Find profession in a special family - GeeksforGeeks

Find profession in a special family - GeeksforGeeks
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 profession of the person.

Examples:

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

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

Source : Oracle Interview

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 profession of parent, then use point 2 above to find profession of current node.

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)

/* 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';

}

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