if you have to find only the repeated element, you can use the below
mentioned code.
for (int i = 1; i < N; i++)
{
array[i] = array[i] ^ array[i-1] ^ i;
}
printf("Answer : %d\n", array[N-1]);
Thinking for the missing no.
On Sun, Sep 11, 2011 at 4:06 PM, ravi maggon <maggonr...@gmail.com> wrote:
>
> 1. Let x be the missing and y be the repeated element
> 3. Calculate the sum of all the elements and sum of squares of all the
> elements of array.
>
> S=Sum of all the elements of array
> S1= Sum of squares of all the elements of array
>
> Sum of first n natural numbers=a= ( n * (n+1) )/2
> Sum of squares of first n natural no=b= n * (n+1) * (2n+1) / 6
>
> a=S+x-y
> b=S1+x^2- y^2
>
> (b-S1)/(a-S)=x+y;
>
> Now we have 2 equations and 2 variables & hence both x and y can be found.
>
> On Sun, Sep 11, 2011 at 3:58 PM, parag khanna <khanna.para...@gmail.com>
> wrote:
>>
>> Q.) Take an array of 100 elements and fill it with the no 1 to 100 such that
>> one no. should skip and one no. should repeated. Find the no. which is
>> repeated and which is skipped.
>>
>>
>>
>>
>>
>>
>> --
>> Parag Khanna
>> B.tech Final Year
>> NIT,Kurukshetra
>>
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>
>
>
> --
> Regards
> Ravi Maggon
> B.E. CSE, Final Year
> Thapar University
>
> www.algorithmguru.com
>
> "Failure is the opportunity to begin again more intelligently"
>
--
Regards
Ravi Maggon
B.E. CSE, Final Year
Thapar University
www.algorithmguru.com
"Failure is the opportunity to begin again more intelligently"
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