Question

java Considering the following code segment, answer the multiple choice questions.

1. public String foo(String input) {

2. String str = input.toLowerCase();

3. return bar(str);

4. }

5. private String bar(String s) {

6. if (s.length() <= 1)   {return "";}

7. else if (s.length() % 2 == 0)   {return "*" + bar(s.substring(2, s.length()));}

8. else if (s.length() % 2 == 1) {return "*" + bar(s+"n");}

9. else {return "";}

10. }

11. Which line(s) indicate the name of the helper method? Which line(s) contains recursive call(s)? How many * are in the return String when we call bar("Dave")?

Answer 1

Ans

bar is helper function of foo. call of bar in foo function indicate that bar is helper of foo.

Below statement in foo indicate helper

return bar(str);

.

.

Recursive call occurs if function call itself.

in bar function there are two recursive calls based on condition in else if.

bar(s.substring(2, s.length())

bar(s+"n");

are recursive calls

.

.

.

bar("Dave)

length of string = 4

in if else ladder

s.length%2 turns out true that concatenate * and call recursively bar("ve") (substring returns string from index 2 to last) and return

length of ve = 2

s.length%2 turns true concatenate * and recursive call to bar("")

for bar("") length =0

so s.length<=1 turns true and return "" to previous 2nd call.

2nd call return * to 1st call

1st call return **

.

Hence two * are returned on call bar("Dave")

.

.

.If any doubt ask in the comments.