## Amazon Interview Question

Software Engineer / Developers**Country:**United States

**Interview Type:**Phone Interview

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We shouldn't be answering in terms of +ve or -ve numbers as there can be cases where all the numbers are either positive or negative. Just find the largest three and the smallest two and then check for which product is bigger. In this case the interviewer may be looking for the most optimum way to find those numbers.

This question definitely needs (2 -ve with 1 +ve) or (3 +ve) numbers.

Lets take the help of data structures to solve this

consider 3 heaps say H1,H2,H3

H1 -> max heap of negative numbers of size 3

H2 -> min heap of negative numbers of size 2

H3 -> max heap of positive numbers of size 3

Iterate through the array of integers and push the integers into their respective heaps (Push 0 in H1 if it comes)

Now H1 and H2 contain our primary results

if (H3 is empty) or (H3 has 2 entries) then take needed values in H1 (This explains all negative values case)

In all the other cases evaluate the required result accordingly

Please note that whatever the value of 'n' is the lookup is finally broken down to 8 numbers at-max

Complexity: It is only one run through the array. so it is O(n)

Note: The heaps solution is proposed as it will be easy to explain and the same can be used for higher problems also (say find 7 integers to find largest product where coding in terms of variables may be difficult)

I can see many alogs above with +ve and -ve number consideration.. but what if all numbers are -ve?

If all are -ve, then the answer is max3*max2*max1 anyway.

-40 -30 -20 -10 -3 -2 -1, then of course -6 is your best shot.

guys.. let it be -ve or +ve numbers..

we need to always find the max1,max2,max3 and min1 and min2

and check whether product of max1,2,3 is greater or min1,min2,max1 is greater..

now regarding DS..

v can keep hash but they have not mentioned anything about range of numbers..

so i feel .. 5 variables or normal linked list will solve the problem..

and no.of comparisons can be easily reduced...

pls correct me if i am wrong..

We only need 3 Max positive values and 2 Min neg values. This can be done in 5 passes at most (we can piggy pos and neg and reduce it to 3 passes). Then we only check whether P1P2P3 or P1N1N2 is greater.

- Anonymous September 23, 2011