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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