of at least one character
+        // in a string doesn't match with the same in the other
+        // string, then both strings together cannot form an anagram
         for(Character c : sMap.keySet()) {
             if(!sMap.get(c).equals(tMap.get(c))) return false;
         }
+
+        // If we haven't returned by now, it means that none of
+        // the previous conditions failed. This means that we
+        // should be able to create an anagram using the given two strings.
+        // Hence, return true.
         return true;
     }
 }