I was told the answer is O(n), O(logn) but I am not sure The solution for the recurrence relationsT(n)=T(n-1) + 0(1)T(n)=T(n/2)+ 0(1)is respectivelya) 0(1), O(n)b) O(n), O(logn)c) O(n^2), O(logn)d) O(logn), O(logn)

A recurrence relation may be defined in this type of way that it's far a mathematical system exists that describes how a series of values is set up by way of expressing each fee within the collection as a result of values.Recurrence Relation: T(n) = T(n-1) + O(1)This recurrence relation represents a linear recursion, in which the hassle of length n is split into subproblems of length n-1. In every step, a steady quantity of labor is executed. the solution to this recurrence is O(n), as the entire work carried out is proportional to the size of the trouble. Recurrence Relation: T(n) = T(n/2) + O(1)This recurrence relation represents a divide-and-conquer state of affairs where the hassle of size n is divided into two subproblems of size n/2. the answer to this recurrence is O(log n) because the problem length is halved in every step. a) O(1), O(n)Incorrect because the first recurrence is O(n), not O(1).c) O(n^2), O(log n)Incorrect. Neither recurrence has a quadratic term (n^2).d) O(log n), O(log n)Incorrect. The first recurrence is O(n), not O(log n).Correct option is b)b) O(n), O(log n)