Practice Questions for Recursion | Set 6

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. 

void abc(char *s)
{
    if(s[0] == '\0')
        return;
 
    abc(s + 1);
    abc(s + 1);
    cout << s[0];    
}

void abc(char *s)
{
    if(s[0] == '\0')
        return;

    abc(s + 1);
    abc(s + 1);
    printf("%c", s[0]);    
}

void abc(String s)
{
    if (s.length() == 0)
        return;

    abc(s.substring(1));
    abc(s.substring(1));
    System.out.print(s.charAt(0));
}

def abc(s):
    if(len(s) == 0):
        return
    
    abc(s[1:])
    abc(s[1:])
    print(s[0],end="")

void abc(string s)
{
    if (s.Length == 0)
        return;

    abc(s.Substring(1));
    abc(s.Substring(1));
    Console.Write(s[0]);
}

function abc(s)
{
    if (s.length == 0)
        return;

    abc(s.substring(1));
    abc(s.substring(1));
    console.log(s[0]);
}

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. 

#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

#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;
}

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

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

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

function abc(s)
{
    if (s.length == 0)
        return;

    abc(s.substring(1));
    abc(s.substring(1));
    console.log(s[0]);
}

abc("xyz");

Question 2 : Guess the output of the following code.

#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

#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;
}

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

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 

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

<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. 

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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Article Tags :

• DSA
• Recursion

Practice Tags :

• Recursion