## McAfee Interview Question

Applications Developers**Country:**United States

**Interview Type:**Phone Interview

Seems like 5k is an estimate. If you estimate the sqrt of 1/2 is 0.5, then 5k is the closest answer.

It varies. In extreme scenarios, it could be from as low as 70 up to as high as 50.000.000!

Bubble sort as we know has O(n^2) worst case performance and O(n) best case (when data are already sorted).

Consider the following extreme scenarios:

a) In the first run data were sorted so bubble sort run in O(n) time, that is, it did 10000 operations in 100 seconds or 100 ops per second. Now assume in second run data were sorted in reverse order so bubble sort run in O(n^2) time. With 100 ops per second it did 5000 ops, enough only to sort sqrt(5000) ~= 70 elements.

b) In the first run data were sorted in reverse order, so bubble sort took O(n^2) time, that means 10.000^2 = 100.000.000 ops or 1.000.000 ops per second. Assume the data are sorted in the second run, now bubble sort, in 50 sec, can process 50.000.000 elements.

Don't pay attention at the exact numbers, I'm talking about orders of magnitude, and it varies a lot.

Here's a simple approach I thought of:

Bubble sort has a complexity of n^2. So the ratio of variation in time taken should be squared of the ratio of variation in data set size. Here, the time reduces by 2, hence the data set size must reduce by 4, i.e., 10000/4=2500 (option c is the correct answer).

As there are 10,000 entries so bubble sort will make at most (10,000 * 5,000) approx comparison .and it takes 100 sec(Given).

100sec-----------------(10,000 * 5,000) approx comparison

1 sec ------------------(10,000 *50 ) approx comparison

so for 50 sec------(2500 * 10,000)approx comparison

Now we have to find out the input size which makes (2500* 10,000) comparison or less...According to options 5000 entries corresponds to approx( 2500* 5000)comparison.

so I would go for 5k...

c*(10k)^2 => 100 sec

- Jack September 02, 20121/2*c*n^2 => 100*1/2 sec => 50 sec

c * (sqrt(1/2)*n)^2 => 50 sec

n1 => input size => sqrt(1/2) * floor(10k)

~ 0.7 * 10 k

=> 7k

= 7000