as long as we are using goto with proper comments to ensure that it won't 
decrease the readability we can use them and ther's no harm in it! 
Secondly the worst case for my algo is o(n) .?..correct me if i am wrong
On Friday, 13 July 2012 18:12:41 UTC+5:30, adarsh kumar wrote:

> Ya I didn't see that part, sorry. And in general, using goto is not 
> advisable. 
> Plus this will exceed O(n) in the worst case, as we may keep visiting the 
> goto again and again. Not sure of its exact time complexity.
> ------------------------------
> From: vindhya chhabra
> Sent: 13-07-2012 17:46
> To: algogeeks@googlegroups.com
> Subject: Re: [algogeeks] Re: Amazon Interview Question
>
> @adarsh
> But i think jatin has asked to check for the number( we achieved in step 
> 1) occuring thrice or not..and in this check 27 will rule out.but i doubt 
> the algo given by Jatin runs in O(n) time. please comment.
>
> On Fri, Jul 13, 2012 at 5:17 PM, adarsh kumar <algog...@gmail.com> wrote:
>
>> @jatin:
>> Your first method may be proved wrong. 
>>  
>> Here is a counter test case:
>>  
>> Suppose the array is:
>>  
>> 27 729 19683 2 3 3 27 3 81 729
>>  
>> Here, 81 occurs once, 19683 occurs once, 2 occurs once,729 occurs twice, 
>> 27 occurs twice, and 3 occurs thrice.
>>  
>> You are supposed to return 3 
>> But as per your method, the product will be computed as
>> 729*19683*2*3*3*27*3*81*729=product(say)
>>  
>> Upon traversing the second time, it will return 27, as product%(27*27*27) 
>> is equal to zero!
>>  
>> regards.
>>  
>>
>>  
>> On Fri, Jul 13, 2012 at 1:29 PM, @jatin <jatinseth...@gmail.com> wrote:
>>
>>> Or we can make a BST from array list in ----o(n)
>>> then traverse it inorder-----o(logn)
>>>  
>>> but this solution requires o(logn) space though. 
>>>
>>> On Friday, 13 July 2012 13:16:50 UTC+5:30, jatin sethi wrote:
>>>  
>>>>  
>>>> 1)Find product of the array and store it in say prod ---- o(n) and o(1) 
>>>> 2)now traverse the array and check if  
>>>>  
>>>> static int i;
>>>> tag:
>>>> while(i<n)
>>>> if( prod %(ar[i]*arr[i]*arr[i] ) ==0)
>>>> break;
>>>> //this may be the required element
>>>> //e-o-while
>>>>  
>>>> //now check is this is the element that is occuring three times ----o(n)
>>>> if(number is not the required one then)
>>>> goto tag;
>>>>
>>>> On Thursday, 12 July 2012 10:55:02 UTC+5:30, algo bard wrote:
>>>>
>>>>> Given an array of integers where some numbers repeat once, some 
>>>>> numbers repeat twice and only one number repeats thrice, how do you find 
>>>>> the number that gets repeated 3 times?
>>>>>
>>>>> Does this problem have an O(n) time and O(1) space solution? 
>>>>> No hashmaps please!
>>>>>
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>
>
>
> -- 
> Vindhya Chhabra
>
>
>
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