Updated solutions

Author: RodneyShag

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

@@ -29,4 +29,7 @@ public static boolean uniqueCharacters(String str) {
 
 // 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
+// Answer: Can do brute-force solution by comparing all pairs
+// Time Complexity: O(n^2), but since strings above NUM_ASCII_CHARS=256 immediately
+//                  return false, time complexity becomes O(1)
+// Space Complexity: O(1)

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

@@ -15,15 +15,16 @@ public static boolean palPerm(String str) {
                 map.merge(ch, 1, Integer::sum);
             }
         }
-
-        // Even length strings: can not have any characters an odd # of times
-        // Odd length strings: can have at most 1 character an odd # of times
-        int exceptions = (str.length() % 2 == 0) ? 0 : 1;
+        
+        //  Odd length strings: Can have at most 1 character an odd # of times
+        // Even length strings: Can have either 0,2,4,6... number of odd characters.
+        //                      Anything above 1 means not a palindrome.
+        int oddCount = 0;
         for (Integer value : map.values()) {
             if (value % 2 != 0) {
-                exceptions--;
+                oddCount++;
             }
-            if (exceptions < 0) {
+            if (oddCount > 1) {
                 return false;
             }
         }

Chp. 01 - Arrays and Strings/_1_5_One_Away/OneAway.java

@@ -13,6 +13,9 @@ public static boolean oneAway(String s1, String s2) {
     }
 
     private static boolean oneEditReplace(String s1, String s2) {
+        if (s1.length() != s2.length()) {
+            return false;
+        }
         int mismatches = 0;
         for (int i = 0; i < s1.length(); i++) {
             if (s1.charAt(i) != s2.charAt(i)) {