Complete the following method, prefix(String s1, String s2), which returns true if s7 is a prefix of $2, s2 is a prefix of s1, or they are equal. Sample runs Input 1: start starp Output 1: false Input 2: espn espn Output 2: true
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- def findOccurrences(s, ch): lst = [] for i in range(0, len(s)): if a==s[i]: lst.append(i) return lst Use the code above instead of enumerate in the code posted below. n=int(input("Number of rounds of Hangman to be played:")) for i in range(0,n): word = input("welcome!") guesses = '' turns = int(input("Enter the number of failed attempts allowed:")) def hangman(word): secrete_word = "-" * len(word) print(" the secrete word " + secrete_word) user_input = input("Guess a letter: ") if user_input in word: occurences = findOccurrences(word, user_input) for index in occurences: secrete_word = secrete_word[:index] + user_input + secrete_word[index + 1:] print(secrete_word) else: user_input = input("Sorry that letter was not found, please try again: ") def findOccurrences(s, ch): return [i for i, letter in enumerate(s) if letter == ch] *** enumerate not discussed in…java: Run length coding is a method to represent a string in a more compact manner. Each character that occurs more than 2 times in a row is represented by the character and a number following it. Two examples are: "abba" → "abba""abcccbbbba" → "abc3b4a"Write a function that calculates how many characters the encoded string is shorter than the original.Q1. Produce a method that reads in a set of values (double) from the user and returns them. Use this header: public static double[] getNumsFromUser (String msg1, String msg2) The implementation of the method should start with a message to input the total number of array elements, followed by a message to enter all the array elements. The method returns the array of elements. The two strings msg1 and msg2 that are passed to the method as parameters will be the two messages that tell the user what to do. In the main class, use the following program outline to test this method: public class Q1 { public static void main(String[] args) { String s1 = "Enter number of students: "; String s2 = "Enter student grades: "; double[] numbers = getNumsFromUser(s1, s2); System.out.println(Arrays.toString(numbers)); public static double[] getNumsFromUser(String msg1, String msg2){ //your code goes here } Sample run Enter number of students: 4 Enter student grades: 12.5 23 11.5 27 [12.5, 23.0, 11.5,…
- public static String reversevowels (String text) Given a text string, create and return a new string of same length where all vowels have been reversed, and all other characters are kept as they were. For simplicity, in this problem only the characters aeiouAEIOU are considered vowels, and y is never a vowel. For example, given the text string "computer science", this method would return "cempetir sceunco". Furthermore, to make this problem more interesting and the result look more palatable, this method must maintain the capitalization of vowels based on the vowel character that was originally in the position that each new vowel character is moved into. For example, "Ilkka Markus" should become "Ulkka Markis" instead of "ulkka MarkIs". Use the handy utility methods in the Character wrapper class to determine whether some particular character is in upper- or lowercase, and convert a character to upper- or lowercase as needed.Write a static recursive method called mrecursiv that displays all of the permutations of the characters in a string passed to the method as its argument. For example, the character sequence abc has the following permutations: acb, bac, bca, cab, cba. Then Write a static method called getInput that get an input string from the user and passed it to the mrecursiv method written above in a method call.def ab_equal(n, k, current): Print out all of the strings of a's and b's of length n so that the number of a's and b's are equal. For n = 2, there's ab and ba. For n = 3 there are no strings since they'd have to have 2 a's and 1 b, or 2 b's a 1 a so not equal. For n = 4, there will be 6 of these strings, and for n = 5, zero again. Hint: use k to track the difference between a's and b's. So for instance if your current is aaabb then k should be equal to either 1 or -1 (your choice depending). When you call the function, you should call it from your main or testing function with the length in the n parameter, 0 should be put into the k parameter, and then an empty string will be passed in for current. What length do you want to run? 4 aabb abab abba baab baba bbaa
- Use for loop in Java language Write the method stringSplosion().* * The method takes one parameter, a non-empty String str (such as "Code") and* returns a String in the form "CCoCodCode". Notice that this includes the* first character of the original String, followed by the first two characters,* and so on until the whole String is used.* * Examples: stringSplosion("Code") returns "CCoCodCode" stringSplosion("abc")* returns "aababc" stringSplosion("x") returns "x"* * @param str the input String to process.* @return a new String as described above.Write a recursive method that displaysa string reversely on the console using the following header: public static void reverseDisplay(String value) For example, reverseDisplay("abcd") displays dcba. Write a test programthat prompts the user to enter a string and displays its reversal.You have a card on which the letter J is written on one side and K on the other. You want to seeall of the possible ways the card will land if you drop it n times. Write a recursive method thatprints each session of dropping the cards with J's and K's. For example if you drop it 4 times in agiven session, all possible ways to drop it are as follows (in exactly the specified order): J J J JJ J J KJ J K JJ J K KJ K J JJ K J KJ K K JJ K K KK J J JK J J KK J K JK J K KK K J JK K J KK K K JK K K K
- Use any loops in Java Language Write the method named withoutString().** Given two strings, base and remove, return a* version of the base string where all instances* of the remove string have been removed (not case* sensitive). You may assume that the remove* string is length 1 or more. Remove only non-overlapping* instances, so with "xxx" removing "xx" leaves "x".** Note: Despite the term "remove" you should actually not* try to remove the characters from a string. You'll continue* to use the "building a string" pattern from lecture.** Examples:* withoutString("Hello there", "llo") returns "He there"* withoutString("Hello there", "e") returns "Hllo thr"* withoutString("Hello there", "x") returns "Hello there"** @param base the base String to modify.* @param remove the substring to remove.* @return the base String with all copies of remove removed.*/// TODO - Write the withoutString method here /*** Write the method named sameEnds().** Given a string, return the longest substring*…solve in java Integer arrays groceryItemPrices and stockRecords are read from input, containing the prices and stock records of each grocery item. Initialize variable count with 0. For each grocery item, increment count if a grocery item meets both of the following conditions: Price is greater than 20. Stock record is less than 20. Lastly, output count followed by a newline. Ex: For the input: 28 14 21 22 7 9 17 24 18 10 20 26 then the output is: 3 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 import java.util.Scanner; public class CombineArrays { publicstaticvoidmain(String[] args) { Scannerscnr=newScanner(System.in); finalintNUM_VALS=6; int[] groceryItemPrices=newint[NUM_VALS]; int[] stockRecords=newint[NUM_VALS]; inti; intcount; for (i=0; i<NUM_VALS; ++i) { groceryItemPrices[i] =scnr.nextInt(); } for (i=0; i<NUM_VALS; ++i) { stockRecords[i] =scnr.nextInt(); } /* Your code goes here */Complete Through this method /** * Count the number of occurrences of substrings "baba" or "mama" * in the input string recursively. They may overlap. * For example, countBabaMama("aba babaa amama ma") → 2, * and countBabaMama("bababamamama") → 4. * @param input is the input string. * @return the number of occurrences. */public static int countBabaMama(String input) { // base case // recursive step }