Consider the following two functions.void fun1(int n) { if(n == 0) return; printf("%d", n); fun2(n - 2); printf("%d", n); } void fun2(int n) { if(n == 0) return; printf("%d", n); fun1(++n); printf("%d", n); }The output printed when fun1(5) is called is

Let us trace the execution of fun1(5) step by step:

1. fun1(5):

   • Prints 5

   • Calls fun2(5 - 2) = fun2(3)

2. fun2(3):

   • Prints 3

   • ++n increments local n from $3$ to $4$, then calls fun1(4)

3. fun1(4):

   • Prints 4

   • Calls fun2(4 - 2) = fun2(2)

4. fun2(2):

   • Prints 2

   • ++n increments local n from $2$ to $3$, then calls fun1(3)

5. fun1(3):

   • Prints 3

   • Calls fun2(3 - 2) = fun2(1)

6. fun2(1):

   • Prints 1

   • ++n increments local n from $1$ to $2$, then calls fun1(2)

7. fun1(2):

   • Prints 2

   • Calls fun2(2 - 2) = fun2(0), which returns immediately because of the base condition n == 0

   • Prints n (which is 2)

   • Returns to fun2(1)

8. Back in fun2(1):

   • Prints n (which was incremented to 2)

   • Returns to fun1(3)

9. Back in fun1(3):

   • Prints n (which is 3)

   • Returns to fun2(2)

10. Back in fun2(2):

    • Prints n (which was incremented to 3)

    • Returns to fun1(4)

11. Back in fun1(4):

    • Prints n (which is 4)

    • Returns to fun2(3)

12. Back in fun2(3):

    • Prints n (which was incremented to 4)

    • Returns to fun1(5)

13. Back in fun1(5):

    • Prints n (which is 5)

    • Returns

Concatenating all printed numbers in chronological order gives: 5 3 4 2 3 1 2 2 2 3 3 4 4 5$\rightarrow$53423122233445.

Thus, option (A) is correct.