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Practice Questions for Recursion | Set 6

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  • Difficulty Level : Medium
  • Last Updated : 21 Aug, 2022
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Question 1 
Consider the following recursive C function. Let len be the length of the string s and num be the number of characters printed on the screen. Give the relation between num and len where len is always greater than 0. 

C++




void abc(char *s)
{
    if(s[0] == '\0')
        return;
  
    abc(s + 1);
    abc(s + 1);
    cout << s[0];   
}
 
// This code is contributed by shubhamsingh10

C




void abc(char *s)
{
    if(s[0] == '\0')
        return;
 
    abc(s + 1);
    abc(s + 1);
    printf("%c", s[0]);   
}

Java




static void abc(char *s)
{
    if(s[0] == '\0')
        return;
   
    abc(s + 1);
    abc(s + 1);
    System.out.print(s[0]);   
}
 
// This code is contributed by shubhamsingh10

Python3




def abc(s):
    if(len(s) == 0):
        return
     
    abc(s[1:])
    abc(s[1:])
    print(s[0])
     
# This code is contributed by shubhamsingh10

C#




static void abc(char *s)
{
    if(s[0] == '\0')
        return;
 
    abc(s + 1);
    abc(s + 1);
    Console.Write(s[0]);
}
 
// This code is contributed by shubhamsingh10

Javascript




<script>
//Javascript Implementation
function abc(s)
{
    if(s.length == 0)
        return;
         
    abc(s.substring(1));
    abc(s.substring(1));
    document.write(s[0]);   
}
 
// This code is contributed by shubhamsingh10
</script>

Time complexity: O(2n)

Auxiliary Space: O(n)
 

The following is the relationship between num and len.  

 num = 2^len-1
s[0] is 1 time printed
s[1] is 2 times printed
s[2] is 4 times printed
s[i] is printed 2^i times
s[strlen(s)-1] is printed 2^(strlen(s)-1) times
total = 1+2+....+2^(strlen(s)-1)
      = (2^strlen(s)) - 1

For example, the following program prints 7 characters. 

C++




#include <bits/stdc++.h>
using namespace std;
 
void abc(char s[])
{
    if(s[0] == '\0')
        return;
 
    abc(s + 1);
    abc(s + 1);
    cout << s[0];
}
 
int main()
{
    abc("xyz");
    return 0;
}
//This code is contributed by shubhamsingh10

C




#include<stdio.h>
 
void abc(char *s)
{
    if(s[0] == '\0')
        return;
 
    abc(s + 1);
    abc(s + 1);
    printf("%c", s[0]);
}
 
int main()
{
    abc("xyz");
    return 0;
}

Java




public class GFG
{
    static void abc(String s)
    {
        if(s.length() == 0)
            return;
      
        abc(s.substring(1));
        abc(s.substring(1));
        System.out.print(s.charAt(0));
    }
 
    public static void main(String[] args) {
        abc("xyz");
    }
}
 
// This code is contributed by divyeh072019

Python3




def abc(s):
    if(len(s) == 0):
        return
     
    abc(s[1:])
    abc(s[1:])
    print(s[0],end="")
 
 
# Driver code
 
abc("xyz")
 
# This code is contributed by shubhamsingh10

C#




using System;
class GFG {
     
    static void abc(string s)
    {
        if(s.Length == 0)
            return;
             
        abc(s.Substring(1));
        abc(s.Substring(1));
        Console.Write(s[0]);
    }
 
  // Driver code
  static void Main() {
    abc("xyz");
  }
}
 
// This code is contributed by divyeshrabadiya07

Javascript




<script>
// Javascript implementation
 
function abc(s)
{
    if(s.length == 0)
        return;
 
    abc(s.substring(1));
    abc(s.substring(1));
    document.write(s[0]);
}
 
abc("xyz");
 
//This code is contributed by shubhamsingh10
</script>

Thanks to bharat nag for suggesting this solution. 
 

Question 2 

C++




#include <iostream>
using namespace std;
 
int fun(int count)
{
    cout << count << endl;
    if(count < 3)
    {
        fun(fun(fun(++count)));
    }
    return count;
}
 
int main()
{
    fun(1);
    return 0;
}
 
// This code is contributed by Shubhamsingh10

C




#include<stdio.h>
int fun(int count)
{
    printf("%d\n", count);
    if(count < 3)
    {
      fun(fun(fun(++count)));
    }
    return count;
}
 
int main()
{
    fun(1);
    return 0;
}

Java




import java.util.*;
  
class GFG{
static int fun(int count)
{
    System.out.println(count);
    if(count < 3)
    {
        fun(fun(fun(++count)));
    }
    return count;
}
 
public static void main(String[] args)
{
    fun(1);
}
}
 
// This code is contributed by Shubhamsingh10

Python3




def fun(count):
    print(count)
    if(count < 3):
        count+=1
        fun(fun(fun(count)))
     
    return count
  
 
if __name__=="__main__"
     
    fun(1)
 
# This code is contributed by Shubhamsingh10

C#




using System;
 
class GFG{
     
    static int fun(int count) 
    
        Console.Write(count+"\n"); 
        if(count < 3) 
        
            fun(fun(fun(++count))); 
        
        return count; 
    
       
    static public void Main ()
    
        fun(1);  
    }
}
 
// This code is contributed by shubhamsingh10

Javascript




<script>
 
    function fun(count)
    {
        document.write(count + "</br>");
        if(count < 3)
        {
            fun(fun(fun(++count)));
        }
        return count;
    }
     
    fun(1);
 
</script>

Output: 

 1
 2
 3
 3
 3
 3
 3

The main() function calls fun(1). fun(1) prints “1” and calls fun(fun(fun(2))). fun(2) prints “2” and calls fun(fun(fun(3))). So the function call sequence becomes fun(fun(fun(fun(fun(3))))). fun(3) prints “3” and returns 3 (note that the count is not incremented and no more functions are called as if the condition is not true for count 3). So the function call sequence reduces to fun(fun(fun(fun(3)))). fun(3) again prints “3” and returns 3. So the function call again reduces to fun(fun(fun(3))) which again prints “3” and reduces it to fun(fun(3)). This continues and we get “3” printed 5 times on the screen. 

Time complexity: O(2n)

Auxiliary Space : O(n), since n extra space has been taken.

Please write comments if you find any of the answers/codes incorrect, or you want to share more information/questions about the topics discussed above.
 


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