Find profession in a special family

Consider a special family of Engineers and Doctors with following rules : 

1. Everybody has two children.
2. First child of an Engineer is an Engineer and second child is a Doctor.
3. First child of an Doctor is Doctor and second child is an Engineer.
4. All generations of Doctors and Engineers start with Engineer.

We can represent the situation using below diagram: 

                E
           /        
          E          D
        /          /  
       E     D     D    E
      /    /    /    / 
     E   D D   E  D  E  E  D

Given level and position of a person in above ancestor tree, find the profession of the person.
Examples :

Input : level = 4, pos = 2
Output : Doctor

Input : level = 3, pos = 4
Output : Engineer

Method 1 (Recursive)
The idea is based on the fact that profession of a person depends on following two.

1. Profession of parent.
2. Position of node : If position of a node is odd, then its profession is same as its parent. Else profession is different from its parent.

We recursively find the profession of parent, then use point 2 above to find the profession of current node. 
Below is implementation of above idea. 

C++

// C++ program to find profession of a person at
// given level and position.
#include<bits/stdc++.h>
using namespace std;

// Returns 'e' if profession of node at given level
// and position is engineer. Else doctor. The function
// assumes that given position and level have valid values.
char findProffesion(int level, int pos)
{
    // Base case
    if (level == 1)
        return 'e';

    // Recursively find parent's profession. If parent
    // is a Doctor, this node will be a Doctor if it is
    // at odd position and an engineer if at even position
    if (findProffesion(level-1, (pos+1)/2) == 'd')
        return (pos%2)? 'd' : 'e';

    // If parent is an engineer, then current node will be
    // an engineer if at add position and doctor if even
    // position.
    return (pos%2)?  'e' : 'd';
}

// Driver code
int main(void)
{
    int level = 4, pos = 2;
    (findProffesion(level, pos) == 'e')? cout << "Engineer"
                                       : cout << "Doctor" ;
    return 0;
}

Java

// Java program to find
// profession of a person
// at given level and position
import java.io.*;

class GFG
{

// Returns 'e' if profession
// of node at given level
// and position is engineer.
// Else doctor. The function
// assumes that given position
// and level have valid values.
static char findProffesion(int level,
                           int pos)
{
    // Base case
    if (level == 1)
        return 'e';

    // Recursively find parent's
    // profession. If parent
    // is a Doctor, this node
    // will be a Doctor if it
    // is at odd position and an
    // engineer if at even position
    if (findProffesion(level - 1,
                      (pos + 1) / 2) == 'd')
        return (pos % 2 > 0) ?
                         'd' : 'e';

    // If parent is an engineer,
    // then current node will be
    // an engineer if at add
    // position and doctor if even
    // position.
    return (pos % 2 > 0) ?
                     'e' : 'd';
}

// Driver code
public static void main (String[] args)
{
    int level = 4, pos = 2;
    if(findProffesion(level,
                      pos) == 'e')
    System.out.println("Engineer");
    else
    System.out.println("Doctor");
}
}

// This code is contributed
// by anuj_67.

Python3

# python 3 program to find profession of a person at
# given level and position.

# Returns 'e' if profession of node at given level
# and position is engineer. Else doctor. The function
# assumes that given position and level have valid values.
def findProffesion(level, pos):
    # Base case
    if (level == 1):
        return 'e'

    # Recursively find parent's profession. If parent
    # is a Doctor, this node will be a Doctor if it is
    # at odd position and an engineer if at even position
    if (findProffesion(level-1, (pos+1)//2) == 'd'):
        if (pos%2):
            return 'd'
        else:
            return 'e'

    # If parent is an engineer, then current node will be
    # an engineer if at add position and doctor if even
    # position.
    if(pos%2):
        return 'e'
    else:
        return 'd'

# Driver code
if __name__ == '__main__':
    level = 3
    pos = 4
    if(findProffesion(level, pos) == 'e'):
        print("Engineer")
    else:
        print("Doctor")

# This code is contributed by
# Surendra_Gangwar

C#

// C# program to find
// profession of a person
// at given level and position
using System;

class GFG
{

// Returns 'e' if profession
// of node at given level
// and position is engineer.
// Else doctor. The function
// assumes that given position
// and level have valid values.
static char findProffesion(int level,
                           int pos)
{
    // Base case
    if (level == 1)
        return 'e';

    // Recursively find parent's
    // profession. If parent
    // is a Doctor, this node
    // will be a Doctor if it
    // is at odd position and an
    // engineer if at even position
    if (findProffesion(level - 1,
                      (pos + 1) / 2) == 'd')
        return (pos % 2 > 0) ?
                         'd' : 'e';

    // If parent is an engineer,
    // then current node will be
    // an engineer if at add
    // position and doctor if even
    // position.
    return (pos % 2 > 0) ?
                     'e' : 'd';
}

// Driver code
public static void Main ()
{
    int level = 4, pos = 2;
    if(findProffesion(level,
                    pos) == 'e')
    Console.WriteLine("Engineer");
    else
    Console.WriteLine("Doctor");
}
}

// This code is contributed
// by anuj_67.

PHP

<?php
// PHP program to find profession
// of a person at given level
// and position.

// Returns 'e' if profession of
// node at given level and position
// is engineer. Else doctor. The
// function assumes that given
// position and level have valid values.
function findProffesion($level, $pos)
{
    // Base case
    if ($level == 1)
        return 'e';

    // Recursively find parent's
    // profession. If parent is
    // a Doctor, this node will
    // be a doctor if it is at
    // odd position and an engineer
    // if at even position
    if (findProffesion($level - 1,
                      ($pos + 1) / 2) == 'd')
        return ($pos % 2) ? 'd' : 'e';

    // If parent is an engineer, then
    // current node will be an engineer
    // if at odd position and doctor
    // if even position.
    return ($pos % 2) ? 'e' : 'd';
}

// Driver code
$level = 4; $pos = 2;
if((findProffesion($level,
                   $pos) == 'e') == true)
    echo "Engineer";
else
    echo "Doctor" ;

// This code is contributed by ajit
?>

Javascript

<script>

// JavaScript program to find
// profession of a person
// at given level and position

// Returns 'e' if profession
// of node at given level
// and position is engineer.
// Else doctor. The function
// assumes that given position
// and level have valid values.
function findProffesion(level,
                           pos)
{
    // Base case
    if (level == 1)
        return 'e';

    // Recursively find parent's
    // profession. If parent
    // is a Doctor, this node
    // will be a Doctor if it
    // is at odd position and an
    // engineer if at even position
    if (findProffesion(level - 1,
                      (pos + 1) / 2) == 'd')
        return (pos % 2 > 0) ?
                         'd' : 'e';

    // If parent is an engineer,
    // then current node will be
    // an engineer if at add
    // position and doctor if even
    // position.
    return (pos % 2 > 0) ?
                     'e' : 'd';
}

// Driver Code

    let level = 4, pos = 2;
    if(findProffesion(level,
                      pos) == 'e')
    document.write("Engineer");
    else
    document.write("Doctor");

</script>

Output :

Doctor

Time Complexity: O(logn)

Auxiliary Space: O(1) 

Method 2 (Using Bitwise Operators)

Level 1: E
Level 2: ED
Level 3: EDDE
Level 4: EDDEDEED
Level 5: EDDEDEEDDEEDEDDE 

Level input isn’t necessary (if we ignore max position limit) because first elements are same.
The result is based on count of 1’s in binary representation of position minus one. If count of 1’s is even then result is Engineer, else then Doctor.
And of course position limit is 2^(Level-1)

C++

// C++ program to find profession of a person at
// given level and position.
#include<bits/stdc++.h>
using namespace std;

/* Function to get no of set bits in binary
   representation of passed binary no. */
int countSetBits(int n)
{
    int count = 0;
    while (n)
    {
      n &= (n-1) ;
      count++;
    }
    return count;
}

// Returns 'e' if profession of node at given level
// and position is engineer. Else doctor. The function
// assumes that given position and level have valid values.
char findProffesion(int level, int pos)
{
    // Count set bits in 'pos-1'
    int c = countSetBits(pos-1);

    // If set bit count is odd, then doctor, else engineer
    return (c%2)?  'd' : 'e';
}

// Driver code
int main(void)
{
    int level = 3, pos = 4;
    (findProffesion(level, pos) == 'e')? cout << "Engineer"
                                       : cout << "Doctor" ;
    return 0;
}

Java

// Java program to find profession of a person at
// given level and position.
class GFG{
/* Function to get no of set bits in binary
representation of passed binary no. */
static int countSetBits(int n)
{
    int count = 0;
    while (n!=0)
    {
    n &= (n-1) ;
    count++;
    }
    return count;
}

// Returns 'e' if profession of node at given level
// and position is engineer. Else doctor. The function
// assumes that given position and level have valid values.
static char findProffesion(int level, int pos)
{
    // Count set bits in 'pos-1'
    int c = countSetBits(pos-1);

    // If set bit count is odd, then doctor, else engineer
    return (c%2 !=0)? 'd' : 'e';
}

// Driver code
public static void main(String [] args)
{
    int level = 3, pos = 4;
    String prof = (findProffesion(level, pos) == 'e')? "Engineer"
                                    : "Doctor"  ;
    System.out.print(prof);
}
}

Python3

# Python3 program to find profession of a person at
# given level and position.

""" Function to get no of set bits in binary
   representation of passed binary no. """
def countSetBits(n):
    count = 0
    while n > 0:
      n &= (n-1)
      count+=1
    return count

# Returns 'e' if profession of node at given level
# and position is engineer. Else doctor. The function
# assumes that given position and level have valid values.
def findProffesion(level, pos):
    # Count set bits in 'pos-1'
    c = countSetBits(pos-1)

    # If set bit count is odd, then doctor, else engineer
    if c%2 == 0:
        return 'e'
    else:
        return 'd'

level, pos = 3, 4
if findProffesion(level, pos) == 'e':
    print("Engineer")
else:
    print("Doctor")

    # This code is contributed by divyeshrabadiya07.

C#

using System;

// c# program to find profession of a person at
// given level and position.
public class GFG
{
/* Function to get no of set bits in binary
representation of passed binary no. */
public static int countSetBits(int n)
{
    int count = 0;
    while (n != 0)
    {
    n &= (n - 1);
    count++;
    }
    return count;
}

// Returns 'e' if profession of node at given level
// and position is engineer. Else doctor. The function
// assumes that given position and level have valid values.
public static char findProffesion(int level, int pos)
{
    // Count set bits in 'pos-1'
    int c = countSetBits(pos - 1);

    // If set bit count is odd, then doctor, else engineer
    return (c % 2 != 0)? 'd' : 'e';
}

// Driver code
public static void Main(string[] args)
{
    int level = 3, pos = 4;
    string prof = (findProffesion(level, pos) == 'e')? "Engineer" : "Doctor";
    Console.Write(prof);
}
}

  // This code is contributed by Shrikant13

PHP

<?php
// PHP program to find profession
// of a person at given level and position.

// Function to get no of set
// bits in binary representation
// of passed binary no.
function countSetBits($n)
{
    $count = 0;
    while ($n)
    {
        $n &= ($n - 1) ;
        $count++;
    }
    return $count;
}

// Returns 'e' if profession of
// node at given level and position
// is engineer. Else doctor. The
// function assumes that given
// position and level have valid values.
function findProffesion($level, $pos)
{
    // Count set bits in 'pos-1'
    $c = countSetBits($pos - 1);

    // If set bit count is odd,
    // then doctor, else engineer
    return ($c % 2) ? 'd' : 'e';
}

// Driver Code
$level = 3;
$pos = 4;
if((findProffesion($level,
                   $pos) == 'e') == true)
        echo "Engineer \n";
    else
        echo "Doctor \n";

// This code is contributed by aj_36
?>

Javascript

<script>

    // Javascript program to find
    // profession of a person at
    // given level and position.

    /* Function to get no of set bits in binary
    representation of passed binary no. */
    function countSetBits(n)
    {
        let count = 0;
        while (n != 0)
        {
          n &= (n - 1);
          count++;
        }
        return count;
    }

    // Returns 'e' if profession of node at given level
    // and position is engineer. Else doctor.
    // The function assumes that given position and
    // level have valid values.
    function findProffesion(level, pos)
    {
        // Count set bits in 'pos-1'
        let c = countSetBits(pos - 1);

        // If set bit count is odd, then doctor,
        // else engineer
        return (c % 2 != 0)? 'd' : 'e';
    }

    let level = 3, pos = 4;
    let prof = (findProffesion(level, pos) == 'e')?
                "Engineer" : "Doctor";
    document.write(prof);

</script>

Output :

Engineer

Time Complexity: O(logn)

Auxiliary Space: O(1)

Thanks to Furkan Uslu for suggesting this method.