Dado un número n . El problema es verificar si cada bit en la representación binaria del número dado está configurado o no. Aquí 0 <= n .
Ejemplos:
Input : 7 Output : Yes (7)10 = (111)2 Input : 14 Output : No
Método 1: si n = 0, la respuesta es ‘No’. De lo contrario, realice las dos operaciones hasta que n se convierta en 0.
While (n > 0)
If n & 1 == 0,
return 'No'
n >> 1
Si el ciclo termina sin devolver ‘No’, entonces todos los bits se establecen en la representación binaria de n .
C++
// C++ implementation to check whether every digit in the
// binary representation of the given number is set or not
#include <bits/stdc++.h>
using namespace std;
// function to check if all the bits are set or not in the
// binary representation of 'n'
string areAllBitsSet(int n)
{
// all bits are not set
if (n == 0)
return "No";
// loop till n becomes '0'
while (n > 0) {
// if the last bit is not set
if ((n & 1) == 0)
return "No";
// right shift 'n' by 1
n = n >> 1;
}
// all bits are set
return "Yes";
}
// Driver program to test above
int main()
{
int n = 7;
cout << areAllBitsSet(n);
return 0;
}
// This code is contributed by Sania Kumari Gupta (kriSania804)
C
// C implementation to check whether every digit in the
// binary representation of the given number is set or not
#include <stdio.h>
// function to check if all the bits are set or not in the
// binary representation of 'n'
void areAllBitsSet(int n)
{
// all bits are not set
if (n == 0)
printf("No");
// loop till n becomes '0'
while (n > 0) {
// if the last bit is not set
if ((n & 1) == 0)
printf("No");
// right shift 'n' by 1
n = n >> 1;
}
// all bits are set
printf("Yes");
}
// Driver program to test above
int main()
{
int n = 7;
areAllBitsSet(n);
return 0;
}
// This code is contributed by Sania Kumari Gupta (kriSania804)
Java
// java implementation to check
// whether every digit in the
// binary representation of the
// given number is set or not
import java.io.*;
class GFG {
// function to check if all the bits
// are setthe bits are set or not
// in the binary representation of 'n'
static String areAllBitsSet(int n)
{
// all bits are not set
if (n == 0)
return "No";
// loop till n becomes '0'
while (n > 0)
{
// if the last bit is not set
if ((n & 1) == 0)
return "No";
// right shift 'n' by 1
n = n >> 1;
}
// all bits are set
return "Yes";
}
// Driver program to test above
public static void main (String[] args) {
int n = 7;
System.out.println(areAllBitsSet(n));
}
}
// This code is contributed by vt_m
Python3
# Python implementation # to check whether every # digit in the binary # representation of the # given number is set or not # function to check if # all the bits are set # or not in the binary # representation of 'n' def areAllBitsSet(n): # all bits are not set if (n == 0): return "No" # loop till n becomes '0' while (n > 0): # if the last bit is not set if ((n & 1) == 0): return "No" # right shift 'n' by 1 n = n >> 1 # all bits are set return "Yes" # Driver program to test above n = 7 print(areAllBitsSet(n)) # This code is contributed # by Anant Agarwal.
C#
// C# implementation to check
// whether every digit in the
// binary representation of the
// given number is set or not
using System;
class GFG
{
// function to check if
// all the bits are set
// or not in the binary
// representation of 'n'
static String areAllBitsSet(int n)
{
// all bits are not set
if (n == 0)
return "No";
// loop till n becomes '0'
while (n > 0)
{
// if the last bit
// is not set
if ((n & 1) == 0)
return "No";
// right shift 'n' by 1
n = n >> 1;
}
// all bits are set
return "Yes";
}
// Driver Code
static public void Main ()
{
int n = 7;
Console.WriteLine(areAllBitsSet(n));
}
}
// This code is contributed by ajit
PHP
<?php
// PHP implementation to check
// whether every digit in the
// binary representation of the
// given number is set or not
// function to check if all the
// bits are set or not in the
// binary representation of 'n'
function areAllBitsSet($n)
{
// all bits are not set
if ($n == 0)
return "No";
// loop till n becomes '0'
while ($n > 0)
{
// if the last bit is not set
if (($n & 1) == 0)
return "No";
// right shift 'n' by 1
$n = $n >> 1;
}
// all bits are set
return "Yes";
}
// Driver Code
$n = 7;
echo areAllBitsSet($n);
// This code is contributed by aj_36
?>
Javascript
<script>
// javascript implementation to check
// whether every digit in the
// binary representation of the
// given number is set or not
// function to check if all the bits
// are setthe bits are set or not
// in the binary representation of 'n'
function areAllBitsSet(n)
{
// all bits are not set
if (n == 0)
return "No";
// loop till n becomes '0'
while (n > 0)
{
// if the last bit is not set
if ((n & 1) == 0)
return "No";
// right shift 'n' by 1
n = n >> 1;
}
// all bits are set
return "Yes";
}
// Driver program to test above
var n = 7;
document.write(areAllBitsSet(n));
// This code contributed by Princi Singh
</script>
Producción :
Yes
Complejidad de tiempo: O(d), donde ‘d’ es el número de bits en la representación binaria de n .
Espacio auxiliar: O(1)
Método 2: Si n = 0, la respuesta es ‘No’. De lo contrario, agregue 1 a n . Sea num = n + 1. Si num & (num – 1) == 0, entonces todos los bits están configurados, de lo contrario, todos los bits no están configurados.
Explicación: si todos los bits en la representación binaria de n están configurados, agregarle ‘1’ producirá un número que será una potencia perfecta de 2. Ahora, verifique si el nuevo número es una potencia perfecta de 2 o no.
C++
// C++ implementation to check whether every
// digit in the binary representation of the
// given number is set or not
#include <bits/stdc++.h>
using namespace std;
// function to check if all the bits are set
// or not in the binary representation of 'n'
string areAllBitsSet(int n)
{
// all bits are not set
if (n == 0)
return "No";
// if true, then all bits are set
if (((n + 1) & n) == 0)
return "Yes";
// else all bits are not set
return "No";
}
// Driver program to test above
int main()
{
int n = 7;
cout << areAllBitsSet(n);
return 0;
}
Java
// JAVA implementation to check whether
// every digit in the binary representation
// of the given number is set or not
import java.io.*;
class GFG {
// function to check if all the
// bits are set or not in the
// binary representation of 'n'
static String areAllBitsSet(int n)
{
// all bits are not set
if (n == 0)
return "No";
// if true, then all bits are set
if (((n + 1) & n) == 0)
return "Yes";
// else all bits are not set
return "No";
}
// Driver program to test above
public static void main (String[] args) {
int n = 7;
System.out.println(areAllBitsSet(n));
}
}
// This code is contributed by vt_m
Python3
# Python implementation to # check whether every # digit in the binary # representation of the # given number is set or not # function to check if # all the bits are set # or not in the binary # representation of 'n' def areAllBitsSet(n): # all bits are not set if (n == 0): return "No" # if true, then all bits are set if (((n + 1) & n) == 0): return "Yes" # else all bits are not set return "No" # Driver program to test above n = 7 print(areAllBitsSet(n)) # This code is contributed # by Anant Agarwal.
C#
// C# implementation to check
// whether every digit in the
// binary representation of
// the given number is set or not
using System;
class GFG
{
// function to check if all the
// bits are set or not in the
// binary representation of 'n'
static String areAllBitsSet(int n)
{
// all bits are not set
if (n == 0)
return "No";
// if true, then all
// bits are set
if (((n + 1) & n) == 0)
return "Yes";
// else all bits are not set
return "No";
}
// Driver Code
static public void Main ()
{
int n = 7;
Console.WriteLine(areAllBitsSet(n));
}
}
// This code is contributed by m_kit
PHP
<?php
// PHP implementation to check
// whether every digit in the
// binary representation of the
// given number is set or not
// function to check if all
// the bits are set or not in
// the binary representation of 'n'
function areAllBitsSet($n)
{
// all bits are not set
if ($n == 0)
return "No";
// if true, then all
// bits are set
if ((($n + 1) & $n) == 0)
return "Yes";
// else all bits
// are not set
return "No";
}
// Driver Code
$n = 7;
echo areAllBitsSet($n);
// This code is contributed by ajit
?>
Javascript
<script>
// javascript implementation to check whether
// every digit in the binary representation
// of the given number is set or not
// function to check if all the
// bits are set or not in the
// binary representation of 'n'
function areAllBitsSet(n)
{
// all bits are not set
if (n == 0)
return "No";
// if true, then all bits are set
if (((n + 1) & n) == 0)
return "Yes";
// else all bits are not set
return "No";
}
// Driver program to test above
var n = 7;
document.write(areAllBitsSet(n));
// This code contributed by Princi Singh
</script>
Producción
Yes
Tiempo Complejidad: O(1)
Espacio Auxiliar: O(1)
Método 3 : simplemente podemos contar los bits establecidos totales presentes en la representación binaria del número y, en función de esto, podemos verificar si el número es igual a pow (2, __builtin_popcount (n)). Si resulta ser igual, devolvemos 1, de lo contrario devolvemos 0;
C++
#include <bits/stdc++.h>
using namespace std;
void isBitSet(int N)
{
if (N == pow(2, __builtin_popcount(N)) - 1)
cout << "Yes\n";
else cout << "No\n";
}
int main()
{
int N = 7;
isBitSet(N);
return 0;
}
Java
import java.util.*;
class GFG{
static void isBitSet(int N)
{
if (N == Math.pow(2, Integer.bitCount(N)) - 1)
System.out.print("Yes\n");
else System.out.print("No\n");
}
public static void main(String[] args)
{
int N = 7;
isBitSet(N);
}
}
// This code is contributed by umadevi9616
Python3
def bitCount(n):
n = n - ((n >> 1) & 0x55555555);
n = (n & 0x33333333) + ((n >> 2) & 0x33333333);
return ((n + (n >> 4) & 0xF0F0F0F) * 0x1010101) >> 24;
def isBitSet(N):
if (N == pow(2, bitCount(N)) - 1):
print("Yes");
else:
print("No");
if __name__ == '__main__':
N = 7;
isBitSet(N);
# This code is contributed by gauravrajput1
C#
using System;
public class GFG{
static int bitCount (int n) {
n = n - ((n >> 1) & 0x55555555);
n = (n & 0x33333333) + ((n >> 2) & 0x33333333);
return ((n + (n >> 4) & 0xF0F0F0F) * 0x1010101) >> 24;
}
static void isBitSet(int N)
{
if (N == Math.Pow(2, bitCount(N)) - 1)
Console.Write("Yes\n");
else Console.Write("No\n");
}
public static void Main(String[] args)
{
int N = 7;
isBitSet(N);
}
}
// This code is contributed by umadevi9616
Javascript
<script>
function bitCount (n) {
n = n - ((n >> 1) & 0x55555555);
n = (n & 0x33333333) + ((n >> 2) & 0x33333333);
return ((n + (n >> 4) & 0xF0F0F0F) * 0x1010101) >> 24;
}
function isBitSet(N) {
if (N == Math.pow(2, bitCount(N)) - 1)
document.write("Yes\n");
else
document.write("No\n");
}
var N = 7;
isBitSet(N);
// This code is contributed by umadevi9616
</script>
Producción:
Yes
Tiempo Complejidad: O(1)
Espacio Auxiliar: O(1)
Referencias:
https://www.careercup.com/question?id=9503107
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Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA