Найдите профессию в особенной семье

Опубликовано: 20 Января, 2022

Рассмотрим особую семью инженеров и врачей со следующими правилами:

  1. У всех двое детей.
  2. Первый ребенок инженера - инженер, а второй ребенок - доктор.
  3. Первый ребенок Доктора - Доктор, а второй ребенок - Инженер.
  4. Все поколения врачей и инженеров начинаются с инженера.

Мы можем представить ситуацию, используя диаграмму ниже:

 E
           /        
          ED
        / /  
       EDDE
      / / / / 
     EDDEDEED

Зная уровень и положение человека в верхнем дереве предков, найдите профессию человека.
Примеры :

 Ввод: уровень = 4, поз = 2
Выход: Доктор

Ввод: уровень = 3, поз = 4
Продукт: Инженер
Рекомендуется: сначала решите эту проблему на «ПРАКТИКЕ», прежде чем переходить к решению.

Метод 1 (рекурсивный)
Идея основана на том, что профессия человека зависит от следующих двух.

  1. Профессия родителя.
  2. Положение узла: если положение узла нечетное, то его профессия такая же, как у его родителя. Другая профессия отличается от своей родительской.

Мы рекурсивно находим профессию родителя, затем используем пункт 2 выше, чтобы найти профессию текущего узла.
Ниже представлена реализация вышеизложенной идеи.

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>

Выход :

 Доктор

Метод 2 (с использованием побитовых операторов)

 Уровень 1: E
Уровень 2: ED
Уровень 3: EDDE
Уровень 4: EDDEDEED
Уровень 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);
}
}

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.
 
Oracle

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