more appropriate O(qn).
since p and q are constant ...q can have maximum constant value can go upto n....so the second part of the recurrence will give higher order .
so the order can be n^n(option b)
Should not option is O(qn). since we dont take heigher value for big-oh.
question is wrong . If it was qn inplace of qn, then C is right .
otherwise O( qn) is more appropriate.
Agree O( 2n) is far far less than O(nn) .
In d link mentioned below. Yeah. :)