Practice Questions for Recursion | Set 3

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 << " * ";  
}  
  
// This code is contributed by shubhamsingh10 

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(" * ");  
}  
   
// This code is contributed by shubhamsingh10 

Python3

def  fun1(n): 
    i = 0 
    if (n > 1): 
        fun1(n - 1)  
    for i in range(n): 
        print(" * ",end="") 
  
# This code is contributed by shubhamsingh10 

C#

static void fun1(int n)  
{  
    int i = 0;  
    if (n > 1)  
        fun1(n-1);  
    for (i = 0; i < n; i++)  
        Console.Write(" * "); 
}  
  
// This code is contributed by shubhamsingh10 

Answer: Total numbers of stars printed is equal to 1 + 2 + …. (n-2) + (n-1) + n, which is n(n+1)/2.

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 <<" "; 
} 
  
// This code is contributed by shubhamsingh10 

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=" ") 
  
# This code is contributed by shubhamsingh10 

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+" "); 
} 
  
// This code is contributed by Shubhamsingh10 

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.

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My Personal Notes

C++

C

Java

Python3

C#

C++

C

Java

Python3

C#

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