Practice Questions for Recursion | Set 3
Last Updated :
22 Dec, 2023
Explain the functionality of below recursive functions.
Question 1
C++
void fun1(int n)
{
int i = 0;
if (n > 1)
fun1(n - 1);
for (i = 0; i < n; i++)
cout << " * ";
}
|
C
void fun1(int n)
{
int i = 0;
if (n > 1)
fun1(n-1);
for (i = 0; i < n; i++)
printf(" * ");
}
|
Java
static void fun1(int n)
{
int i = 0;
if (n > 1)
fun1(n - 1);
for (i = 0; i < n; i++)
System.out.print(" * ");
}
|
Python3
def fun1(n):
i = 0
if (n > 1):
fun1(n - 1)
for i in range(n):
print(" * ",end="")
|
C#
static void fun1(int n)
{
int i = 0;
if (n > 1)
fun1(n-1);
for (i = 0; i < n; i++)
Console.Write(" * ");
}
|
Javascript
<script>
function fun1(n)
{
let i = 0;
if (n > 1)
fun1(n - 1);
for(i = 0; i < n; i++)
document.write(" * ");
}
</script>
|
Answer: Total numbers of stars printed is equal to 1 + 2 + …. (n-2) + (n-1) + n, which is n(n+1)/2.
Time complexity: O(n2)
Auxiliary Space: O(n), due to recursion call stack
Question 2
C++
#define LIMIT 1000
void fun2(int n)
{
if (n <= 0)
return;
if (n > LIMIT)
return;
cout << n <<" ";
fun2(2*n);
cout << n <<" ";
}
|
C
#define LIMIT 1000
void fun2(int n)
{
if (n <= 0)
return;
if (n > LIMIT)
return;
printf("%d ", n);
fun2(2*n);
printf("%d ", n);
}
|
Java
int LIMIT = 1000;
void fun2(int n)
{
if (n <= 0) return;
if (n > LIMIT) return;
System.out.print(String.format("%d ", n));
fun2(2 * n);
System.out.print(String.format("%d ", n));
}
|
Python3
LIMIT = 1000
def fun2(n):
if (n <= 0):
return
if (n > LIMIT):
return
print(n, end=" ")
fun2(2 * n)
print(n, end=" ")
|
C#
int LIMIT = 1000
void fun2(int n)
{
if (n <= 0)
return;
if (n > LIMIT)
return;
Console.Write(n+" ");
fun2(2*n);
Console.Write(n+" ");
}
|
Javascript
<script>
let LIMIT = 1000;
function fun2(n)
{
if (n <= 0)
return;
if (n > LIMIT)
return;
document.write(n + " "));
fun2(2 * n);
document.write(n + " "));
}
</script>
|
Answer: For a positive n, fun2(n) prints the values of n, 2n, 4n, 8n … while the value is smaller than LIMIT. After printing values in increasing order, it prints same numbers again in reverse order. For example fun2(100) prints 100, 200, 400, 800, 800, 400, 200, 100.
If n is negative, the function is returned immediately.
Time complexity: O(n)
Auxiliary Space: O(log2n), due to recursion call stack
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