Practice Questions for Recursion | Set 2

Difficulty Level : Easy
Last Updated : 13 Jul, 2021

Explain the functionality of following functions.

Question 1

C

/* Assume that n is greater than or equal to 1 */
int fun1(int n)
{
    if (n == 1)
        return 0;
    else
        return 1 + fun1(n / 2);
}

Java

/* Assume that n is greater than or equal to 1 */
static int fun1(int n)
{
    if (n == 1)
        return 0;
    else
        return 1 + fun1(n / 2);
}
 
// This code is contributed by shubhamsingh10

Python3

# Assume that n is greater than or equal to 1 */
def fun1(n):
    if(n == 1):
        return 0
    else:
        return 1 + fun1(n//2)
 
# This code is contributed by shubhamsingh10

C#

/* Assume that n is greater than or equal to 1 */
static int fun1(int n)
{
    if (n == 1)
        return 0;
    else
        return 1 + fun1(n / 2);
}
 
// This code is contributed by shubhamsingh10

Javascript

<script>
 
/* Assume that n is greater than or equal to 1 */
function fun1(n)
{
    if (n == 1)
        return 0
    else
        return 1 + fun1(n / 2)
}
 
// This code is contributed by gottumukkalabobby
 
</script>

C++

/* Assume that n is greater than or equal to 1 */
int fun1(int n)
{
    if (n == 1)
        return 0;
    else
        return 1 + fun1(n / 2);
}
 
// This code is contributed by shubhamsingh10
// Improved by Adwitiya Mourya

Answer: The function calculates and returns

log2floor

. 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++

/* Assume that n is greater than or equal to 0 */
void fun2(int n)
{
if(n == 0)
    return;
 
fun2(n/2);
cout << n%2 << endl;
}
 
//This code is contributed by shubhamsingh10

C

/* Assume that n is greater than or equal to 0 */
void fun2(int n)
{
  if(n == 0)
    return;
 
  fun2(n/2);
  printf("%d", n%2);
} 

Java

/* Assume that n is greater than or equal to 1 */
static void fun2(int n)
{
if(n == 0)
    return;
  
fun2(n/2);
System.out.println(n%2);
}
 
// This code is contributed by Shubhamsingh10

Python3

# Assume that n is greater than or equal to 0 */
def fun2(n):
    if(n == 0):
        return
     
    fun2(n / 2)
    print(n % 2, end="")
 
# This code is contributed by shubhamsingh10

C#

void fun2(int n)
{
if(n == 0)
    return;
   
fun2(n/2);
Console.Write(n%2);
}
// This code is contributed by shubhamsingh10

Javascript

<script>
 
// Assume that n is greater than or equal to 1
function fun2(n)
{
    if (n == 0)
        return;
     
    fun2(n / 2);
    document.write(n % 2 + " ")
}
 
// This code is contributed by gottumukkalabobby
 
</script>

Answer: The function fun2() prints binary equivalent of n. For example, if n is 21 then fun2() prints 10101.
Note that 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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