diff --git a/README.md b/README.md
index d96b4a6..2a8f739 100644
--- a/README.md
+++ b/README.md
@@ -1,4 +1,4 @@
-# complexity-practice
+# complexity-practice (Alston Dsouza)
 
 An exercise in analyzing complexity, writing efficient methods, and choosing appropriate data structures.
 
diff --git a/src/Main.java b/src/Main.java
index 1ca5e5b..2776c48 100644
--- a/src/Main.java
+++ b/src/Main.java
@@ -6,7 +6,7 @@
 public class Main {
 
   // The time complexity is:
-  // YOUR ANSWER HERE
+  // 0(x^2)
   public static void timesTable(int x) {
     for(int i = 1; i <= x; i++) {
         for(int j = 1; j <= x; j++) {
@@ -17,7 +17,7 @@ public static void timesTable(int x) {
   }
 
   // The time complexity is:
-  // YOUR ANSWER HERE
+  // 0(n) where n is the length of the string
   public static void printLetters(String word) {
     char[] letters = word.toCharArray();
 
@@ -27,7 +27,7 @@ public static void printLetters(String word) {
   }
 
   // The time complexity is:
-  // YOUR ANSWER HERE
+  // 0(m) where m is the number of banned passwords
   public static boolean isBanned(String password) {
     String[] bannedPasswords = {"password", "hello", "qwerty"};
     boolean banned = false;
@@ -41,7 +41,7 @@ public static boolean isBanned(String password) {
 
 
   // The time complexity is:
-  // YOUR ANSWER HERE
+  // 0(n) where n is the length of the array nums
   public static int computeProduct(int[] nums) {
     int total = 1;
     for(int num : nums) {
@@ -51,7 +51,7 @@ public static int computeProduct(int[] nums) {
   }
 
   // The time complexity is:
-  // YOUR ANSWER HERE
+  // 0(n), same as computeProduct
   public static void describeProduct(int[] nums) {
     System.out.println("About to compute the product of the array...");
     int product = computeProduct(nums);
@@ -60,17 +60,18 @@ public static void describeProduct(int[] nums) {
 
 
   // The time complexity is:
-  // YOUR ANSWER HERE
+  // 0(n) where n is the input value
   public static int computeFactorial(int n) {
     int result = 1;
     for(int i = 1; i <= n; i++) {
-        result *= n;
+        result *= i;
     }
     return result;
   }
 
   // Assume that the largest number is no bigger than the length
   // of the array
+  // 0(n * k) where n is the length of nums, and k is the value of the largest number
   public static void computeAllFactorials(int[] nums) {
     for(int num : nums) {
         int result = computeFactorial(num);
@@ -80,7 +81,7 @@ public static void computeAllFactorials(int[] nums) {
 
 
   // The time complexity is:
-  // YOUR ANSWER HERE
+  // 0(n) where n is the size of the array list
   public static void checkIfContainedArrayList(ArrayList<String> arr, String target) {
     if (arr.contains(target)) {
         System.out.println(target + " is present in the list");
@@ -92,7 +93,7 @@ public static void checkIfContainedArrayList(ArrayList<String> arr, String targe
 
   // assume n = wordsA.length = wordsB.length
   // The time complexity is:
-  // YOUR ANSWER HERE
+  // 0(n^2) where n is the lenght of the arrays
   public static boolean containsOverlap(String[] wordsA, String[] wordsB) {
     for(String wordA : wordsA) {
         for(String wordB : wordsB) {
@@ -105,7 +106,7 @@ public static boolean containsOverlap(String[] wordsA, String[] wordsB) {
   }
 
   // The time complexity is:
-  // YOUR ANSWER HERE
+  // 0(n) where n is the length of wordsA
   public static boolean containsOverlap2(String[] wordsA, String[] wordsB) {
     Set<String> wordsSet = new HashSet<>();
     for(String word : wordsA) {
@@ -122,7 +123,7 @@ public static boolean containsOverlap2(String[] wordsA, String[] wordsB) {
   }
 
   // The time complexity is:
-  // YOUR ANSWER HERE
+  // 0(n) where n is the length of the chars array
   public static void printCharacters(char[] chars) {
     for (int i = 0; i < chars.length; i++) {
       char character = chars[i];
@@ -130,12 +131,12 @@ public static void printCharacters(char[] chars) {
     }
   }
   // The time complexity is:
-  // YOUR ANSWER HERE
+  // 0(1) since only two variables are involved
   public static double computeAverage(double a, double b) {
     return (a + b) / 2.0;
   }
   // The time complexity is:
-  // YOUR ANSWER HERE
+  // 0(1) since HashSet operations (contains) are 0(1) on average
   public static void checkIfContainedHashSet(HashSet<String> set, String target)
   {
     if (set.contains(target)) {
@@ -150,7 +151,7 @@ public static void checkIfContainedHashSet(HashSet<String> set, String target)
   // A queryName is given, and this method returns the corresponding email if it is found
   // Otherwise, it returns "Person not found"
   // What is the time complexity of this method?
-  // YOUR ANSWER HERE
+  // 0(n) where n is the length of the names array
   public static String emailLookup(String[] names, String[] emails, String queryName) {
     for(int i = 0; i < names.length; i++) {
       if (names[i].equals(queryName)) {
@@ -165,14 +166,14 @@ public static String emailLookup(String[] names, String[] emails, String queryNa
   // keys are names and the values are emails.
   // Write this method to efficiently return the corresponding email or "Person not found" if appropriate
   // What is the time complexity of your solution?
-  // YOUR ANSWER HERE
+  // 0(1) since HashMap operations get are 0(1) on average
   public static String emailLookupEfficient(HashMap<String, String> namesToEmails, String queryName) {
     return null;
   }
 
   // What is the time complexity of this method?
   // (assume the set and list have the same number of elements)
-  // YOUR ANSWER HERE
+  // 0(n * m) where n is the size of the set and m is the size of the list
   public static boolean hasCommon(HashSet<String> wordSet, ArrayList<String> wordList) {
     for(String word : wordSet) {
       if(wordList.contains(word)) {
@@ -184,7 +185,7 @@ public static boolean hasCommon(HashSet<String> wordSet, ArrayList<String> wordL
   // Rewrite hasCommon so it does the same thing as hasCommon, but with a better time complexity.
   // Do not change the datatype of wordSet or wordList.
   // What is the time complexity of your new solution?
-  // YOUR ANSWER HERE
+  // 0(n + m), where n is the size of the set and m is the size of the size of the list
   public static boolean hasCommonEfficient(HashSet<String> wordSet, ArrayList<String> wordList) {
     return false;
   }
@@ -194,14 +195,16 @@ public static boolean hasCommonEfficient(HashSet<String> wordSet, ArrayList<Stri
   // The prices will be updated frequently throughout the day, and you need to efficiently update
   // and access the current price for each stock. The order of the ticker symbols is not important.
   // What would be a good choice of data structure?
-  // YOUR ANSWER HERE
+  // A good choice of data structure for stock prices would be: HashMap<String, Double>
+  // HashMap allows for efficient update and lookups (0)1) on average.
 
   // Suppose you are building a music player application where users can create playlists.
   // Songs can be added to the end of the playlist in the order the user chooses, and the user can
   // skip to the next or previous song. Most operations involve adding songs and accessing them by
   // their position in the playlist.
   // What would be a good choice of data structure?
-  // YOUR ANSWER HERE
+  // A good choice of data structure for playlists would be: LinkedList<String>
+  // LinkedList allows for efficient insertion at the end and traversal (0)1) insertion, 0(n) access).
 
   // Suppose you are developing a search feature that keeps track of the user's
   // recent search queries. You want to store the queries in the order they were made,
@@ -209,5 +212,6 @@ public static boolean hasCommonEfficient(HashSet<String> wordSet, ArrayList<Stri
   // relatively small, and it is more important to preserve the order of the searches than
   // to optimize for fast lookups or deletions.
   // What would be a good choice of data structure?
-  // YOUR ANSWER HERE
+  // A good choice of data structure for recent search queries would be: ArrayList<String>
+  // ArrayList allows for fast iteration and maintains order of insertion.
 }
\ No newline at end of file