@ashish aggarwal
good solution
but my problem is
ex.
i have n array of element 1,2,3,3,4;
XOR of (1,2,3,3,4)with (1,2,3,4,5). finally i have
XOR of 3 nd 5 (3^5) is 110

now what
pls explan me in detail
On Thu, Sep 2, 2010 at 1:30 PM, ashish agarwal <ashish.cooldude...@gmail.com
> wrote:

> In this method overflow will be there..if number is just bigger...so by
> doing XOR we can get missing number and repeated number .
> take xor of all element of array and take this xor with array[1...n]
> So we get xor of two numbers.
> now get set bit of this xor and proceed.
>
> On Thu, Sep 2, 2010 at 12:36 AM, bittu <shashank7andr...@gmail.com> wrote:
>
>> @luckyzoner
>>
>>  can post the c program of what u ave said above..
>>
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