## Bloomberg LP Interview Question

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

**Interview Type:**Written Test

Actually, one can use logarithmic series to solve this problem.

Check this out

```
T(n) = n(T(n/2)^2)
Log(T(n)) = Log(n(T(n/2)^2)) => Log(n) + 2Log(T(n/2))
Log(T(n)) = Log(n) + 2(Log(n/2) + 2Log(T(n/4)) and so on which you can write as
Log(T(n)) = Sum(i = 0 to k)(2^i * Log(n/2^i) where k is Log(n)
```

Now, If you see the values of this sequence varies from Log(n) to n.

Worst case would be that the sum is nLog(n) only, considering every value to n, and i varies from 0 to Log(n), so that sum.

```
Log(T(n)) = nLog(n)
T(n) = 2^(nLog(n)) = n2^n.
```

and so this does not have asymtotic bound.

Now we know that the sum would always be more than n, which itself make the

`T(n) = 2^n`

and so, this does not have any asymptotic lower bound, and so it does not have tight asympotatic bound.

T(n) = 4^(n-1)/n

Brute-force exact solution:

```
T(n) = n*(T(n/2)^2)
ln(T(n)) = f(n) = 2*f(n/2) + ln(n)
f(2^x) = 2*f(2^(x-1)) + ln2*x
f(2^x) = g(x) = 2*g(x-1) + ln2*x
g(x) = h(x)*2^x = 2*h(x-1)*2^(x-1) + ln2*x
h(x) = h(x-1) + ln2*x/2^x = ln2*(x/2^x + (x-1)/2^(x-1) +...) = ln2*(2 - (x+2)/2^x)
//
h(n) = 1/2+2/4 +3/8+...+n/2^n = 2 - (n+2)/2^n
h1(x) = 1 + x/2 +...+ (x/2)^n = [1-(x/2)^(n+1)]/[1-(x/2)]
h(n) = h1'(x=1) = -2*[(n+1)*(1/2)^(n+1)] + 2*[1-(1/2)^(n+1)] = 2 - (n+2)/2^n
//
g(x) = ln2*(2 - (x+2)/2^x)*2^x = ln2*(2^(x+1) - (x+2))
--check: ln2*(2^(x+1) - (x+2)) = 2*ln2*(2^x - (x+1)) + ln2*x : OK!!
f(x) = f(2^lg2(x)) = g(lg2(x)) = ln2*(2*x - (lg2(x)+2))
T(n) = exp(ln2*(2*n - (lg2(n)+2))) = 2^(2*n - (lg2(n)+2)) = 4^(n-1)/n
--check: 4^(n-1)/n = n*(4^(n/2-1)/(n/2))^2 = 4*n*4^(n-2)/n^2 ...OK!
```

No one asks these types of questions in interviews. We are not in college.

- John October 11, 2016Please take your homework elsewhere.