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]
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]
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
Related questions
Question
![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]
}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdd97c0ef-a3a9-4c4a-b75b-2f4b31d21cac%2F2b1bae09-1a83-4039-82ed-96ba9e8d20c4%2Fraj1uv_processed.jpeg&w=3840&q=75)
Transcribed Image 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]
}
![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') {](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdd97c0ef-a3a9-4c4a-b75b-2f4b31d21cac%2F2b1bae09-1a83-4039-82ed-96ba9e8d20c4%2Fxiugce_processed.jpeg&w=3840&q=75)
Transcribed Image 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') {
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