ohh its pretty easy i cudnt make it in the written..
nywayz thanx @nishant and @anurag

On Sun, Jul 17, 2011 at 4:35 PM, Anurag Aggarwal <anurag19aggar...@gmail.com
> wrote:

> take two extra arrays b[] and c[]
> in b[] store the following thing
> b[0]=1;
> b[i]=b[i-1]*a[i-1];
>
>
> in c[] store following things
> c[n-1]=1;
> c[i]=c[i+1]*a[i+1]   (i>n-1)
> fill the c[] array in reverse order i.e. start from n-1 then go to 0;
>
> now output[] would be
> output[i]=b[i]*c[i];
>
>
>
>
> On Sun, Jul 17, 2011 at 4:28 PM, geek forgeek <geekhori...@gmail.com>wrote:
>
>> given an array a[0......n-1]  .required to find the output array out
>> [0.........n-1] such that out [i] is the product of all the numbers a[0] to
>> a[n-1] excluding a[i]
>> for ex out[2]=a[0]*a[1]*a[3]*a[4]....a[n-1]
>> constraint is not using division operator
>>
>> how to do this in O(n)??
>>
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>
>
> Anurag Aggarwal
>
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