Given array A.
Compute array B such that
B[0] = 1;
for(i = 1; i < n; i++)
B[i] = B[i-1]*A[i-1]
now,
mul = 1;
for (i = n-2; i >=0; i--){
mul = mul*A[i];
B[i] = B[i]*mul;
}
On Fri, Sep 30, 2011 at 2:18 AM, Hatta <[email protected]> wrote:
> are the algorithm instance always a sequence incremented by one?
>
>
>
> On Thu, Sep 29, 2011 at 8:26 AM, raju <[email protected]> wrote:
> > Given an integer array. { 1,2,3,4,5 }
> > Compute array containing elements
> > 120,60,40,30,24 (2*3*4*5,1*3*4*5, 1*2*4*5, 1*2*3*5, 1*2*3*4)
> > We shouldn't use division operator( / )
> > Time complexity O(n) .. Space complexity O(1)
> >
> > ~raju
> >
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> Hatta
>
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