Dado un número n, encuentre la mayor potencia de 2 que sea menor o igual que n.
Ejemplos:
Input : n = 10 Output : 8 Input : n = 19 Output : 16 Input : n = 32 Output : 32
Una solución simple es comenzar a verificar desde n y seguir disminuyendo hasta encontrar una potencia de 2.
C++
// C++ program to find highest power of 2 smaller
// than or equal to n.
#include <bits/stdc++.h>
using namespace std;
int highestPowerof2(int n)
{
int res = 0;
for (int i = n; i >= 1; i--) {
// If i is a power of 2
if ((i & (i - 1)) == 0) {
res = i;
break;
}
}
return res;
}
// Driver code
int main()
{
int n = 10;
cout << highestPowerof2(n);
return 0;
}
// This code is contributed by Sania Kumari Gupta
// (kriSania804)
C
// C program to find highest power of 2 smaller
// than or equal to n.
#include <stdio.h>
int highestPowerof2(int n)
{
int res = 0;
for (int i = n; i >= 1; i--) {
// If i is a power of 2
if ((i & (i - 1)) == 0) {
res = i;
break;
}
}
return res;
}
// Driver code
int main()
{
int n = 10;
printf("%d", highestPowerof2(n));
return 0;
}
// This code is contributed by Sania Kumari Gupta
// (kriSania804)
Java
// Java program to find highest power of
// 2 smaller than or equal to n.
class GFG{
static int highestPowerof2(int n)
{
int res = 0;
for(int i = n; i >= 1; i--)
{
// If i is a power of 2
if ((i & (i-1)) == 0)
{
res = i;
break;
}
}
return res;
}
// Driver code
public static void main(String[] args)
{
int n = 10;
System.out.print(highestPowerof2(n));
}
}
// This code is contributed by 29AjayKumar
Python3
# Python3 program to find highest # power of 2 smaller than or # equal to n. def highestPowerof2(n): res = 0; for i in range(n, 0, -1): # If i is a power of 2 if ((i & (i - 1)) == 0): res = i; break; return res; # Driver code n = 10; print(highestPowerof2(n)); # This code is contributed by mits
C#
// C# code to find highest power
// of 2 smaller than or equal to n.
using System;
class GFG
{
public static int highestPowerof2(int n)
{
int res = 0;
for (int i = n; i >= 1; i--)
{
// If i is a power of 2
if ((i & (i - 1)) == 0)
{
res = i;
break;
}
}
return res;
}
// Driver Code
static public void Main ()
{
int n = 10;
Console.WriteLine(highestPowerof2(n));
}
}
// This code is contributed by ajit
PHP
<?php
// PHP program to find highest
// power of 2 smaller than or
// equal to n.
function highestPowerof2($n)
{
$res = 0;
for ($i = $n; $i >= 1; $i--)
{
// If i is a power of 2
if (($i & ($i - 1)) == 0)
{
$res = $i;
break;
}
}
return $res;
}
// Driver code
$n = 10;
echo highestPowerof2($n);
// This code is contributed by m_kit
?>
Javascript
<script>
// JavaScript program to find highest power
// of 2 smaller than or equal to n.
function highestPowerof2(n)
{
let res = 0;
for (let i = n; i >= 1; i--)
{
// If i is a power of 2
if ((i & (i - 1)) == 0)
{
res = i;
break;
}
}
return res;
}
// Driver code
let n = 10;
document.write(highestPowerof2(n));
</script>
Producción
8
Complejidad temporal : O(n). En el peor de los casos, el ciclo se ejecuta piso(n/2) veces. El peor caso ocurre cuando n es de la forma 2 x – 1.
Espacio auxiliar : O (1) ya que solo se usa espacio constante para variables
Una solución eficiente es utilizar el operador de desplazamiento a la izquierda bit a bit para encontrar todas las potencias de 2 a partir de 1. Para cada potencia, compruebe si es menor o igual que n o no. A continuación se muestra la implementación de la idea.
C++
// C++ program to find highest power of 2 smaller
// than or equal to n.
#include <bits/stdc++.h>
using namespace std;
int highestPowerof2(unsigned int n)
{
// Invalid input
if (n < 1)
return 0;
int res = 1;
// Try all powers starting from 2^1
for (int i = 0; i < 8 * sizeof(unsigned int); i++) {
int curr = 1 << i;
// If current power is more than n, break
if (curr > n)
break;
res = curr;
}
return res;
}
// Driver code
int main()
{
int n = 10;
cout << highestPowerof2(n);
return 0;
}
// This code is contributed by Sania Kumari Gupta
C
// C program to find highest power of 2 smaller
// than or equal to n.
#include <stdio.h>
int highestPowerof2(unsigned int n)
{
// Invalid input
if (n < 1)
return 0;
int res = 1;
// Try all powers starting from 2^1
for (int i = 0; i < 8 * sizeof(unsigned int); i++) {
int curr = 1 << i;
// If current power is more than n, break
if (curr > n)
break;
res = curr;
}
return res;
}
// Driver code
int main()
{
int n = 10;
printf("%d", highestPowerof2(n));
return 0;
}
// This code is contributed by Sania Kumari Gupta
Java
// Java program to find
// highest power of 2 smaller
// than or equal to n.
import java.io.*;
class GFG
{
static int highestPowerof2(int n)
{
// Invalid input
if (n < 1)
return 0;
int res = 1;
// Try all powers
// starting from 2^1
for (int i = 0; i < 8 * Integer.BYTES; i++)
{
int curr = 1 << i;
// If current power is
// more than n, break
if (curr > n)
break;
res = curr;
}
return res;
}
// Driver code
public static void main(String[] args)
{
int n = 10;
System.out.println(highestPowerof2(n));
}
}
// This code is contributed aj_36
python3
# Python3 program to find highest power of 2 smaller # than or equal to n. import sys def highestPowerof2( n): # Invalid input if (n < 1): return 0 res = 1 #Try all powers starting from 2^1 for i in range(8*sys.getsizeof(n)): curr = 1 << i # If current power is more than n, break if (curr > n): break res = curr return res # Driver code if __name__ == "__main__": n = 10 print(highestPowerof2(n))
C#
// C# program to find
// highest power of 2 smaller
// than or equal to n.
using System;
class GFG
{
static int highestPowerof2(int n)
{
// Invalid input
if (n < 1)
return 0;
int res = 1;
// Try all powers
// starting from 2^1
for (int i = 0; i < 8 * sizeof(uint); i++)
{
int curr = 1 << i;
// If current power is
// more than n, break
if (curr > n)
break;
res = curr;
}
return res;
}
// Driver code
static public void Main ()
{
int n = 10;
Console.WriteLine(highestPowerof2(n));
}
}
// This code is contributed ajit
PHP
<?php
// PHP program to find highest
// power of 2 smaller
// than or equal to n.
function highestPowerof2($n)
{
// Invalid input
if ($n < 1)
return 0;
$res = 1;
// Try all powers starting
// from 2^1
for ($i = 0; $i < 8 * PHP_INT_SIZE; $i++)
{
$curr = 1 << $i;
// If current power is
// more than n, break
if ($curr > $n)
break;
$res = $curr;
}
return $res;
}
// Driver code
$n = 10;
echo highestPowerof2($n);
// This code is contributed
// by m_kit
?>
Javascript
<script>
function highestPowerof2(n)
{
// Invalid input
if (n < 1)
return 0;
let res = 1;
// Try all powers starting from 2^1
for (let i=0; i<8; i++)
{
let curr = 1 << i;
// If current power is more than n, break
if (curr > n)
break;
res = curr;
}
return res;
}
// Driver code
let n = 10;
document.write(highestPowerof2(n));
</script>
Producción
8
Complejidad del tiempo: O(32)
Espacio Auxiliar: O(1)
Una solución usando Log(n)
Gracias a Anshuman Jha por sugerir esta solución.
C++
// C++ program to find highest power of 2 smaller
// than or equal to n.
#include<bits/stdc++.h>
using namespace std;
int highestPowerof2(int n)
{
int p = (int)log2(n);
return (int)pow(2, p);
}
// Driver code
int main()
{
int n = 10;
cout << highestPowerof2(n);
return 0;
}
Java
// Java program to find
// highest power of 2
// smaller than or equal to n.
import java.io.*;
class GFG
{
static int highestPowerof2(int n)
{
int p = (int)(Math.log(n) /
Math.log(2));
return (int)Math.pow(2, p);
}
// Driver code
public static void main (String[] args)
{
int n = 10;
System.out.println(highestPowerof2(n));
}
}
// This code is contributed
// by m_kit
Python3
# Python3 program to find highest # power of 2 smaller than or # equal to n. import math def highestPowerof2(n): p = int(math.log(n, 2)); return int(pow(2, p)); # Driver code n = 10; print(highestPowerof2(n)); # This code is contributed by mits
C#
// C# program to find
// highest power of 2
// smaller than or equal to n.
using System;
class GFG
{
static int highestPowerof2(int n)
{
int p = (int)(Math.Log(n) /
Math.Log(2));
return (int)Math.Pow(2, p);
}
// Driver code
static public void Main ()
{
int n = 10;
Console.WriteLine(highestPowerof2(n));
}
}
// This code is contributed
// by ajit
PHP
<?php
// PHP program to find highest
// power of 2 smaller than or
// equal to n.
function highestPowerof2($n)
{
$p = (int)log($n, 2);
return (int)pow(2, $p);
}
// Driver code
$n = 10;
echo highestPowerof2($n);
// This code is contributed by ajit
?>
Javascript
<script>
// Javascript program to find
// highest power of 2
// smaller than or equal to n.
function highestPowerof2(n)
{
let p = parseInt(Math.log(n) / Math.log(2), 10);
return Math.pow(2, p);
}
let n = 10;
document.write(highestPowerof2(n));
// This code is contributed by divyeshrabadiya07.
</script>
Producción
8
Complejidad de tiempo: O (logn)
Espacio Auxiliar: O(1)
Solución usando máscaras de bits:
C++
// C++ program to find highest power of 2 smaller
// than or equal to n.
#include <iostream>
using namespace std;
unsigned highestPowerof2(unsigned x)
{
// check for the set bits
x |= x >> 1;
x |= x >> 2;
x |= x >> 4;
x |= x >> 8;
x |= x >> 16;
// Then we remove all but the top bit by xor'ing the
// string of 1's with that string of 1's shifted one to
// the left, and we end up with just the one top bit
// followed by 0's.
return x ^ (x >> 1);
}
int main()
{
int n = 10;
cout << highestPowerof2(n) << "\n";
return 0;
}
// This code is contributed by Rudrakshi.
Java
// Java program to find highest power of 2 smaller
// than or equal to n.
import java.io.*;
class GFG
{
static int highestPowerof2(int x)
{
// check for the set bits
x |= x >> 1;
x |= x >> 2;
x |= x >> 4;
x |= x >> 8;
x |= x >> 16;
// Then we remove all but the top bit by xor'ing the
// string of 1's with that string of 1's shifted one to
// the left, and we end up with just the one top bit
// followed by 0's.
return x ^ (x >> 1);
}
// Driver code
public static void main (String[] args)
{
int n = 10;
System.out.println(highestPowerof2(n));
}
}
// This code is contributed by avanitrachhadiya2155
Python3
# Python3 program to find highest power of 2 smaller than or equal to n. def highestPowerof2(x): # check for the set bits x |= x >> 1 x |= x >> 2 x |= x >> 4 x |= x >> 8 x |= x >> 16 # Then we remove all but the top bit by xor'ing the # string of 1's with that string of 1's shifted one to # the left, and we end up with just the one top bit # followed by 0's. return x ^ (x >> 1) n = 10 print(highestPowerof2(n)) # This code is contributed by divyesh072019.
C#
// C# program to find highest power of 2 smaller
// than or equal to n.
using System;
public class GFG
{
static int highestPowerof2(int x)
{
// check for the set bits
x |= x >> 1;
x |= x >> 2;
x |= x >> 4;
x |= x >> 8;
x |= x >> 16;
// Then we remove all but the top bit by xor'ing the
// string of 1's with that string of 1's shifted one to
// the left, and we end up with just the one top bit
// followed by 0's.
return x ^ (x >> 1);
}
// Driver code
public static void Main(String[] args)
{
int n = 10;
Console.WriteLine(highestPowerof2(n));
}
}
// This code is contributed by umadevi9616
Javascript
<script>
// Javascript program to find highest power of 2 smaller
// than or equal to n.
function highestPowerof2(x)
{
// check for the set bits
x |= x >> 1;
x |= x >> 2;
x |= x >> 4;
x |= x >> 8;
x |= x >> 16;
// Then we remove all but the top bit by xor'ing the
// string of 1's with that string of 1's shifted one to
// the left, and we end up with just the one top bit
// followed by 0's.
return x ^ (x >> 1);
}
let n = 10;
document.write(highestPowerof2(n))
// This code is contributed by rag2127
</script>
Producción
8
Complejidad de tiempo: O(1)
Espacio auxiliar : O (1) ya que solo se usa espacio constante para variables
Una solución usando MSB
Si el número dado es la potencia de dos, entonces es el número requerido; de lo contrario, establezca solo el bit más significativo que nos da el número requerido.
C++
// C++ program to find
// smallest power of 2
// smaller than or equal to n
#include <iostream>
using namespace std;
long long highestPowerof2(long long N)
{
// if N is a power of two simply return it
if (!(N & (N - 1)))
return N;
// else set only the most significant bit
return 0x8000000000000000 >> (__builtin_clzll(N));
}
// Driver Code
int main()
{
long long n = 5;
cout << highestPowerof2(n);
return 0;
}
// This code is contributed by phasing17
Producción
4
Complejidad de tiempo : O (1) ya que contar ceros a la izquierda puede causar una complejidad de tiempo de O (64) como máximo.
Espacio Auxiliar: O(1)
Problema de aplicación:
Algunas personas están haciendo cola. Un proceso de selección sigue una regla en la que se seleccionan personas que se encuentran en posiciones pares. De las personas seleccionadas se forma una cola y nuevamente de estas solo se seleccionan las personas en posiciones pares. Esto continúa hasta que nos quedamos con una persona. Averigüe la posición de esa persona en la cola original.
Imprime la posición (cola original) de esa persona que queda.
Ejemplos:
Input: n = 10
Output:8
Explanation :
1 2 3 4 5 6 7 8 9 10 ===>Given queue
2 4 6 8 10
4 8
8
Input: n = 17
Input: 16
Explanation :
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 ===>Given queue
2 4 6 8 10 12 14 16
4 8 12 16
8 16
16
Artículo relacionado:
Potencia de 2 mayor o igual a un número dado.
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Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA