Explain the functionality of the following functions.
Question 1
C++
int fun1( int n)
{
if (n == 1)
return 0;
else
return 1 + fun1(n / 2);
}
|
Java
static int fun1( int n)
{
if (n == 1 )
return 0 ;
else
return 1 + fun1(n / 2 );
}
|
Python3
def fun1(n):
if (n = = 1 ):
return 0
else :
return 1 + fun1(n / / 2 )
|
C#
static int fun1( int n)
{
if (n == 1)
return 0;
else
return 1 + fun1(n / 2);
}
|
Javascript
<script>
function fun1(n)
{
if (n == 1)
return 0
else
return 1 + fun1(Math.floor(n / 2))
}
</script>
|
Answer: The function calculates and returns 
For example, if n is between 8 and 15 then fun1() returns 3. If n is between 16 to 31 then fun1() returns 4.
Question 2
C++
void fun2( int n)
{
if (n == 0)
return ;
fun2(n/2);
cout << n%2 << endl;
}
|
C
void fun2( int n)
{
if (n == 0)
return ;
fun2(n/2);
printf ( "%d" , n%2);
}
|
Java
static void fun2( int n)
{
if (n == 0 )
return ;
fun2(n/ 2 );
System.out.println(n% 2 );
}
|
Python3
def fun2(n):
if (n = = 0 ):
return
fun2(n / / 2 )
print (n % 2 , end = "")
|
C#
void fun2( int n)
{
if (n == 0)
return ;
fun2(n/2);
Console.Write(n%2);
}
|
Javascript
<script>
function fun2(n)
{
if (n == 0)
return ;
fun2(Math.floor(n / 2));
document.write(n % 2 + " " )
}
</script>
|
Auxiliary Space: O(log2N), due to recursion call stack
Time Complexity: O(log N)
Answer: The function fun2() prints the binary equivalent of n. For example, if n is 21 then fun2() prints 10101.
Note: Above functions are just for practicing recursion, they are not the ideal implementation of the functionality they provide.
Please write comments if you find any of the answers/codes incorrect.
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Last Updated :
25 Apr, 2023
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