In my_func(a), Recurrence Relation will be

$T(n) = \begin{cases} 4T\bigg(\frac{n}{2}\bigg)+{n} & \quad \text{if } n \geq 4\\ 1 & \quad \text{if } n <4 \end{cases} $

In new_func(a), Recurrence Relation will be

$T(n) = \begin{cases} 3T\bigg(\frac{n}{2}\bigg)+{n} & \quad \text{if } n \geq 4\\ 1 & \quad \text{if } n <4 \end{cases} $

You can solve Both of these Recurrence Relations using Master Theorem as explained in link.

The Time Complexity of my_func(a) will be $\theta(n^2)$ since $\log_24 = 2$

The Time Complexity of new_func(a) will be $\theta(n^{1.5849})$ since $\log_23 = 1.5849$

You can solve both of these questions by Substitution Method, which is Time Consuming. One of the Example using this method is attached.

The new_func(a) will be slower than Merge Sort, and faster than my_func(a).

In my_func(a), Recurrence Relation will be

$T(n) = \begin{cases} 4T\bigg(\frac{n}{2}\bigg)+{n} & \quad \text{if } n \geq 4\\ 1 & \quad \text{if } n <4 \end{cases} $

In new_func(a), Recurrence Relation will be

$T(n) = \begin{cases} 3T\bigg(\frac{n}{2}\bigg)+{n} & \quad \text{if } n \geq 4\\ 1 & \quad \text{if } n <4 \end{cases} $

You can solve Both of these Recurrence Relations using Master Theorem as explained in link.

The Time Complexity of my_func(a) will be $\theta(n^2)$ since $\log_24 = 2$

The Time Complexity of new_func(a) will be $\theta(n^{1.5849})$ since $\log_23 = 1.5849$

You can solve both of these questions by Substitution Method, which is Time Consuming. One of the Example using this method is attached.

The new_func(a) will be slower than Merge Sort, and faster than my_func(a).

In my_func(a), Recurrence Relation will be

$T(n) = \begin{cases} 4T\bigg(\frac{n}{2}\bigg)+{n} & \quad \text{if } n \geq 4\\ 1 & \quad \text{if } n <4 \end{cases} $

In new_func(a), Recurrence Relation will be

$T(n) = \begin{cases} 3T\bigg(\frac{n}{2}\bigg)+{n} & \quad \text{if } n \geq 4\\ 1 & \quad \text{if } n <4 \end{cases} $

You can solve Both of these Recurrence Relations using Master Theorem as explained in link.

The Time Complexity of my_func(a) will be $\theta(n^2)$ since $\log_24 = 2$

The Time Complexity of new_func(a) will be $\theta(n^{1.5849})$ since $\log_23 = 1.5849$

In my_func(a), Recurrence Relation will be

$T(n) = \begin{cases} 4T\bigg(\frac{n}{2}\bigg)+{n} & \quad \text{if } n \geq 4\\ 1 & \quad \text{if } n <4 \end{cases} $

In new_func(a), Recurrence Relation will be

$T(n) = \begin{cases} 3T\bigg(\frac{n}{2}\bigg)+{n} & \quad \text{if } n \geq 4\\ 1 & \quad \text{if } n <4 \end{cases} $

You can solve Both of these Recurrence Relations using Master Theorem as explained in link.

The Time Complexity of my_func(a) will be $\theta(n^2)$ since $\log_24 = 2$

The Time Complexity of new_func(a) will be $\theta(n^{1.5849})$ since $\log_23 = 1.5849$

You can solve both of these questions by Substitution Method, which is Time Consuming. One of the Example using this method is attached.

The new_func(a) will be slower than Merge Sort, and faster than my_func(a).