Skip to content

Answered the time complex questions as best as I can #24

New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

Sign up for GitHub

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

Open
Z1springer wants to merge 1 commit into
base: main
Choose a base branch
from
Open

Answered the time complex questions as best as I can #24

Changes from all commits
Commits
Show all changes
1 commit
Select commit
File filter

Filter by extension

Filter by extension
Conversations
Failed to load comments.
Loading
Jump to
Jump to file
Failed to load files.
Loading
Diff view
Diff view
152 changes: 81 additions & 71 deletions src/Main.java
View file
Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -6,138 +6,135 @@
public class Main {

// The time complexity is:
// YOUR ANSWER HERE
// O(n^2)
public static void timesTable(int x) {
for(int i = 1; i <= x; i++) {
for(int j = 1; j <= x; j++) {
System.out.print(i*j + " ");
}
System.out.println();
for (int i = 1; i <= x; i++) {
for (int j = 1; j <= x; j++) {
System.out.print(i * j + " ");
}
System.out.println();
}
}

// The time complexity is:
// YOUR ANSWER HERE
// O(n)
public static void printLetters(String word) {
char[] letters = word.toCharArray();

for (char letter : letters) {
System.out.println(letter);
System.out.println(letter);
}
}

// The time complexity is:
// YOUR ANSWER HERE
// O(n^2)
public static boolean isBanned(String password) {
String[] bannedPasswords = {"password", "hello", "qwerty"};
String[] bannedPasswords = { "password", "hello", "qwerty" };
boolean banned = false;
for(String bannedPassword : bannedPasswords) {
if(password.equals(bannedPassword)) {
banned = true;
}
for (String bannedPassword : bannedPasswords) {
if (password.equals(bannedPassword)) {
banned = true;
}
}
return banned;
}


// The time complexity is:
// YOUR ANSWER HERE
// O(n)
public static int computeProduct(int[] nums) {
int total = 1;
for(int num : nums) {
total *= num;
for (int num : nums) {
total *= num;
}
return total;
}

// The time complexity is:
// YOUR ANSWER HERE
// O(1)
public static void describeProduct(int[] nums) {
System.out.println("About to compute the product of the array...");
int product = computeProduct(nums);
System.out.println("The product I found was " + product);
}


// The time complexity is:
// YOUR ANSWER HERE
// O(n)
public static int computeFactorial(int n) {
int result = 1;
for(int i = 1; i <= n; i++) {
result *= n;
for (int i = 1; i <= n; i++) {
result *= n;
}
return result;
}

// Assume that the largest number is no bigger than the length
// of the array
// O(n^2)
public static void computeAllFactorials(int[] nums) {
for(int num : nums) {
int result = computeFactorial(num);
System.out.println("The factorial of " + num + " is " + result);
for (int num : nums) {
int result = computeFactorial(num);
System.out.println("The factorial of " + num + " is " + result);
}
}


// The time complexity is:
// YOUR ANSWER HERE
// O(1)
public static void checkIfContainedArrayList(ArrayList<String> arr, String target) {
if (arr.contains(target)) {
System.out.println(target + " is present in the list");
System.out.println(target + " is present in the list");
} else {
System.out.println(target + " is not present in the list");
System.out.println(target + " is not present in the list");
}
}


// assume n = wordsA.length = wordsB.length
// The time complexity is:
// YOUR ANSWER HERE
// O(n^2)
public static boolean containsOverlap(String[] wordsA, String[] wordsB) {
for(String wordA : wordsA) {
for(String wordB : wordsB) {
if(wordA.equals(wordB)) {
return true;
}
for (String wordA : wordsA) {
for (String wordB : wordsB) {
if (wordA.equals(wordB)) {
return true;
}
}
}
return false;
}

// The time complexity is:
// YOUR ANSWER HERE
// O(n^2)
public static boolean containsOverlap2(String[] wordsA, String[] wordsB) {
Set<String> wordsSet = new HashSet<>();
for(String word : wordsA) {
wordsSet.add(word);
for (String word : wordsA) {
wordsSet.add(word);
}

for(String word : wordsB) {
if(wordsSet.contains(word)) {
return true;
}
for (String word : wordsB) {
if (wordsSet.contains(word)) {
return true;
}
}

return false;
}

// The time complexity is:
// YOUR ANSWER HERE
// O(1)
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(n)
public static double computeAverage(double a, double b) {
return (a + b) / 2.0;
}

// The time complexity is:
// YOUR ANSWER HERE
public static void checkIfContainedHashSet(HashSet<String> set, String target)
{
// O(n)
public static void checkIfContainedHashSet(HashSet<String> set, String target) {
if (set.contains(target)) {
System.out.println(target + " is present in the set");
} else {
Expand All @@ -147,12 +144,13 @@ public static void checkIfContainedHashSet(HashSet<String> set, String target)

// emailLookup attempts to find the email associated with a name.
// The name at index i of names corresponds to the email at index i of emails
// A queryName is given, and this method returns the corresponding email if it is found
// 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^2)
public static String emailLookup(String[] names, String[] emails, String queryName) {
for(int i = 0; i < names.length; i++) {
for (int i = 0; i < names.length; i++) {
if (names[i].equals(queryName)) {
return emails[i];
}
Expand All @@ -163,50 +161,62 @@ public static String emailLookup(String[] names, String[] emails, String queryNa
// Suppose that emailLookupEfficient performs the same task as emailLookup
// However, instead of two arrays it is passed a map where the
// keys are names and the values are emails.
// Write this method to efficiently return the corresponding email or "Person not found" if appropriate
// 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)
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
// O(n)
public static boolean hasCommon(HashSet<String> wordSet, ArrayList<String> wordList) {
for(String word : wordSet) {
if(wordList.contains(word)) {
for (String word : wordSet) {
if (wordList.contains(word)) {
return true;
}
}
return false;
}
// Rewrite hasCommon so it does the same thing as hasCommon, but with a better time complexity.

// 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)
public static boolean hasCommonEfficient(HashSet<String> wordSet, ArrayList<String> wordList) {
return false;
}

// Suppose you are building a dashboard that displays real-time stock prices.
// You want to keep track of the current price of each stock, with the stock's ticker symbol as the key.
// 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.
// You want to keep track of the current price of each stock, with the stock's
// ticker symbol as the key.
// 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

// 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
// O(n^2)

// 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

// 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
// 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
Expand Down