LINE1 V [ Choose ] if (s.length() == 0) { if (s.charAt(0) == '0') { LINE2 return countOnes(s.substring(0)); return '1'; if (s.length == 1) { LINE3 if (s.length() : 1) { } else { if (s.charAt[0] ==...

To whomever keeps answering this: I appreciate the answers you’ve already given, but could you (or someone else) write the code using the options from the drop-down list shown in the second photo? Thank you!

Extracted text: LINE1 V [ Choose ] if (s.length() == 0) { if (s.charAt(0) == '0') { LINE2 return countOnes(s.substring(0)); return '1'; if (s.length == 1) { LINE3 if (s.length() : 1) { } else { if (s.charAt[0] == '0') { LINE4 } return ""; return 0; LINE5 return countOnes(s.substring(1); return 1+ countOnes(s.substring(1)); if (s.charAt(0) == '1') { LINE6 { return countOnes(s.substring(2)); if (s.length == 0) { LINE7 return 1+ countOnes(s.substring(2)); return '0'; return 1; LINE8 return 1+ countOnes(s.substring(0)); if (s.charAt[0] == '1') {


Extracted text: Counting the number of 1 bits in a bit string s can be accomplished in Java by first initializing an integer counter n to 0. Next, we implement a loop which iterates over each character c of s and when c is '1' we add 1 to n. When c is '0' we do nothing. Here is the implementation of that algorithm: private int countOnes(String s) { int n = 0; for (int i = 0; i < s.length(); ++i) { if (s.charat(i) == '1') { ++n; } return n; this problem is also amenable to being solved using a recursive method rather than an iterative method (one that employs a loop). your job is to select the correct pieces of code from those available and arrange them in proper order to implement a recursive method with the same method signature. private int countones(string s) { [line1] [line2] [line3] [line4] [line5] [line6] [line7] [line8] [line9] } s.length();="" ++i)="" {="" if="" (s.charat(i)="=" '1')="" {="" ++n;="" }="" return="" n;="" this="" problem="" is="" also="" amenable="" to="" being="" solved="" using="" a="" recursive="" method="" rather="" than="" an="" iterative="" method="" (one="" that="" employs="" a="" loop).="" your="" job="" is="" to="" select="" the="" correct="" pieces="" of="" code="" from="" those="" available="" and="" arrange="" them="" in="" proper="" order="" to="" implement="" a="" recursive="" method="" with="" the="" same="" method="" signature.="" private="" int="" countones(string="" s)="" {="" [line1]="" [line2]="" [line3]="" [line4]="" [line5]="" [line6]="" [line7]="" [line8]="" [line9]="">