complete the missing methods of this RecursiveMethods java file:

import java.util.List; import java.util.ArrayList;

/** * A utility class containing several recursive methods (Lab 7, W2018) * *

 * * For all methods in this API, you are forbidden to use any loops, nor * String or List based methods such as "contains", or methods that use regular expressions * 

* * @author * */ public final class RecursiveMethods {

private RecursiveMethods() {}

/** * Given a string, recursively compress all sets of repeating adjacent chars * within an existing string to a single char. For example, "xyyzzz" yields "xyz". * *

 * * removeRepeats("fffaaar Ouuut") returns "far Out" * removeRepeats("nooooo wooorrriiies") returns "no worries" * removeRepeats("Tomorrow") returns "Tomorow" * * 

* * @param str a string of characters * @return a version of the original string with all repeating adjacent sequences of * the same character, reduced to a single character */ public static String removeRepeats(String str) { }

/** * Return the decimal (integer) representation of a given binary number * (represented as a string). * * *

* E.g. "110" = 1*2^2 + 1*2^1 + 0*2^0 = 4 + 2 + 0 = 6. *

* This method should compute the integer value recursively (no loops allowed). * In this method, the input binary string may have leading zeros (e.g. "00011" = 3). *

* *

 * * binaryToInt("0") returns 0 * binaryToInt("101") returns 5 * binaryToInt("11011") returns 27 * binaryToInt("111") returns 7 * binaryToInt("0111") returns 7 * * 

* * @param b a non-empty, binary string * @return an integer that represents the decimal value of b */ public static int binaryToInt( String b ) { } /** * Return the binary representation of a given number (see Lab 7 URL) * *

* Note that the binary representation of 0 should be 0, and, apart from this case, there * should be no leading 0's in the binary representation. I.e. 2 = "10" not "0010" *

* *

 * * intToBinary(7) returns "111" * intToBinary(2) returns "10" * intToBinary(13) returns "1101" * intToBinary(0) returns "0" * * 

* * @param value an integer to be converted to a binary string * @return a string of binary digits representing the integer "value" */ public static String intToBinary( int value ) { }

/** * Returns the index of the element equal to the integer "target" in * the specified sorted list. * *

* Assumptions: * The list t is not modified by this method. * If target exists it exists only once in the list. *

*

 * * indexOfTarget([3,2,1], 1 ) gives index = 2; * indexOfTarget([3,2,1], 2 ) gives index = 1; * indexOfTarget([3,2,1], 3 ) gives index = 0; * indexOfTarget([3,2,1], 10 ) gives index = -1; * * 

* * @param t a list sorted in ascending order * @param target the integer we want to find the index of * @return the index of the element in t that is equal to the target, or -1 if the target is not present in the list * */ public static int findTarget(List t, int target) {

} /** * Given an array of ints, is it possible to divide the ints into two * groups, so that the sum of one group is a multiple of 12, and the sum of * the other group is odd. Every int must be in one group or the other (never in both). * *

* Hint: Write a recursive helper method (a method that is called from twoGroups and * will actually do the recursion). This helper method can take whatever arguments you like, * although no loops are allowed! *

*

 * * twoGroups([6, 6, 5]) returns true * twoGroups([5, 5, 6]) returns false * twoGroups([5, 5, 6, 1]) returns true * twoGroups([6, 5, 6, 1]) returns false * * 

* * @param t an array of integers * @return a boolean indicating true if there exists a non empty subset of integers in nums * whose sum is a multiple of 12, where the sum of the remaining non empty subset of integers is odd */ public static boolean twoGroups(List t) {

}

}