e Courses Organizations Help QUESTION 1 Let s be a String of length 1 consisting of only characters 'O' and 1. We define a parity character p to be a 0 or '1' such that when p is appended to the end of s to form a String t, the number of 1' characters in t will be even. ek) For example, let s be the String "00101". p can be either 0' or '1'.Suppose we let p be '1. Then when we append p onto the end of s to form t we will arrive at t "001011". The number of '1' characters in t is odd, but is supposed to be even. Therefore, we made a wrong choice: p should be '0' so t would be "001010, which now contains two (an even number) '1' characters. The parity character can be determined thusly. Note: to get the character at index n in a string s we call the charAtfn) method on s, e.g. to get the first character of s we would write s.charAt(0). 1. If the length of s is 1, then p is just the first character of s. For example, if s is "0 then p is 'O' and if s is "1* then p is '1 2. If the first character of s matches the parity character of s.substring 1) then p is 'o' 3. If the first character of s does not match the parity character of s.substring(1) then p is '1 Note that s.substringl1) is the Java String class method that returns a substring of s starting at index 1 and continuing to the end of s. For example, if s is "Fred" then s.substring(1) returns the string "red" For this question, complete the code below which is a recursive method named char parity(String s) which determines and returns the parity character p for String s. Note: we are not asking you to return the String t, we want you to return the parity character p. For example, if we call parity0 with the specified strings, we will get back the associated parity character char pl-parity(") char p2 parity("i) char p3-parity("0011001010m): 17 Rule 2: t would be "00110010100" so p3 is assigned 0 char p4 parity("1011011") Rule 3: t would be 10110111" so p4 is assigned 1 // Rule 1: t would be -00" so pl is assigned ". /I Rule 1: t would be 1" so p2 is assigned 1" char parity (String s) if ( #) { // Implement Rule 1 ) else iE #} { // Implenent Rule 2 Save All Answers