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

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

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 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 doctar, this node will be a doctal 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 enginner 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 doctar, this node      // will be a doctal 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 enginner 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 doctar, this node will be a doctal 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 enginner if at add position and doctor if even     # position.     if(pos%2):         return 'e'     else:         return 'd'    # Driver code if __name__ == '__main__':     level = 4     pos = 2     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 doctar, this node      // will be a doctal 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 enginner 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



Output :

Doctor

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 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); } }

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



Output :

Engineer

Thanks to Furkan Uslu for suggesting this method.

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