## InMobi Interview Question

Software Engineer / Developers**Team:**InMobi

**Country:**India

**Interview Type:**In-Person

E(i) = e yi{xi+1} if e > 1 and xi+1 > 1 then yi = *

= if e > 1 and 0 >= xi+1 <1 then yi = /

= if e > 1 and 0< xi+1 then yi = +

= if 0 > e < 1 and xi +1 > 1 then yi = *

= if 0 > e < 1 and 0>= xi +1 < 1 then yi = /

= if 0> e < 1 and 0 < xi+1 then yi = +

= if 0 < e and xi +1 > 1 then yi =+

= if 0 < e and xi +1 < 1 then yi =-

= if 0 < e and xi +1 < 0 then yi = *

Is it expected that this will be done in any more efficient way than evaluating all possibilities?

- Anonymous February 19, 2012