Write the method catDog
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Use for loops in Java Language
Write the method catDog.
*
* Given a String str, return true if the words "cat" and "dog" appear the same
* number of times in the string.
*
* Examples: catDog("catdog") returns true catDog("catcat") returns false
* catDog("1cat1cadodog") returns true
*
* @param str the String to examine.
* @return true if there are the same number of cats and dogs.
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- Use while loop in Java Language Write the method xyzMiddle().** Given a String str, does "xyz" appear in the* "middle" of the string. To define middle, we'll* say that the number characters to the left and* right of the "xyz" must differ by, at most, one.** Examples:* xyzMiddle("AAxyzBB") returns true* xyzMiddle("AxyzBB") returns true* xyzMiddle("AxyzBBB") returns false** @param str the String to examine.* @return true if xyz is in the "middle" of str.*/// TODO - Write the method xyZMiddle here. /*** Write the method named repeatSeparator().** Given two String inputs, word and separator,* along with a third int input count, return a* big String consisting of count copies of word,* each separated by separator.** Note: This is a very common algorithm, called the* fencepost algorithm, because just like building a* fence, you need 11 fenceposts to hold up 10 sections* of fence.** Examples:* repeatSeparator("Word", "X", 3) returns "WordXWordXWord"* repeatSeparator("This", "And", 2) returns…JAVA Write a java program that takes a string as input from the user and finds out if the string is a palindrome or not. A string is a palindrome if it reads the same from forward as well as from backward. For example, the string “madam” is a palindrome. Use a for loop to solve this problem. Sample inputs and outputs: Input: madam Output: palindrome Input: Dad Output: palindrome Input: hello Output: not a palindromeThere is two method use loops in java language Write the method mirrorEnds().* * Given a String str, look for a mirror image (backwards) string * at both the beginning and end of the given string. In other words, * zero or more characters at the very beginning of the given string, * and at the very end of the string in reverse order (possibly * overlapping). For example, the string "abXYZba" has the mirror * end "ab".* * Examples:* mirrorEnds("abXYZba") returns "ab"* mirrorEnds("abca") returns "a"* mirrorEnds("aba") returns "aba"* * @param str the String to search.* @return the string that is mirrored at the end.*/// TODO - Write the method mirrorEnds here. /*** Write the method named gHappy().** We'll say that a lowercase 'g' in a string is "happy" * if there is another 'g' immediately to its left or right. * Return true if all the g's in the given string are happy.** Examples:* gHappy("xxggxx") returns true* gHappy("xxgxx") returns false* gHappy("xxggyygxx") returns false** @param…
- in java String inputWord is read from input. Write a while loop that iterates until inputWord is equal to "Finale". In each iteration of the loop: Read integer clothingNumber from input. If clothingNumber is less than 30, output the value of inputWord, followed by ": low on stock" and a newline. Otherwise, output the value of inputWord, followed by ": well stocked" and a newline. Read string inputWord from input. Click here for example 1 2 3 4 5 6 7 8 9 10 11 12 13 14 import java.util.Scanner; public class PairedDataProcessing { publicstaticvoidmain(String[] args) { Scannerscnr=newScanner(System.in); StringinputWord; intclothingNumber; inputWord=scnr.next(); /* Your code goes here */ } }JAVA: Replace the for loop below with a code using the "While" loop: int s = 0; int c = 0; boolean valid = true; for(int i = 4; i <= 20 || valid; i = i+2) { if(i%4 != 0) { s = s + i; c++; } if(i > 20) { valid = false; } } System.out.println(s / c);Help me solve this problem /*** Using the Counter-controlled while Loop.** @author* @version*/public class PS06B{/*** Write the method xyzMiddle().** Given a String str, does "xyz" appear in the* "middle" of the string. To define middle, we'll* say that the number characters to the left and* right of the "xyz" must differ by, at most, one.** Examples:* xyzMiddle("AAxyzBB") returns true* xyzMiddle("AxyzBB") returns true* xyzMiddle("AxyzBBB") returns false** @param str the String to examine.* @return true if xyz is in the "middle" of str.*/// TODO - Write the method xyZMiddle here. /*** Write the method named repeatSeparator().** Given two String inputs, word and separator,* along with a third int input count, return a* big String consisting of count copies of word,* each separated by separator.** Note: This is a very common algorithm, called the* fencepost algorithm, because just like building a* fence, you need 11 fenceposts to hold up 10 sections* of fence.** Examples:*…
- Modify this JAVA Program Take out while (true) { from the program and use something else. Instead use a recursive method that doesn't copy the string over and over again. import java.util.Scanner;public class PalindromeChecker { public static void main(String[] args) { Scanner scanner = new Scanner(System.in); while (true) { System.out.print("Please enter a string to test for palindrome or type QUIT to exit:\n"); String input = scanner.nextLine().toLowerCase(); if (input.equals("quit")) { break; } if (isPalindrome(input)) { System.out.println("The input is a palindrome."); } else { System.out.println("The input is not a palindrome."); } } } private static boolean isPalindrome(String str) { str = str.replaceAll("[^a-zA-Z0-9]", ""); int left = 0; int right = str.length() - 1; while (left < right) {…In java language has three different method Write the method xyzMiddle().** Given a String str, does "xyz" appear in the* "middle" of the string. To define middle, we'll* say that the number characters to the left and* right of the "xyz" must differ by, at most, one.** Examples:* xyzMiddle("AAxyzBB") returns true* xyzMiddle("AxyzBB") returns true* xyzMiddle("AxyzBBB") returns false** @param str the String to examine.* @return true if xyz is in the "middle" of str.*/// TODO - Write the method xyZMiddle here. /*** Write the method named repeatSeparator().** Given two String inputs, word and separator,* along with a third int input count, return a* big String consisting of count copies of word,* each separated by separator.** Note: This is a very common algorithm, called the* fencepost algorithm, because just like building a* fence, you need 11 fenceposts to hold up 10 sections* of fence.** Examples:* repeatSeparator("Word", "X", 3) returns "WordXWordXWord"* repeatSeparator("This", "And",…Write a java program that reads in three strings and sorts them alphabetically. You can ask to enter 3 strings in three separate lines. It doesn't have to be one line. (USE IF STATMENTS OR SWITCH STATMENTS, NO LOOPS OR ARRAYS) Ex: Enter three strings: Charlie Able BakerStrings in order: Able Baker Charlie
- Write a Java program using a for loop or a while loop that reads a word and prints the word in reverse. For example, if the user provides the input "Harry", the program prints "yrraH".USING JAVA Implement two methods that: Capitalize the first letter of each word in the given text. Reverse the characters in the words that are in the odd positions of the original text. Add a test program that exercises these methods for a string with at least tenwords in it. For example: The original text is "learning javafx is fun! javafx is a powerful library." The result and reversed text will be "Learning xfavaJ Is !nuF Javafx sI A lufrewoP Library."In Java Integer in is read from input. Write a while loop that iterates until in is less than or equal to 0. In each iteration: Update in with the quotient of in and 7. Output the updated in, followed by a newline. Click here for exampleEx: If the input is 2401, then the output is: 343 49 7 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 import java.util.Scanner; public class QuotientCalculator { publicstaticvoidmain(String[] args) { Scannerscnr=newScanner(System.in); intin; in=scnr.nextInt(); /your code here/ } }