Explanation : For sufficiently large n,
T(n)=Tn−1)+T(n−2)−T(n−3).
If the order of the algorithm for which above recurrence is applicable for the time complexity, is a constant,
T(n)=T(n−1).
⟹T(n)=T(n)+T(n−2)−T(n−3)
⟹T(n−2)=T(n−3)
Going like this, we must have T(1)=T(2)=T(3) which is option A.