@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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