asked on

has22 challenge

Hi,

I am working on below challenge

http://codingbat.com/prob/p121853
I wrote my code as below

public boolean has22(int[] nums) {
 
  
  boolean result=false;

  int sum=0;
  for(int n:nums){
   if(n==2)
   result=true;
  
  }
   return result;




}

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I am not passing all tests

Expected      Run
has22([1, 2, 2]) → true      true      OK
has22([1, 2, 1, 2]) → false      true      X
has22([2, 1, 2]) → false      true      X
has22([2, 2, 1, 2]) → true      true      OK
has22([1, 3, 2]) → false      true      X
has22([1, 3, 2, 2]) → true      true      OK
has22([2, 3, 2, 2]) → true      true      OK
has22([4, 2, 4, 2, 2, 5]) → true      true      OK
has22([1, 2]) → false      true      X
has22([2, 2]) → true      true      OK
has22([2]) → false      true      X
has22([]) → false      false      OK
has22([3, 3, 2, 2]) → true      true      OK
has22([5, 2, 5, 2]) → false      true      X
other tests
X

How to improve my design, approach, code? please advise

ASKER CERTIFIED SOLUTION

CPColin

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See answer

SOLUTION

krakatoa

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See answer

gudii9

ASKER

let me check

krakatoa

Is this snippet relevant or not in this question:

if(nums[y]==2&&nums[y+1]==2)

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gudii9

ASKER

public boolean has22(int[] nums) {
 
  
  boolean result=false;

  int sum=0;
  int len=nums.length;
  for(int i=0;i<len-1;i++){
   if(nums[i]==2&&nums[i+1]==2)
   result=true;
  
  }
   return result;




}

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above passed all tests? any improvements, suggestions, comments?

CPColin

You could return true as soon as you find two 2's next to each other and save a little processing time. (If you have an array that's a thousand elements long and the first two elements are 2's, you don't need to check the rest of the array, since you already know you'll be returning true.)

gudii9

ASKER

You could return true as soon as you find two 2's next to each other and save a little processing time.

how to do this?

CPColin

return true;

gudii9

ASKER

public boolean has22(int[] nums) {
 
  
 // boolean result=false;

  int sum=0;
  int len=nums.length;
  for(int i=0;i<len-1;i++){
   if(nums[i]==2&&nums[i+1]==2)
  return true;
  
  }
  // return result;
return false;



}

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i see . this is very powerful to use return here.

CPColin

Yep! You can't always break out of a loop early or return from a loop early, but it can be nice, when you can.

krakatoa

i see . this is very powerful to use return here.

I ALREADY posted code with this return clause in it !

Did you even *look* at that ???

You ARE a troll, are you not ????