diff --git a/src/Main.java b/src/Main.java
index 036c766..a5b15c9 100644
--- a/src/Main.java
+++ b/src/Main.java
@@ -5,8 +5,9 @@
 
 public class Main {
 
-  // The time complexity is:
-  // YOUR ANSWER HERE
+  // The time complexity is: 
+  // YOUR ANSWER HERE 
+  // O(x^2) because there are two nested loops that each run x time.
   public static void timesTable(int x) {
     for(int i = 1; i <= x; i++) {
         for(int j = 1; j <= x; j++) {
@@ -18,6 +19,7 @@ public static void timesTable(int x) {
 
   // The time complexity is:
   // YOUR ANSWER HERE
+  // O(n) because we loop through each letter in the word once. where n is the length of the word in letters array.
   public static void printLetters(String word) {
     char[] letters = word.toCharArray();
 
@@ -28,12 +30,13 @@ public static void printLetters(String word) {
 
   // The time complexity is:
   // YOUR ANSWER HERE
+  // 0(n) where n is the number of bannedPassword
   public static boolean isBanned(String password) {
     String[] bannedPasswords = {"password", "hello", "qwerty"};
     boolean banned = false;
     for(String bannedPassword : bannedPasswords) {
         if(password.equals(bannedPassword)) {
-            banned = true;
+            banned = true; // O(1)
         }
     }
     return banned;
@@ -42,6 +45,7 @@ public static boolean isBanned(String password) {
 
   // The time complexity is:
   // YOUR ANSWER HERE
+  // 0(n) where n is the num of elements in the nums array
   public static int computeProduct(int[] nums) {
     int total = 1;
     for(int num : nums) {
@@ -52,6 +56,7 @@ public static int computeProduct(int[] nums) {
 
   // The time complexity is:
   // YOUR ANSWER HERE
+  // O(n) where n is the lngth of the nums array 
   public static void describeProduct(int[] nums) {
     System.out.println("About to compute the product of the array...");
     int product = computeProduct(nums);
@@ -61,6 +66,7 @@ public static void describeProduct(int[] nums) {
 
   // The time complexity is:
   // YOUR ANSWER HERE
+  // o(n) where n is value of n and we use in for loop
   public static int computeFactorial(int n) {
     int result = 1;
     for(int i = 1; i <= n; i++) {
@@ -71,6 +77,7 @@ public static int computeFactorial(int n) {
 
   // Assume that the largest number is no bigger than the length
   // of the array
+  // The time complexity is: O(n^2) where n is the legnth of the nums array.
   public static void computeAllFactorials(int[] nums) {
     for(int num : nums) {
         int result = computeFactorial(num);
@@ -81,7 +88,7 @@ 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) where n is the number of element in arr -> until target is found or done w arr length
   public static void checkIfContainedArrayList(ArrayList<String> arr, String target) {
     if (arr.contains(target)) {
         System.out.println(target + " is present in the list");
@@ -94,7 +101,7 @@ public static void checkIfContainedArrayList(ArrayList<String> arr, String targe
   // assume n = wordsA.length = wordsB.length
   // assume that each String is bounded by a constant length
   // The time complexity is:
-  // YOUR ANSWER HERE
+  // O(n^2) where n is num of elem in wordsA and wordsB and wordsA = wordsB
   public static boolean containsOverlap(String[] wordsA, String[] wordsB) {
     for(String wordA : wordsA) {
         for(String wordB : wordsB) {
@@ -108,7 +115,8 @@ 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) where n is worldsA.length and wordB.length.
+  // asumme that wordsA.length = wordsB.length like the last one.
   public static boolean containsOverlap2(String[] wordsA, String[] wordsB) {
     Set<String> wordsSet = new HashSet<>();
     for(String word : wordsA) {
@@ -125,7 +133,7 @@ public static boolean containsOverlap2(String[] wordsA, String[] wordsB) {
   }
 
   // The time complexity is:
-  // YOUR ANSWER HERE
+  // O(n) where n is chars,length
   public static void printCharacters(char[] chars) {
     for (int i = 0; i < chars.length; i++) {
       char character = chars[i];
@@ -133,14 +141,14 @@ public static void printCharacters(char[] chars) {
     }
   }
   // The time complexity is:
-  // YOUR ANSWER HERE
+  // O(1)
   public static double computeAverage(double a, double b) {
     return (a + b) / 2.0;
   }
 
   // assume that each String is bounded by a constant length
   // The time complexity is:
-  // YOUR ANSWER HERE
+  // O(1) -> call once and set.contains
   public static void checkIfContainedHashSet(HashSet<String> set, String target)
   {
     if (set.contains(target)) {
@@ -156,7 +164,7 @@ public static void checkIfContainedHashSet(HashSet<String> set, String target)
   // Otherwise, it returns "Person not found"
   // assume that each String is bounded by a constant length
   // What is the time complexity of this method?
-  // YOUR ANSWER HERE
+  // O(n) where n = names.length 
   public static String emailLookup(String[] names, String[] emails, String queryName) {
     for(int i = 0; i < names.length; i++) {
       if (names[i].equals(queryName)) {
@@ -172,7 +180,7 @@ public static String emailLookup(String[] names, String[] emails, String queryNa
   // Write this method to efficiently return the corresponding email or "Person not found" if appropriate
   // assume that each String is bounded by a constant length
   // What is the time complexity of your solution?
-  // YOUR ANSWER HERE
+  // O(1) we can use the query for HashMap 
   public static String emailLookupEfficient(HashMap<String, String> namesToEmails, String queryName) {
     return null;
   }
@@ -180,7 +188,7 @@ public static String emailLookupEfficient(HashMap<String, String> namesToEmails,
   // 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) where n is worldList and wordSet size 
   public static boolean hasCommon(HashSet<String> wordSet, ArrayList<String> wordList) {
     for(String word : wordSet) {
       if(wordList.contains(word)) {
@@ -193,8 +201,13 @@ public static boolean hasCommon(HashSet<String> wordSet, ArrayList<String> wordL
   // Do not change the datatype of wordSet or wordList.
   // assume that each String is bounded by a constant length
   // What is the time complexity of your new solution?
-  // YOUR ANSWER HERE
+  // O(n) where n is wordList size so we down from O(n^2) to O(n)
   public static boolean hasCommonEfficient(HashSet<String> wordSet, ArrayList<String> wordList) {
+    for (String word : wordList) {
+      if (wordSet.contains(word)) {
+        return true;
+      }
+    }
     return false;
   }
 
@@ -203,14 +216,16 @@ 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
+  // O(n) - Using HashMap 
+  // keys -> Ticker symboys 
+  // values -> current price of each stock
 
   // 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
+  // Array List to use so you can access by position array index or add songs where you want in the length.
 
   // 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,
@@ -218,5 +233,5 @@ 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
+  // TreeMap so it can keep the order like they want and they priotize order over searches
 }
\ No newline at end of file