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Daniel and Mathew time complexity #3

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5b71dae
completed time complexity problems
dangrabo Feb 6, 2025
c32ad4e
updated
dangrabo Feb 6, 2025
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65 changes: 33 additions & 32 deletions src/Main.java
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Original file line number Diff line number Diff line change
Expand Up @@ -5,8 +5,7 @@

public class Main {

// The time complexity is:
// YOUR ANSWER HERE
// The time complexity is: o(x^2)
public static void timesTable(int x) {
for(int i = 1; i <= x; i++) {
for(int j = 1; j <= x; j++) {
Expand All @@ -16,8 +15,7 @@ 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 word
public static void printLetters(String word) {
char[] letters = word.toCharArray();

Expand All @@ -26,8 +24,7 @@ public static void printLetters(String word) {
}
}

// The time complexity is:
// YOUR ANSWER HERE
// The time complexity is: o(1)
public static boolean isBanned(String password) {
String[] bannedPasswords = {"password", "hello", "qwerty"};
boolean banned = false;
Expand All @@ -40,8 +37,7 @@ public static boolean isBanned(String password) {
}


// The time complexity is:
// YOUR ANSWER HERE
// The time complexity is: o(n) where n is the size of the array
public static int computeProduct(int[] nums) {
int total = 1;
for(int num : nums) {
Expand All @@ -50,17 +46,15 @@ public static int computeProduct(int[] nums) {
return total;
}

// The time complexity is:
// YOUR ANSWER HERE
// The time complexity is: o(n)
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)
public static int computeFactorial(int n) {
int result = 1;
for(int i = 1; i <= n; i++) {
Expand All @@ -69,6 +63,7 @@ public static int computeFactorial(int n) {
return result;
}

// The time complexity is: o(n*m) where m is the length of the array
// Assume that the largest number is no bigger than the length
// of the array
public static void computeAllFactorials(int[] nums) {
Expand All @@ -80,8 +75,7 @@ public static void computeAllFactorials(int[] nums) {


// assume that each String is bounded by a constant length
// The time complexity is:
// YOUR ANSWER HERE
// The time complexity is: o(n) where n is the length of the array
public static void checkIfContainedArrayList(ArrayList<String> arr, String target) {
if (arr.contains(target)) {
System.out.println(target + " is present in the list");
Expand All @@ -93,8 +87,7 @@ 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
// 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 @@ -107,8 +100,7 @@ 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
// The time complexity is: o(n+m) where n is the length of wordsA[] and m is the length of wordsB[]
public static boolean containsOverlap2(String[] wordsA, String[] wordsB) {
Set<String> wordsSet = new HashSet<>();
for(String word : wordsA) {
Expand All @@ -124,23 +116,21 @@ 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 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;
}

// assume that each String is bounded by a constant length
// The time complexity is:
// YOUR ANSWER HERE
// The time complexity is: o(1)
public static void checkIfContainedHashSet(HashSet<String> set, String target)
{
if (set.contains(target)) {
Expand All @@ -156,7 +146,7 @@ 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 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 @@ -172,29 +162,40 @@ 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)
public static String emailLookupEfficient(HashMap<String, String> namesToEmails, String queryName) {
return null;
if (namesToEmails.containsKey(queryName)) {
return namesToEmails.get(queryName);
}
return "Person not found";
}

// 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 length of the array
public static boolean hasCommon(HashSet<String> wordSet, ArrayList<String> wordList) {
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.
// 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)
public static boolean hasCommonEfficient(HashSet<String> wordSet, ArrayList<String> wordList) {
for (String word : wordList) {
if (wordSet.contains(word)) {
return true;
}
}

return false;
}

Expand All @@ -203,20 +204,20 @@ 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
// HashMap

// 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
// ArrayList or TreeMap

// 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
// ArrayList
}