## Sunday, June 6, 2010

Question:
There is an array A[N] of N numbers. You have to compose an array Output[N] such that Output[i] will be equal to multiplication of all the elements of A[N] except A[i]. For example Output[0] will be multiplication of A[1] to A[N-1] and Output[1] will be multiplication of A[0] and from A[2] to A[N-1]. Solve it without division operator and in O(n).

void mult_of_other_no_div(int* data, int* out, int size)
{
//O(n)
int multi = 1;
int i;

for(i = 0; i < size; i++)
{
out[i] = multi;
multi *= data[i];
}

multi = 1;
for(i = size-1; i >= 0; i--)
{
out[i] *= multi;
multi  *= data[i];
}
}