Find profession in a special family

3.4

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

Source : Oracle Interview

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 C++ implementation of above idea.

// 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 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;
}

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++ 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;
}

Output:

Engineer

Thanks to Furkan Uslu for suggesting this method.

This article is contributed by Harsh Parikh. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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