Want to protect your cyber security and still get fast solutions? Ask a secure question today.Go Premium

x
  • Status: Solved
  • Priority: Medium
  • Security: Public
  • Views: 251
  • Last Modified:

bunnyEars2 challenge

Hi,

I am going through below challenge.

http://codingbat.com/prob/p107330


I have not understood below description

Recursion-1 > bunnyEars2
prev  |  next  |  chance
We have bunnies standing in a line, numbered 1, 2, ... The odd bunnies (1, 3, ..) have the normal 2 ears. The even bunnies (2, 4, ..) we'll say have 3 ears, because they each have a raised foot. Recursively return the number of "ears" in the bunny line 1, 2, ... n (without loops or multiplication).

bunnyEars2(0) → 0
bunnyEars2(1) → 2
bunnyEars2(2) → 5
So even bunnies has 2 ears and odd ones have 3 ears and we have to return total ears?
please advise
0
gudii9
Asked:
gudii9
  • 3
  • 2
2 Solutions
 
CPColinSenior Java ArchitectCommented:
You should figure out the first challenge before you continue to its sequel. You've essentially just posted ten questions that all boil down to, "I don't understand recursion." Figure out one challenge first, so you get the hang of how recursion works, before you proceed to a pile of others.
0
 
awking00Commented:
In the bunnyEars challenge, I tried to show you how if you keep passing in (bunnies - 1) to your method and adding 2 you eventually will reach the base case. You need to get that method to work, then the modification for this challenge simply becomes adding 2 for an even number of bunnies and 3 for an odd number. Keep in mind that when you pass in (bunnies - 1), you're changing an odd number to an even number and vice versa.
1
 
gudii9Author Commented:
pseudeo code:
1. if o bunnies return ears 0;
2. if one bunny ears 0 i.e 2+bunnyEars(1-1)
3. if n bunnies and odd numbered 2+bunneyEars(n-1)
4. if n bunnies and even numbered 3+bunneyEars(n-1)
5 return ear count
0
Industry Leaders: We Want Your Opinion!

We value your feedback.

Take our survey and automatically be enter to win anyone of the following:
Yeti Cooler, Amazon eGift Card, and Movie eGift Card!

 
gudii9Author Commented:
public int bunnyEars2(int bunnies) {
  int result=0;
 if(bunnies==0){
   result=0;
 }
 else{
   if(bunnies%2==1){
   result=2+bunnyEars2(bunnies-1);
   }
    else if(bunnies%2==0){
   result=3+bunnyEars2(bunnies-1);
   }
 }
 return result;
}

Open in new window


above passed all tests. any improvement or alternate approaches?
0
 
awking00Commented:
No need for the result variable -
public int bunnyEars2(int bunnies) {
 if(bunnies==0){
   return 0;
 }
 if(bunnies%2==1) {
   return 2+bunnyEars2(bunnies-1);
 } else  {
   return 3+bunnyEars2(bunnies-1);
 }
}

15:}
0
 
gudii9Author Commented:
public int bunnyEars2(int bunnies) {
 if(bunnies==0){
   return 0;
 } 
 if(bunnies%2==1) {
   return 2+bunnyEars2(bunnies-1);
 } else  {
   return 3+bunnyEars2(bunnies-1);
 }
}

Open in new window

i like above approach much better without result
0

Featured Post

Free Tool: IP Lookup

Get more info about an IP address or domain name, such as organization, abuse contacts and geolocation.

One of a set of tools we are providing to everyone as a way of saying thank you for being a part of the community.

  • 3
  • 2
Tackle projects and never again get stuck behind a technical roadblock.
Join Now