diff --git a/src/Main.java b/src/Main.java
index 036c766..e420049 100644
--- a/src/Main.java
+++ b/src/Main.java
@@ -7,6 +7,8 @@ public class Main {
 
   // The time complexity is:
   // YOUR ANSWER HERE
+  // O(x^2)
+  // X = Size of the integer x
   public static void timesTable(int x) {
     for(int i = 1; i <= x; i++) {
         for(int j = 1; j <= x; j++) {
@@ -18,6 +20,8 @@ public static void timesTable(int x) {
 
   // The time complexity is:
   // YOUR ANSWER HERE
+  // O(n)
+  // n is the length of the string
   public static void printLetters(String word) {
     char[] letters = word.toCharArray();
 
@@ -28,6 +32,8 @@ public static void printLetters(String word) {
 
   // The time complexity is:
   // YOUR ANSWER HERE
+  //O(n)
+  // n is the fixed elements within the banned passwords array
   public static boolean isBanned(String password) {
     String[] bannedPasswords = {"password", "hello", "qwerty"};
     boolean banned = false;
@@ -42,6 +48,8 @@ public static boolean isBanned(String password) {
 
   // The time complexity is:
   // YOUR ANSWER HERE
+  // O(n)
+  // Where n equals the length of nums
   public static int computeProduct(int[] nums) {
     int total = 1;
     for(int num : nums) {
@@ -52,6 +60,8 @@ public static int computeProduct(int[] nums) {
 
   // The time complexity is:
   // YOUR ANSWER HERE
+  //O(n)
+  // n is the length of nums
   public static void describeProduct(int[] nums) {
     System.out.println("About to compute the product of the array...");
     int product = computeProduct(nums);
@@ -61,6 +71,8 @@ public static void describeProduct(int[] nums) {
 
   // The time complexity is:
   // YOUR ANSWER HERE
+  // O(n)
+  // Where n is the length
   public static int computeFactorial(int n) {
     int result = 1;
     for(int i = 1; i <= n; i++) {
@@ -71,6 +83,8 @@ public static int computeFactorial(int n) {
 
   // Assume that the largest number is no bigger than the length
   // of the array
+  //O(n)
+  // n is the length and values of nums
   public static void computeAllFactorials(int[] nums) {
     for(int num : nums) {
         int result = computeFactorial(num);
@@ -82,6 +96,8 @@ public static void computeAllFactorials(int[] nums) {
   // assume that each String is bounded by a constant length
   // The time complexity is:
   // YOUR ANSWER HERE
+  //O(n)
+  // contains is o(n) time complexity
   public static void checkIfContainedArrayList(ArrayList<String> arr, String target) {
     if (arr.contains(target)) {
         System.out.println(target + " is present in the list");
@@ -95,6 +111,8 @@ public static void checkIfContainedArrayList(ArrayList<String> arr, String targe
   // assume that each String is bounded by a constant length
   // The time complexity is:
   // YOUR ANSWER HERE
+  // O(n^2)
+  // Where n is the length of arr words
   public static boolean containsOverlap(String[] wordsA, String[] wordsB) {
     for(String wordA : wordsA) {
         for(String wordB : wordsB) {
@@ -109,6 +127,8 @@ public static boolean containsOverlap(String[] wordsA, String[] wordsB) {
   // assume that each String is bounded by a constant length
   // The time complexity is:
   // YOUR ANSWER HERE
+  //O(N)
+  // .add is O(1) .contains is O(N) 
   public static boolean containsOverlap2(String[] wordsA, String[] wordsB) {
     Set<String> wordsSet = new HashSet<>();
     for(String word : wordsA) {
@@ -126,6 +146,8 @@ public static boolean containsOverlap2(String[] wordsA, String[] wordsB) {
 
   // The time complexity is:
   // YOUR ANSWER HERE
+  // O(n)
+  // Where n is length of chars
   public static void printCharacters(char[] chars) {
     for (int i = 0; i < chars.length; i++) {
       char character = chars[i];
@@ -134,6 +156,8 @@ public static void printCharacters(char[] chars) {
   }
   // The time complexity is:
   // YOUR ANSWER HERE
+   //O(1)
+ //Returns the same time complexity value everytime, because (a) and (b) will always be a constant value.
   public static double computeAverage(double a, double b) {
     return (a + b) / 2.0;
   }
@@ -141,6 +165,8 @@ public static double computeAverage(double a, double b) {
   // assume that each String is bounded by a constant length
   // The time complexity is:
   // YOUR ANSWER HERE
+   //O(1)
+ //Because (.contains) returns a time complexity of O(1) and the target will always be a constant value, also a O(1) complexity
   public static void checkIfContainedHashSet(HashSet<String> set, String target)
   {
     if (set.contains(target)) {
@@ -157,6 +183,8 @@ public static void checkIfContainedHashSet(HashSet<String> set, String target)
   // assume that each String is bounded by a constant length
   // What is the time complexity of this method?
   // YOUR ANSWER HERE
+  //O(n) 
+//Because the size of how many names and emails can differ, causing the code to check each and every element, making it scale in size.
   public static String emailLookup(String[] names, String[] emails, String queryName) {
     for(int i = 0; i < names.length; i++) {
       if (names[i].equals(queryName)) {
@@ -173,14 +201,24 @@ public static String emailLookup(String[] names, String[] emails, String queryNa
   // assume that each String is bounded by a constant length
   // What is the time complexity of your solution?
   // YOUR ANSWER HERE
+  // O(1) 
+  // get is an o(1) time complexity.
   public static String emailLookupEfficient(HashMap<String, String> namesToEmails, String queryName) {
-    return null;
+    String result = namesToEmails.get(queryName);
+
+    if(result == null){
+      result = "Person not found.";
+    }
+    
+    return result;
   }
 
   // What is the time complexity of this method?
   // assume that each String is bounded by a constant length
   // (assume the set and list have the same number of elements)
   // YOUR ANSWER HERE
+  // O(n^2)
+  // because wordSet and wordList are the same size the act as n * n
   public static boolean hasCommon(HashSet<String> wordSet, ArrayList<String> wordList) {
     for(String word : wordSet) {
       if(wordList.contains(word)) {
@@ -194,7 +232,14 @@ public static boolean hasCommon(HashSet<String> wordSet, ArrayList<String> wordL
   // assume that each String is bounded by a constant length
   // What is the time complexity of your new solution?
   // YOUR ANSWER HERE
+  // O(n) 
+  // swap operations now .contains is O(1) and the total is O
   public static boolean hasCommonEfficient(HashSet<String> wordSet, ArrayList<String> wordList) {
+    for (String word : wordList) {
+      if (wordSet.contains(word)) {
+        return true;
+      }
+    }
     return false;
   }
 
@@ -205,6 +250,9 @@ public static boolean hasCommonEfficient(HashSet<String> wordSet, ArrayList<Stri
   // What would be a good choice of data structure?
   // YOUR ANSWER HERE
 
+  //HashMap 
+  //The HashMap can store keys and values, making it easy to store the stocks and the ticker symbol.
+
   // 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
@@ -212,6 +260,9 @@ public static boolean hasCommonEfficient(HashSet<String> wordSet, ArrayList<Stri
   // What would be a good choice of data structure?
   // YOUR ANSWER HERE
 
+  //List
+  //The list can store and access by position. 
+
   // 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,
   // so you can display them to the user for quick access. The number of recent searches is
@@ -219,4 +270,7 @@ public static boolean hasCommonEfficient(HashSet<String> wordSet, ArrayList<Stri
   // to optimize for fast lookups or deletions.
   // What would be a good choice of data structure?
   // YOUR ANSWER HERE
+
+  //HashMap
+  //HashMap would allow for fast lookups and deletion with an O(1) time complexity.
 }
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