Give the solution for the following recurrences T(n) = 2T(n/2) + nlogn

Recurrence Relation:

$$T(n) = 2T(n/2) + n \log n$$

Solution:

Step 1: Guess the solution:

Based on the recurrence relation, we can guess that the solution is of the form:

$$T(n) = a n \log n + b n + c$$

where (a), (b), and (c) are constants.

Step 2: Prove the guess:

To prove the guess, we need to show that the function (T(n) = a n \log n + b n + c) satisfies the recurrence relation.

Substituting (T(n) = a n \log n + b n + c) into the recurrence relation, we get:

$$a n \log n + b n + c = 2(a (n/2) \log (n/2) + b (n/2) + c) + n \log n$$

Simplifying the equation, we get:

$$a n \log n + b n + c = a n \log n + b n + 2c + n \log n$$

Equating the coefficients of (n \log n), (n), and the constants, we get:

$$a = a$$

$$b = b$$

$$c = 2c$$

This shows that the function (T(n) = a n \log n + b n + c) satisfies the recurrence relation.

Step 3: Find the values of the constants:

To find the values of the constants (a), (b), and (c), we can use the initial condition (T(1) = 1).

Substituting (n = 1) into the function (T(n)), we get:

$$T(1) = a (1) \log 1 + b (1) + c = 1$$

Simplifying the equation, we get:

$$b + c = 1$$

Since we don't have enough information to find the values of (a), (b), and (c) uniquely, we can assign (a = 1), which gives us:

$$b + c = 0$$

Therefore, the solution to the recurrence relation is:

$$T(n) = n \log n$$

Complexity Analysis:

The time complexity of the solution (T(n) = n \log n) is (O(n \log n)). This is because the recursive calls divide the problem size by 2, and the work done at each level is (O(n)).