Finished the Homework, Time Complexity Analysis

src/Main.java

@@ -6,7 +6,7 @@
 public class Main {
 
   // The time complexity is:
-  // YOUR ANSWER HERE
+  // O(N^2) -> n*n = n^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
+  // O(N) -> n = length of the word
   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
+  // O(1) -> n = a fixed 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
+  // O(N) -> n = the length of the int[] array
   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
+  // O(N) -> n = the length of the int[] array 
   public static void describeProduct(int[] nums) {
     System.out.println("About to compute the product of the array...");
     int product = computeProduct(nums);
@@ -60,7 +60,7 @@ public static void describeProduct(int[] nums) {
 
 
   // The time complexity is:
-  // YOUR ANSWER HERE
+  // O(N) -> simplified to n due to for loop complexity
   public static int computeFactorial(int n) {
     int result = 1;
     for(int i = 1; i <= n; i++) {
@@ -71,6 +71,7 @@ public static int computeFactorial(int n) {
 
   // Assume that the largest number is no bigger than the length
   // of the array
+  // O(N^2) -> n*n = n^2, double loops
   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
+  // O(N) -> n = size of the ArrayList
   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
+  // O(N^2) -> n*n = n^2
   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
+  // O(N) -> n = length of wordsA
   public static boolean containsOverlap2(String[] wordsA, String[] wordsB) {
     Set<String> wordsSet = new HashSet<>();
     for(String word : wordsA) {
@@ -122,20 +123,20 @@ public static boolean containsOverlap2(String[] wordsA, String[] wordsB) {
   }
 
   // The time complexity is:
-  // YOUR ANSWER HERE
+  // O(N) -> n = the length of the chars[] array
   public static void printCharacters(char[] chars) {
     for (int i = 0; i < chars.length; i++) {
       char character = chars[i];
       System.out.println("The character at index " + i + " is " + character);
     }
   }
   // The time complexity is:
-  // YOUR ANSWER HERE
+  // O(1) -> constant 
   public static double computeAverage(double a, double b) {
     return (a + b) / 2.0;
   }
   // The time complexity is:
-  // YOUR ANSWER HERE
+  // O(1) -> simplifies and an average case for HashSet 
   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
+  // O(N) -> n = 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
+  // O(1) -> average case for HashMap
   public static String emailLookupEfficient(HashMap<String, String> namesToEmails, String queryName) {
-    return null;
+    return namesToEmails.getOrDefault(queryName, "Person not found");
   }
 
   // What is the time complexity of this method?
   // (assume the set and list have the same number of elements)
-  // YOUR ANSWER HERE
+  // O(N^2) -> n*n = n^2
   public static boolean hasCommon(HashSet<String> wordSet, ArrayList<String> wordList) {
     for(String word : wordSet) {
       if(wordList.contains(word)) {
@@ -184,8 +185,13 @@ 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
+  // O(N) -> flipped the order of the ArrayList and HashSet to improve time complexity
   public static boolean hasCommonEfficient(HashSet<String> wordSet, ArrayList<String> wordList) {
+    for (String word: wordList) {
+      if(wordSet.contains(word)) {
+        return true;
+      }
+    }
     return false;
   }
 
@@ -194,20 +200,23 @@ 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
+  // I would use a HashMap since it's efficient for frequent updates and provides an easy lookups using stock's ticket symbol as the key. 
+  // The average time complexity for HashMap is O(1), so it's fast. 
 
   // 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
+  // For this scenario, I would use an ArrayList because we need to be able to access the items by an index and maintain the order. 
+  // The average time complexity for ArrayList is O(1) or O(N) which is not slow. 
 
   // 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
   // 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
+  // I would use a LinkedList because it has an order of insertion, which is significant in this situation. 
+  // It allows to insert new queries at the end of the list, while maintaining the order.
 }