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- Write code that creates an ArrayList that can hold String objects. Add the names of three cars to the ArrayList, and then display the contents of the ArrayList.public static String pancakeScramble(String text) This nifty little problem is taken from the excellent Wolfram Challenges problem site where you can also see examples of what the result should be for various arguments. Given a text string, construct a new string by reversing its first two characters, then reversing the first three characters of that, and so on, until the last round where you reverse your entire current string.This problem is an exercise in Java string manipulation. For some mysterious reason, the Java String type does not come with a reverse method. The canonical way to reverse a Java string str is to first convert it to mutable StringBuilder, reverse its contents, and convert the result back to an immutable string, that is,str = new StringBuilder(str).reverse().toString(); Here's the tester it must pass: @Test public void testPancakeScramble() throws IOException {// Explicit test casesassertEquals("", P2J3.pancakeScramble(""));assertEquals("alu",…public static String pancakeScramble(String text) This nifty little problem is taken from the excellent Wolfram Challenges problem site where you can also see examples of what the result should be for various arguments. Given a text string, construct a new string by reversing its first two characters, then reversing the first three characters of that, and so on, until the last round where you reverse your entire current string. This problem is an exercise in Java string manipulation. For some mysterious reason, the Java String type does not come with a reverse method. The canonical way to reverse a Java string str is to first convert it to mutable StringBuilder, reverse its contents, and convert the result back to an immutable string, that is, str = new StringBuilder(str).reverse().tostring(); A bit convoluted, but does what is needed without fuss or muss. Maybe one day the Java strings will come with the reverse method built in, just like the string data types of all sensible…
- URLify: Write a method to replace all spaces in a string with '%20'. You may assume that the string has sufficient space at the end to hold the additional characters, and that you are given the "true" length of the string. (Note: if implementing in Java, please use a character array so that you can perform this operation in place.)EXAMPLEInput: "Mr John Smith ", 13Output: "Mr%20John%20Smith"Unfair hurdles are hurdles which are either too high, or way too close together. Create a function which takes in an array of strings, representing hurdles, and returns whether or not they are unfair. For the purposes of this challenge, unfair hurdles are: • At least 4 characters tall. • Strictly less than 4 spaces apart. Examples // Hurdle are good distance apart but are way too tall. isUnfairHurdle ([ "# "# "# ]) → true ## // Hurdles are a fine height but are way too close together. isUnfairHurdle ([ "# # # #", "# # # #" "# # ]) → true "# "# ]) → false #" // These hurdles are mighty splendid. isUnfairHurdle([ #strobogrammatic number is a number that looks the same when rotated 180 degrees (looked at upside down). Find all strobogrammatic numbers that are of length = n. For example, Given n = 2, return ["11","69","88","96"]. def gen_strobogrammatic(n): Given n, generate all strobogrammatic numbers of length n. :type n: int :rtype: List[str] return helper(n, n) def helper(n, length): if n == 0: return [™"] if n == 1: return ["1", "0", "8"] middles = helper(n-2, length) result = [] for middle in middles: if n != length: result.append("0" + middle + "0") result.append("8" + middle + "8") result.append("1" + middle + "1") result.append("9" + middle + "6") result.append("6" + middle + "9") return result def strobogrammatic_in_range(low, high): :type low: str :type high: str :rtype: int res = count = 0 low_len = len(low) high_len = len(high) for i in range(low_len, high_len + 1): res.extend(helper2(i, i)) for perm in res: if len(perm) == low_len and int(perm) int(high): continue count += 1 return…
- Computer Science This is a codingbat problem. Please answer in Java and do not change the starter code. You have been asked to write the game of tic tac toe. In the first version of the game, we will only be comparing horizontally for a win. A win is determine by all of the elements in a row either being an X or an O. You need to return a String with the winner or TIE if there is no winner. ticTacToe(["X", "O", "O"], ["X", "O", "O"], ["X", "X", "X"]) → "X"ticTacToe(["X", "O", "X"], ["X", "X", "O"], ["O", "O", "O"]) → "O"ticTacToe(["X", "O", "X"], ["X", "X", "O"], ["O", "X", "O"]) → "TIE" public String ticTacToe(String[] top, String[] middle, String[] bottom) { }public static String pancakeScramble(String text) This nifty little problem is taken from the excellent Wolfram Challenges problem site where you can also see examples of what the result should be for various arguments. Given a text string, construct a new string by reversing its first two characters, then reversing the first three characters of that, and so on, until the last round where you reverse your entire current stringdef q5(sentence): Assumes sentence is a string. Returns a list of all the words in sentence, where a word is a token separated by white space, (you can use sentence.split()), and then for each word, make it lowercase and remove any character that is not alpha-numeric (a-z or 8-9). For example, I q5("The Plague" (French: "La Peste"), 1947, by Albert CamUs.') should return ['the', 'plague', 'french', 'la', 'peste', '1947', 'by', 'albert', 'camus'] q5('Red@Dragon....ca is great!') should return ['reddragonca', 'is', 'great'] pass
- In computational geometry, often you need to find the rightmost lowest point in a set of points. Write the following function that returns the rightmost lowest point in a set of points:# Return a list of two values for a pointdef getRightmostLowestPoint(points):Write a test program that prompts the user to enter the coordinates of six points and displays the rightmost lowest point.Given: String[] name={"John", "Mary", "Jim", "Sam"}; Which one is correct: name[1]="Mary" name[1]="John" name.get(1)="Mary" name.get(1)="John"public static String[] getTokens(String raw, String delimiter){ // Line #1 goes here... int numTokens = st.countTokens(); // Line #3 goes here... for(int i = 0; i < numTokens; i++){ // Line #5 goes here } // Line #7 goes here } Using the code above, which one of these should be the code for "Line #7": ret[i] = st.nextToken(); return ret; String[] ret = new String[numTokens]; StringTokenizer st = new StringTokenizer(raw, delimiter);
