Hi,

just compare the 0th position of the string with length-1 position of the
string, 2nd position of the string with length-2 postion of the string and
so.on

eg: 12132, in first comparison itself, will fail.

Thanks
Prasanna M

On Mon, Oct 13, 2014 at 4:47 AM, Carl Barton <[email protected]>
wrote:

> You don't need to reverse anything. You reverse half the number and then
> compare positions, why not just compare things straight away? Also note
> that your solution is not n/2. Should the length be n it would be at least
> n operations. n/2 to reverse half the string and then n/2 comparisons.
> However, your method is depedent on the number of digits in a number and a
> number n does not have n digits in it, it will be approximately log n. See
> here for more information
> http://math.stackexchange.com/questions/231742/proof-how-many-digits-does-a-number-have-lfloor-log-10-n-rfloor-1
>
>
> On 12 October 2014 21:06, Rishav Mishra <[email protected]> wrote:
>
>> Hi everyone,
>>
>> I was recently asked a question to find the most optimal solution to
>> finding if a given number 'n' is a palindrome or not.
>>
>> I suggested reversing the first half of the number and comparing it with
>> the second half, giving complexity O (n/2).
>>
>> He still seemed unsatisfied and wanted me to further optimize it. Any
>> clues on how to optimize this simple question further!?
>>
>>
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Thanks
Prasanna M

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