Power of time

There are $n$ processes to be completed by you. All the processes have a unique number assigned to them from $1$ to $n$. You are given two things:

• The calling order in which all the processes are called.

• The ideal order in which all the processes should have been executed.

If the process in the calling order cannot be executed (does not fit the ideal order), then it is put to the end of the calling order queue for execution.

Now, let us demonstrate this by an example. Let's say that there are $3$ processes, the calling order of the processes is: $3$ - $2$ - $1$. The ideal order is: $1$ - $3$ - $2$, i.e. process number $3$ will only be executed after process number $1$ has been completed; process number $2$ will only be executed after process number $3$ has been executed.

• Iteration #$1$: Since the ideal order has process #$1$ to be executed firstly, the calling order is changed, i.e., the first element must be pushed to the last place. Changing the position of the element takes $1$ unit of time. The new calling order is: $2$ - $1$ - $3$. Time taken in step #$1$: $1$.

• Iteration #$2$: Since the ideal order has process #$1$ to be executed firstly, the calling order has to be changed again, i.e., the first element must be pushed to the last place. The new calling order is: $1$ - $3$ - $2$. Time taken in step #$2$: $1$.

• Iteration #$3$: Since the first element of the calling order is same as the ideal order, that process will be executed. And it will be thus popped out. Time taken in step #$3$: $1$.

• Iteration #$4$: Since the new first element of the calling order is same as the ideal order, that process will be executed. Time taken in step #$4$: $1$.

• Iteration #$5$: Since the last element of the calling order is same as the ideal order, that process will be executed. Time taken in step #$5$: $1$.

Total time taken: $5$ units.

Executing a process takes $1$ unit of time. Changing the position takes $1$ unit of time.