Dados dos números, diga a y b. Imprima su XOR después de igualar las longitudes de su representación binaria agregando ceros finales a la representación binaria de uno más pequeño.
Ejemplos:
Input : a = 13, b = 5 Output : 7 Explanation : Binary representation of 13 is 1101 and of 5 is 101. As the length of "101" is smaller, so add a '0' to it making it "1010', to make the length of binary representations equal. XOR of 1010 and 1101 gives 0111 which is 7. Input : a = 7, b = 5 Output : 2 Explanation : Since the length of binary representations of 7 i.e, 111 and 5 i.e, 101 are same, hence simply print XOR of a and b.
Enfoque: cuente el número de bits en representación binaria del número más pequeño de a y b. Si el número de bits en un número más pequeño (digamos a) excede al número más grande (digamos b), entonces aplique el desplazamiento a la izquierda al número más pequeño por el número de bits en exceso, es decir, a = a<<(bits en exceso). Después de aplicar el desplazamiento a la izquierda, se agregarán ceros finales al final de la representación binaria del número más pequeño para que la cantidad de bits en la representación binaria de ambos números sea igual. XOR ambas representaciones binarias para obtener el resultado final.
A continuación se muestra la implementación del método anterior:
C++
// C++ implementation to return
// XOR of two numbers after making
// length of their binary representation same
#include <bits/stdc++.h>
using namespace std;
// function to count the number
// of bits in binary representation
// of an integer
int count(int n)
{
// initialize count
int c = 0;
// count till n is non zero
while (n)
{
c++;
// right shift by 1
// i.e, divide by 2
n = n>>1;
}
return c;
}
// function to calculate the xor of
// two numbers by adding trailing
// zeros to the number having less number
// of bits in its binary representation.
int XOR(int a, int b)
{
// stores the minimum and maximum
int c = min(a,b);
int d = max(a,b);
// left shift if the number of bits
// are less in binary representation
if (count(c) < count(d))
c = c << ( count(d) - count(c) );
return (c^d);
}
// driver code to check the above function
int main()
{
int a = 13, b = 5;
cout << XOR(a,b);
return 0;
}
Java
// Java implementation to return
// XOR of two numbers after making
// length of their binary representation same
import java.io.*;
class GFG {
// function to count the number
// of bits in binary representation
// of an integer
static int count(int n)
{
// initialize count
int c = 0;
// count till n is non zero
while (n != 0)
{
c++;
// right shift by 1
// i.e, divide by 2
n = n >> 1;
}
return c;
}
// function to calculate the xor of
// two numbers by adding trailing
// zeros to the number having less number
// of bits in its binary representation.
static int XOR(int a, int b)
{
// stores the minimum and maximum
int c = Math.min(a, b);
int d = Math.max(a, b);
// left shift if the number of bits
// are less in binary representation
if (count(c) < count(d))
c = c << ( count(d) - count(c) );
return (c ^ d);
}
// driver code to check the above function
public static void main(String args[])
{
int a = 13, b = 5;
System.out.println(XOR(a, b));
}
}
// This code is contributed by Nikita Tiwari.
Python3
# Python 3 implementation to return XOR # of two numbers after making length # of their binary representation same # Function to count the number of bits # in binary representation of an integer def count(n) : # initialize count c = 0 # count till n is non zero while (n != 0) : c += 1 # right shift by 1 # i.e, divide by 2 n = n >> 1 return c # Function to calculate the xor of # two numbers by adding trailing # zeros to the number having less number # of bits in its binary representation. def XOR(a, b) : # stores the minimum and maximum c = min(a, b) d = max(a, b) # left shift if the number of bits # are less in binary representation if (count(c) < count(d)) : c = c << ( count(d) - count(c) ) return (c^d) # Driver Code a = 13; b = 5 print(XOR(a, b)) # This code is contributed by Nikita Tiwari.
C#
// C# implementation to return XOR of two
// numbers after making length of their
// binary representation same
using System;
class GFG {
// function to count the number
// of bits in binary representation
// of an integer
static int count(int n)
{
// initialize count
int c = 0;
// count till n is non zero
while (n != 0)
{
c++;
// right shift by 1
// i.e, divide by 2
n = n >> 1;
}
return c;
}
// function to calculate the xor of
// two numbers by adding trailing
// zeros to the number having less number
// of bits in its binary representation.
static int XOR(int a, int b)
{
// stores the minimum and maximum
int c = Math.Min(a, b);
int d = Math.Max(a, b);
// left shift if the number of bits
// are less in binary representation
if (count(c) < count(d))
c = c << ( count(d) - count(c) );
return (c ^ d);
}
// driver code to check the above function
public static void Main()
{
int a = 13, b = 5;
Console.WriteLine(XOR(a, b));
}
}
// This code is contributed by vt_m.
PHP
<?php
// php implementation to return XOR
// of two numbers after making
// length of their binary
// representation same
// function to count the number
// of bits in binary representation
// of an integer
function count1($n)
{
// initialize count
$c = 0;
// count till n is
// non zero
while ($n)
{
$c++;
// right shift by 1
// i.e, divide by 2
$n = $n>>1;
}
return $c;
}
// function to calculate the xor of
// two numbers by adding trailing
// zeros to the number having less number
// of bits in its binary representation.
function XOR1($a, $b)
{
// stores the minimum
// and maximum
$c = min($a,$b);
$d = max($a,$b);
// left shift if the number of bits
// are less in binary representation
if (count1($c) < count1($d))
$c = $c << ( count1($d) - count1($c) );
return ($c^$d);
}
// Driver Code
$a = 13;
$b = 5;
echo XOR1($a, $b);
// This code is contributed by mits
?>
Javascript
<script>
// JavaScript program to return
// XOR of two numbers after making
// length of their binary representation same
// function to count the number
// of bits in binary representation
// of an integer
function count(n)
{
// initialize count
let c = 0;
// count till n is non zero
while (n != 0)
{
c++;
// right shift by 1
// i.e, divide by 2
n = n >> 1;
}
return c;
}
// function to calculate the xor of
// two numbers by adding trailing
// zeros to the number having less number
// of bits in its binary representation.
function XOR(a, b)
{
// stores the minimum and maximum
let c = Math.min(a, b);
let d = Math.max(a, b);
// left shift if the number of bits
// are less in binary representation
if (count(c) < count(d))
c = c << ( count(d) - count(c) );
return (c ^ d);
}
// Driver code
let a = 13, b = 5;
document.write(XOR(a, b));
</script>
Producción :
7
Complejidad de Tiempo : O(log 2 n)
Espacio Auxiliar : O(1)