Practice Recursion - Easy

Question 1 : Explain the functionality of the following functions.

C++

using namespace std;
int fun1(int x, int y)
{
    if (x == 0)
        return y;
    else
        return fun1(x - 1, x + y);
}

int main()
{
    cout << fun1(5, 2) << endl;  
    return 0;
}

int fun1(int x, int y)
{
    if (x == 0)
        return y;
    else
        return fun1(x - 1, x + y);
}

// Driver code
int main()
{
    printf("%d\n", fun1(5, 2));   
    return 0;
}

Java

static int fun1(int x, int y)
{
    if (x == 0)
        return y;
    else
        return fun1(x - 1, x + y);
}

// Driver code
public static void main(String[] args)
{
    System.out.println(fun1(5, 2)); 
}

Python

def fun1(x, y) :

    if (x == 0) :
        return y
    else :
        return fun1(x - 1, x + y)

# Driver code
print(fun1(5, 2))

using System;

class GFG {

    static int fun1(int x, int y)
    {
        if (x == 0)
            return y;
        else
            return fun1(x - 1, x + y);
    }

    // Driver code
    static void Main(string[] args)
    {
        Console.WriteLine(fun1(5, 2));  
    }
}

JavaScript

function fun1(x, y)
{
    if (x == 0) 
        return y
    else 
        return fun1(x - 1, x + y)
}

// Driver code

console.log(fun1(5, 2));

Answer: This function adds the sum of first x natural numbers to y. It repeatedly decreases x and adds its value to y until x becomes 0. Final result: y + (1 + 2 + ... + x) = y + x(x + 1) / 2. For example, if x is 5 and y is 2, then fun should return 15 + 2 = 17.

Question 2 : Explain the functionality of the following functions.

C++

using namespace std;

/* 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);
}

// Driver code
int main() {

    cout << fun1(8);
    return 0;
}

Java

class GFG {

    /* 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);
    }
    
    // Driver code
    public static void main(String[] args)
    {
        System.out.println(fun1(8));
    }
}

Python

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

# Driver code
print(fun1(8))

using System;

// Assume that n is greater than or equal to 1 
public class GFG {

    static int fun1(int n)
    {
        if (n == 1)
            return 0;
        else
            return 1 + fun1(n / 2);
    }
    
    //  Driver code
    static public void Main()
    {
        Console.WriteLine(fun1(8));
    }
}

JavaScript

// Assume that n is greater than or equal to 1
function fun1(n)
{
    if (n <= 1)
        return 0
    else
        return 1 + fun1(Math.floor(n / 2))
}

// Driver code
console.log(fun1(8))

The above function calculates and returns ⌊log₂(n)⌋. 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 3

C++

using namespace std;

/* Assume that n is greater than or equal to 0 */
void fun2(int n)
{
    if (n == 0)
        return;

    fun2(n / 2);
    cout << n % 2;
}

// Driver code

int main()
{
    fun2(21);
    return 0;
}

#include <stdio.h>

void fun2(int n)
{
  if(n == 0)
    return;

  fun2(n/2);
  printf("%d", n%2);
}

// Driver code

int main() {

    fun2(21);
    return 0;
}

Java

class GFG {

    /* Assume that n is greater than or equal to 1 */
    static void fun2(int n)
    {
        if (n == 0)
            return;

        fun2(n / 2);
        System.out.print(n % 2);
    }
    
    // Driver code
    
    public static void main(String[] args) { fun2(21); }
}

Python

def fun2(n):
    if(n == 0):
        return
    
    fun2(n // 2)
    print(n % 2, end="")

# Driver code

fun2(21)

using System;

public class GFG {

    static void fun2(int n)
    {
        if (n == 0)
            return;

        fun2(n / 2);
        Console.Write(n % 2);
    }
    // Driver code
    static public void Main()
    {
        fun2(21);
    }
}

JavaScript

// Assume that n is greater than or equal to 1 
function fun2(n)
{
    if (n == 0)
        return;
    
    fun2(Math.floor(n / 2));
    console.log(n % 2 + " ")
}

// Driver code
fun2(21)

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

Question 4 : Predict the output of the following program.

C++

#include <iostream>
using namespace std;

void fun(int x) 
{ 
    if(x > 0) 
    { 
        fun(--x); 
        cout << x <<" "; 
        fun(--x); 
    } 
} 

// Driver code
int main() 
{ 
    int a = 4; 
    fun(a); 
    return 0; 
}

#include<stdio.h>
void fun(int x)
{
  if(x > 0)
  {
     fun(--x);
     printf("%d\t", x);
     fun(--x);
  }
}

// Driver code
int main()
{
  int a = 4;
  fun(a);
  getchar();
  return 0;
}

Java

import java.io.*;

class GFG {

    static void fun(int x) 
    { 
        if(x > 0) 
        { 
            fun(--x); 
            System.out.print(x + " "); 
            fun(--x); 
        } 
    } 
    
    // Driver code
    public static void main (String[] args)
    { 
        int a = 4; 
        fun(a); 
    } 
}

Python

def fun(x):
    
    if(x > 0):
        x -= 1
        fun(x) 
        print(x , end=" ")
        x -= 1
        fun(x) 
        
# Driver code
a = 4 
fun(a)

using System;

class GFG{

    static void fun(int x) 
    { 
        if(x > 0) 
        { 
            fun(--x); 
            Console.Write(x + " "); 
            fun(--x); 
        } 
    }  
    
    // Driver code
    static public void Main () 
    { 
        int a = 4; 
        fun(a);  
    } 
}

JavaScript

function fun(x)
{
    if (x > 0)
    {
        x -= 1
        fun(x)
        console.log(x + " ");
        x -= 1
        fun(x)
    }    
}

// Driver code
let a = 4;
fun(a)

                   fun(4);
                   /
                fun(3), print(3), fun(2)(prints 0 1)
               /
           fun(2), print(2), fun(1)(prints 0)
           /
       fun(1), print(1), fun(0)(does nothing)
       /
    fun(0), print(0), fun(-1) (does nothing)

Explain the functionality of below recursive functions.

Question 5 : Predict the output of the following program.

C++

using namespace std;
void fun1(int n) 
{ 
   int i = 0;   
   if (n > 1) 
     fun1(n - 1); 
   for (i = 0; i < n; i++) 
     cout << " * "; 
} 

// Driver code
int main()
{
    fun1(4);
}

void fun1(int n)
{
   int i = 0;  
   if (n > 1)
     fun1(n-1);
   for (i = 0; i < n; i++)
     printf(" * ");
}

// Driver code
void main()
{
    fun1(4);
}

Java

class GFG {

    static void fun1(int n)
    {
        int i = 0;
        if (n > 1)
            fun1(n - 1);
        for (i = 0; i < n; i++)
            System.out.print(" * ");
    }
    
    // Driver code
    public static void main(String[] args) { fun1(4); }
}

Python

def fun1(n):
    i = 0
    if (n > 1):
        fun1(n - 1)
    for i in range(n):
        print(" * ", end="")

# Driver code
fun1(4)

using System;
class GFG {
    static void fun1(int n)
    {
        int i = 0;
        if (n > 1)
            fun1(n - 1);
        for (i = 0; i < n; i++)
            Console.Write(" * ");
    }

    // Driver code
    static void Main(string[] args) { fun1(4); }
}

JavaScript

function fun1(n)
{
    let i = 0;

    if (n > 1)
        fun1(n - 1);

    for (i = 0; i < n; i++)
        console.log(" * ");
}

// Driver code
fun1(4)

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

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Practice Recursion - Easy

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