Primer número que deja un resto impar después de una división repetitiva por 2

Dados dos números enteros A y B , la tarea es imprimir el número entero entre los dos, que se convertirá en un número impar por un número menor de divisiones por 2. Si ambos números se convierten en un número impar después del mismo número de operaciones, imprimir -1. 

Ejemplos:  

Entrada: A = 10 y B = 8 
Salida: 10 
Explicación: 
Paso 1: A/2 = 5, B/2 = 4 
Por lo tanto, A es el primer número que se convierte en un número entero impar.

Entrada: A = 20 y B = 12 
Salida: -1 
Explicación: 
Paso 1: A/2 = 10, B/2 = 6 
Paso 2: A/2 = 5, B/2 = 3 
Por lo tanto, A y B se convierten a un entero impar al mismo tiempo. 
 

Enfoque ingenuo: 
el enfoque más simple para resolver el problema es el siguiente: 

• Comprueba si alguno de los dos números dados es par o impar. Si uno de ellos es impar, imprime ese número.
• Si ambos son impares, imprima -1.
• De lo contrario, siga dividiendo ambos por 2 en cada paso y verifique si alguno de ellos se convierte en un número entero impar o no. Si uno de ellos se convierte, imprime el valor inicial de ese número. Si ambos se convierten al mismo tiempo, imprima -1.

A continuación se muestra la implementación del enfoque anterior: 

// C++ program to find the first
// number to be converted to an odd
// integer by repetitive division by 2
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the first number
// to to be converted to an odd value
int odd_first(int a, int b)
{
     
    // Initial values
    int true_a = a;
    int true_b = b;
 
    // Perform repetitive divisions by 2
    while(a % 2 != 1 && b % 2 != 1)
    {
        a = a / 2;
        b = b / 2;
    }
 
    // If both become odd at same step
    if (a % 2 == 1 && b % 2 == 1)
        return -1;
 
    // If a is first to become odd
    else if (a % 2 == 1)
        return true_a;
 
    // If b is first to become odd
    else
        return true_b;
}
         
// Driver code
int main()
{
    int a = 10;
    int b = 8;
 
    cout << odd_first(a, b);
}
 
// This code is contributed by code_hunt

// Java program to find the first
// number to be converted to an odd
// integer by repetitive division by 2
import java.util.*;
import java.lang.Math;
import java.io.*;
 
class GFG{
     
// Function to return the first number
// to to be converted to an odd value
static int odd_first(int a, int b)
{
     
    // Initial values
    int true_a = a;
    int true_b = b;
 
    // Perform repetitive divisions by 2
    while(a % 2 != 1 && b % 2 != 1)
    {
        a = a / 2;
        b = b / 2;
    }
 
    // If both become odd at same step
    if (a % 2 == 1 && b % 2 == 1)
        return -1;
 
    // If a is first to become odd
    else if (a % 2 == 1)
        return true_a;
 
    // If b is first to become odd
    else
        return true_b;
}
     
// Driver code
public static void main(String[] args)
{
    int a = 10;
    int b = 8;
 
    System.out.print(odd_first(a, b));
}
}
 
// This code is contributed by code_hunt

# Python3 program to find the first
# number to be converted to an odd
# integer by repetitive division by 2
 
# Function to return the first number
# to to be converted to an odd value
def odd_first(a, b):
 
  # Initial values
  true_a = a
  true_b = b
 
  # Perform repetitive divisions by 2
  while(a % 2 != 1 and b % 2 != 1):
    a = a//2
    b = b//2
 
  # If both become odd at same step
  if a % 2 == 1 and b % 2 == 1:
    return -1
 
  # If a is first to become odd
  elif a % 2 == 1:
    return true_a
 
  # If b is first to become odd
  else:
    return true_b
 
# Driver Code
a, b = 10, 8
print(odd_first(a, b))

// C# program to find the first
// number to be converted to an odd
// integer by repetitive division by 2
using System;
 
class GFG{
     
// Function to return the first number
// to to be converted to an odd value
static int odd_first(int a, int b)
{
     
    // Initial values
    int true_a = a;
    int true_b = b;
 
    // Perform repetitive divisions by 2
    while(a % 2 != 1 && b % 2 != 1)
    {
        a = a / 2;
        b = b / 2;
    }
 
    // If both become odd at same step
    if (a % 2 == 1 && b % 2 == 1)
        return -1;
 
    // If a is first to become odd
    else if (a % 2 == 1)
        return true_a;
 
    // If b is first to become odd
    else
        return true_b;
}
     
// Driver code
public static void Main()
{
    int a = 10;
    int b = 8;
 
    Console.Write(odd_first(a, b));
}
}
 
// This code is contributed by sanjoy_62

<script>
 
// Javascript program to find the first
// number to be converted to an odd
// integer by repetitive division by 2
 
// Function to return the first number
// to to be converted to an odd value
function odd_first(a, b)
{
     
    // Initial values
    var true_a = a;
    var true_b = b;
 
    // Perform repetitive divisions by 2
    while(a % 2 != 1 && b % 2 != 1)
    {
        a = a / 2;
        b = b / 2;
    }
 
    // If both become odd at same step
    if (a % 2 == 1 && b % 2 == 1)
        return -1;
 
    // If a is first to become odd
    else if (a % 2 == 1)
        return true_a;
 
    // If b is first to become odd
    else
        return true_b;
}
 
// Driver code
var a = 10, b = 8;
 
document.write(odd_first(a, b));
 
// This code is contributed by Ankita saini
    
</script>

10

Complejidad de tiempo: O(log(min(a, b))) 
Espacio auxiliar: O(1)

Enfoque eficiente: 
siga los pasos a continuación para optimizar el enfoque anterior:  

• Dividir un número por 2 es equivalente a realizar el desplazamiento a la derecha en ese número.
• Por lo tanto, el número con el bit de conjunto menos significativo entre los dos será el primero en convertirse en un número entero impar.

A continuación se muestra la implementación del enfoque anterior.  

// C++ Program to implement the
// above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the position
// least significant set bit
int getFirstSetBitPos(int n)
{
    return log2(n & -n) + 1;
}
 
// Function return the first number
// to be converted to an odd integer
int oddFirst(int a, int b)
{
 
    // Stores the positions of the
    // first set bit
    int steps_a = getFirstSetBitPos(a);
    int steps_b = getFirstSetBitPos(b);
 
    // If both are same
    if (steps_a == steps_b) {
        return -1;
    }
 
    // If A has the least significant
    // set bit
    if (steps_a > steps_b) {
        return b;
    }
 
    // Otherwise
    if (steps_a < steps_b) {
        return a;
    }
}
 
// Driver code
int main()
{
    int a = 10;
    int b = 8;
 
    cout << oddFirst(a, b);
} 

// Java program implementation
// of the approach
import java.util.*;
import java.lang.Math;
import java.io.*;
 
class GFG{
     
// Function to return the position
// least significant set bit
static int getFirstSetBitPos(int n)
{
    return (int)(Math.log(n & -n) /
                 Math.log(2));
}
 
// Function return the first number
// to be converted to an odd integer
static int oddFirst(int a, int b)
{
     
    // Stores the positions of the
    // first set bit
    int steps_a = getFirstSetBitPos(a);
    int steps_b = getFirstSetBitPos(b);
 
    // If both are same
    if (steps_a == steps_b)
    {
        return -1;
    }
 
    // If A has the least significant
    // set bit
    else if (steps_a > steps_b)
    {
        return b;
    }
 
    // Otherwise
    else
    {
        return a;
    }
}
     
// Driver code
public static void main(String[] args)
{
    int a = 10;
    int b = 8;
 
    System.out.print(oddFirst(a, b));
}
}
 
// This code is contributed by code_hunt

# Python3 program to implement the
# above approach
from math import log
 
# Function to return the position
# least significant set bit
def getFirstSetBitPos(n):
    return log(n & -n, 2) + 1
 
# Function return the first number
# to be converted to an odd integer
def oddFirst(a, b):
 
    # Stores the positions of the
    # first set bit
    steps_a = getFirstSetBitPos(a)
    steps_b = getFirstSetBitPos(b)
 
    # If both are same
    if (steps_a == steps_b):
        return -1
 
    # If A has the least significant
    # set bit
    if (steps_a > steps_b):
        return b
 
    # Otherwise
    if (steps_a < steps_b):
        return a
 
# Driver code
if __name__ == '__main__':
     
    a = 10
    b = 8
     
    print(oddFirst(a, b))
 
# This code is contributed by mohit kumar 29

// C# program implementation
// of the approach
using System;
 
class GFG{
     
// Function to return the position
// least significant set bit
static int getFirstSetBitPos(int n)
{
    return (int)(Math.Log(n & -n) /
                 Math.Log(2));
}
 
// Function return the first number
// to be converted to an odd integer
static int oddFirst(int a, int b)
{
     
    // Stores the positions of the
    // first set bit
    int steps_a = getFirstSetBitPos(a);
    int steps_b = getFirstSetBitPos(b);
 
    // If both are same
    if (steps_a == steps_b)
    {
        return -1;
    }
 
    // If A has the least significant
    // set bit
    else if (steps_a > steps_b)
    {
        return b;
    }
 
    // Otherwise
    else
    {
        return a;
    }
}
     
// Driver code
public static void Main()
{
    int a = 10;
    int b = 8;
 
    Console.Write(oddFirst(a, b));
}
}
 
// This code is contributed by sanjoy_62

<script>
 
// Javascript program implementation
// of the approach
 
// Function to return the position
// least significant set bit
function getFirstSetBitPos(n)
{
    return (Math.log(n & -n) /
                 Math.log(2));
}
  
// Function return the first number
// to be converted to an odd integer
function oddFirst(a, b)
{
      
    // Stores the positions of the
    // first set bit
    let steps_a = getFirstSetBitPos(a);
    let steps_b = getFirstSetBitPos(b);
  
    // If both are same
    if (steps_a == steps_b)
    {
        return -1;
    }
  
    // If A has the least significant
    // set bit
    else if (steps_a > steps_b)
    {
        return b;
    }
  
    // Otherwise
    else
    {
        return a;
    }
}
   
 
// Driver Code
     
       let a = 10;
    let b = 8;
  
    document.write(oddFirst(a, b));
         
</script>

10

Complejidad temporal: O(1) 
Espacio auxiliar: O(1)