Your solution to 1 works fine. I hope you get the job. But it needs
O(N) additional storage for the stack. You can also do with constant
additional storage.
#include <stdio.h>
int main(void)
{
#define N (sizeof a / sizeof a[0])
int a[] = {7, 9, 4, 8, 2};
int result[N], i, product;
for (i = 0, product = 1; i < N; product *= a[i++])
result[i] = product;
for (i = N - 1, product = 1; i >= 0; product *= a[i--])
result[i] *= product;
for (i = 0; i < N; i++) printf("%d ", result[i]);
printf("\n");
return 0;
}
On Aug 3, 7:08 am, Poised~ <dip10c...@gmail.com> wrote:
> I am not looking for answer. Just sharing these Section 2 questions:
>
> 1. Given an array arr[] of n integers, construct a Product Array prod[] (of
> same size) such that prod[i] is equal to the product of all the elements of
> arr[] except arr[i]. Solve it without division operator. Give an efficient
> code.
> (if you are interested, here is my solution:http://ideone.com/EaTUF,
> developed at the test time itself).
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