take 2 arrays before and after... before[i] contains the product of all the
numbers before i and after[i] contains product of all the numbers after i
something like this
before[0]=1;
after[n-1]=1;
for(i=1;i<n;i++)
{
before[i]=a[i-1]*before[i-1];
after[n-i-1]=after[n-i]*a[n-i];
}
for(i=0;i<n;i++)
printf("%d\n",before[i]*after[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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