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gudii9Flag for United States of America

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
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CPColin
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SOLUTION
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Avatar of gudii9

ASKER

let me check
Is this snippet relevant or not in this question:

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

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Avatar of 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?
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.)
Avatar of 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?
return true;
Avatar of 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.
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.
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 ????