Updated 2 solutions

Author: RodneyShag

Chp. 01 - Arrays and Strings/_1_1_Is_Unique/IsUnique.java

@@ -1,17 +1,9 @@
-/** Tips **/
-
-// Should ask interviewer if String is ASCII or Unicode (We assume ASCII)
-
-/** Algorithm **/
-
-// Use a HashMap to store characters and detect duplicates.
-
-/** Solution **/
-
 package _1_1_Is_Unique;
 
 import java.util.HashSet;
 
+// Should ask interviewer if String is ASCII or Unicode (We assume ASCII)
+
 public class IsUnique {
     private static final int NUM_CHARS = 256; // number of ASCII characters
 
@@ -31,17 +23,10 @@ public static boolean uniqueCharacters(String str) {
     }
 }
 
-/** Time/Space Complexity **/
-
 //  Time Complexity: O(1)
 // Space Complexity: O(1)
 // Checking for str.length() > 256 lowered our time/space complexity from O(n) to O(1)
 
-/** Follow-up Question **/
-
-// What if we're not allowed to use additional data structures?
-
-/** Follow-up Answer **/
-
-// Do a brute-force solution using nested loops to compare all pairs.
-// This will be O(n^2) time complexity and O(1) space complexity.
+// Follow-up Question: What if we're not allowed to use additional data structures?
+//
+// Answer: Can do brute-force O(n^2) runtime O(1) space solution by comparing all pairs

Chp. 01 - Arrays and Strings/_1_4_Palindrome_Permutation/PalindromePermutation.java

@@ -4,7 +4,6 @@
 
 public class PalindromePermutation {
     public static boolean palPerm(String str) {
-        /* Create HashMap to count characters in String */
         str = str.toLowerCase().replaceAll("\\s", "");
         HashMap<Character, Integer> map = new HashMap<>(26);
         for (int i = 0; i < str.length(); i++) {