Question

There is an array of integers, lets say 3,5,7,9. You are supposed to create another array and populate it such that the second array’s 0th position should be a product of all numbers from the first array excluding the number at its 0th position, meaning it should be 5x7x9(excluding 3), number at the index 1 of the second array will be product of 3x7x9 (excluding 5).

The first answer that came up to my mind was having 2 for loops which will lead to a time complexity of O(n2). Later I figured this:

Multiplying all the numbers in the first array(3x5x7x9), and while populating the second array I will divide this product by the number at that position. divide by 3 if I am populating the 0th position, divide by 5 if I am populating the 1st position and so on. This would bring down the complexity from O(n2) to probably O(2n).

But the interviewer says division is not allowed. I could not think of anything else but storing the different possible multiples in some kind of a data structure and using it while populating. I gave up, but when asked for the answer he said he would maintain 2 arrays of forward and backward multiples. When asked about the space complexity issue of the solution, he said it could be optimized.

Answer 1

I’m not sure if the question is about the space or about the solution itself. Since everyone has been providing solutions it leads me to think they didn’t understand the interviewer’s solution, so here is an explanation:

We maintain two arrays. The first, the running product of the numbers of the original array. So the element at place i will be the product of the first i elements in the original array (no latex, but the i^th entry has value it’s pi{j=0 to i} j). And the second array is simply the reverse direction of that, so the i^th entry will have value: pi{j=N-i to N} j.

To construct the final array required, we simply run along each of our two arrays and multiply entries. So the i^th value in the final array, will be the product of all the entries, which is the product of 0 to i-1 times the product of i+1 to N, which is the product of the first array’s i-1 entry and the second array’s i+1 entry.

Total time and space O(n)