It can be done using one extra array only that is the output array out[]
*int out = (int *)malloc(sizeof(n)); //n is the number of elements in a
int i,temp = 1;
for(i=0;i<n;i++)
{
out[i] = temp;
temp*=a[i];
}
temp =1;
for(i=n-1;i>=0;i--)
{
out[i] *= temp;
temp*=a[i];
}*
On Sun, Jul 17, 2011 at 4:43 PM, manish patel <manispatel...@gmail.com>wrote:
> thanx!!!! the question was asked by MS today in MNNIT.
>
> 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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>
>
>
> --
> With Regards
>
> Manish Patel
> BTech 2nd Year
> Computer Science And Engineering
> National Institute of Technology -Allahabad
>
>
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