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116 changes: 38 additions & 78 deletions src/Main.java
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Original file line number Diff line number Diff line change
Expand Up @@ -2,11 +2,9 @@
import java.util.HashMap;
import java.util.HashSet;
import java.util.Set;

public class Main {

// The time complexity is:
// YOUR ANSWER HERE
// The time complexity is O(n^2), where n is the input 'x'
public static void timesTable(int x) {
for(int i = 1; i <= x; i++) {
for(int j = 1; j <= x; j++) {
Expand All @@ -16,8 +14,7 @@ public static void timesTable(int x) {
}
}

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

Expand All @@ -26,8 +23,7 @@ public static void printLetters(String word) {
}
}

// The time complexity is:
// YOUR ANSWER HERE
// The time complexity is O(m), where m is the number of banned passwords
public static boolean isBanned(String password) {
String[] bannedPasswords = {"password", "hello", "qwerty"};
boolean banned = false;
Expand All @@ -39,48 +35,40 @@ public static boolean isBanned(String password) {
return banned;
}


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

// The time complexity is:
// YOUR ANSWER HERE

// The time complexity isO(n), where n is the length 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);
System.out.println("The product I found was " + product);
}


// The time complexity is:
// YOUR ANSWER HERE
// The time complexity is O(n), where n is the input number 'n'
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
// The time complexity is O(n)
public static void computeAllFactorials(int[] nums) {
for(int num : nums) {
int result = computeFactorial(num);
System.out.println("The factorial of " + num + " is " + result);
}
}


// The time complexity is:
// YOUR ANSWER HERE
// The time complexity is O(n), where n is the length 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");
Expand All @@ -89,10 +77,7 @@ public static void checkIfContainedArrayList(ArrayList<String> arr, String targe
}
}


// assume n = wordsA.length = wordsB.length
// The time complexity is:
// YOUR ANSWER HERE
// The time complexity is: O(n^2), where n is the length of the arrays
public static boolean containsOverlap(String[] wordsA, String[] wordsB) {
for(String wordA : wordsA) {
for(String wordB : wordsB) {
Expand All @@ -104,8 +89,7 @@ public static boolean containsOverlap(String[] wordsA, String[] wordsB) {
return false;
}

// The time complexity is:
// YOUR ANSWER HERE
// The time complexity is: O(n), where n is the total length of the two arrays
public static boolean containsOverlap2(String[] wordsA, String[] wordsB) {
Set<String> wordsSet = new HashSet<>();
for(String word : wordsA) {
Expand All @@ -121,36 +105,29 @@ public static boolean containsOverlap2(String[] wordsA, String[] wordsB) {
return false;
}

// The time complexity is:
// YOUR ANSWER HERE
// The time complexity is O(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];
System.out.println("The character at index " + i + " is " + character);
}
}
// The time complexity is:
// YOUR ANSWER HERE

// The time complexity is O(1)
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)
{

// The time complexity is O(1), assuming the set's `contains` method is O(1)
public static void checkIfContainedHashSet(HashSet<String> set, String target) {
if (set.contains(target)) {
System.out.println(target + " is present in the set");
} else {
System.out.println(target + " is not present in the set");
}
}

// 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
// Otherwise, it returns "Person not found"
// What is the time complexity of this method?
// YOUR ANSWER HERE
// The time complexity is O(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)) {
Expand All @@ -160,19 +137,12 @@ public static String emailLookup(String[] names, String[] emails, String queryNa
return "Person not found";
}

// 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
// What is the time complexity of your solution?
// YOUR ANSWER HERE
// The time complexity is O(1), since a HashMap lookup is O(1) on average
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
// The time complexity is O(n^2), where n is the length of the set/list
public static boolean hasCommon(HashSet<String> wordSet, ArrayList<String> wordList) {
for(String word : wordSet) {
if(wordList.contains(word)) {
Expand All @@ -181,33 +151,23 @@ public static boolean hasCommon(HashSet<String> wordSet, ArrayList<String> wordL
}
return false;
}
// 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

// The time complexity is O(n), where n is the total length of the set and list
public static boolean hasCommonEfficient(HashSet<String> wordSet, ArrayList<String> wordList) {
for(String word : wordList) {
if(wordSet.contains(word)) {
return true;
}
}
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.
// 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
// 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
// to optimize for fast lookups or deletions.
// What would be a good choice of data structure?
// YOUR ANSWER HERE
}
// The best choice of data structure is: HashMap<String, Double>
// A HashMap is ideal for keeping track of stock prices, providing efficient O(1) time complexity for both updates and lookups.

// The best choice of data structure is: ArrayList<String>
// An ArrayList allows efficient adding of songs to the end and accessing songs by their position, making it a good fit for this scenario.

// The best choice of data structure is: LinkedList<String>
// A LinkedList preserves the order of recent search queries, which is ideal when displaying the user's recent searches.
}