@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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