Smallest power of 2 greater than or equal to n

Write a function that, for a given no n, finds a number p which is greater than or equal to n and is the smallest power of 2. 

Examples :

Input: n = 5
Output: 8     

Input: n = 17
Output: 32     

Input  : n = 32
Output: 32     

Method 1: Using log₂(number) 

• Calculate the log₂(N) and store it into a variable say ‘a’
• Check if pow(2, a) is equals to N
  • Return, N
• Otherwise, return pow(2, a + 1)

Below is the implementation of the above approach:

C++

// C++ program to find
// smallest power of 2
// greater than or equal to n
#include <bits/stdc++.h>
using namespace std;

long long nearestPowerOf2(long long N)
{
    long long a = log2(N);

    if (pow(2, a) == N)
        return N;

    return pow(2, a + 1);
}

// Driver Code
int main()
{
    unsigned int n = 5;
    cout << nearestPowerOf2(n);
    return 0;
}

// This code is contributed by hkdass001

Java

/*package whatever //do not write package name here */

import java.io.*;

public class GFG {

    public static long nearestPowerOf2(long N)
    {
        long a = (int)(Math.log(N) / Math.log(2));

        if (Math.pow(2, a) == N)
            return N;

        return (long) Math.pow(2, a + 1);
    }

    // Driver Code
    public static void main (String[] args)  {
        long n = 5;
        System.out.println(nearestPowerOf2(n));
    }
}
// This code is contributed by Ajax

Python3

#Python program to find
#smallest power of 2
#greater than or equal to n
import math

# Function to find the smallest power of 2
# greater than or equal to n
def nearestPowerOf2(N):
    # Calculate log2 of N
    a = int(math.log2(N))

    # If 2^a is equal to N, return N
    if 2**a == N:
        return N

    # Return 2^(a + 1)
    return 2**(a + 1)

# Main function
if __name__ == "__main__":
    # Input number
    n = 5
    # Call the nearestPowerOf2 function
    print(nearestPowerOf2(n))

C#

using System;

public class GFG {

  public static long nearestPowerOf2(long N)
  {
    long a = (int)(Math.Log(N) / Math.Log(2));

    if (Math.Pow(2, a) == N)
      return N;

    return (long) Math.Pow(2, a + 1);
  }

  // Driver Code
  public static void Main (String[] args)  {
    long n = 5;
    Console.WriteLine(nearestPowerOf2(n));
  }
}

// This code contributed by SRJ

Javascript

// Function to find the smallest power of 2
// greater than or equal to N
function nearestPowerOf2(N)
{

  // Calculate log2 of N
  var a = Math.floor(Math.log2(N));

  // If 2^a is equal to N, return N
  if (Math.pow(2, a) === N) {
    return N;
  }

  // Return 2^(a + 1)
  return Math.pow(2, a + 1);
}

// Main function

  // Input number
  var n = 5;

  // Call the nearestPowerOf2 function
  console.log(nearestPowerOf2(n));

Output

8

Time Complexity: O(1)
Auxiliary Space: O(1)

Example :

    Let us try for 17
            pos = 5
            p   = 32    

Method 2 (By getting the position of only set bit in result )

    /* If n is a power of 2 then return n */
    1  If (n & !(n&(n-1))) then return n 
    2  Else keep right shifting n until it becomes zero 
        and count no of shifts
        a. Initialize: count = 0
        b. While n ! = 0
                n = n>>1
                count = count + 1

     /* Now count has the position of set bit in result */
    3  Return (1 << count)  

Example :

    Let us try for 17
                 count = 5
                 p     = 32   

C++

// C++ program to find
// smallest power of 2
// greater than or equal to n
#include <bits/stdc++.h>
using namespace std;

unsigned int nextPowerOf2(unsigned int n)
{
    unsigned count = 0;

    // First n in the below condition
    // is for the case where n is 0
    if (n && !(n & (n - 1)))
        return n;

    while( n != 0)
    {
        n >>= 1;
        count += 1;
    }

    return 1 << count;
}

// Driver Code
int main()
{
    unsigned int n = 0;
    cout << nextPowerOf2(n);
    return 0;
}

// This code is contributed by rathbhupendra

C

#include<stdio.h>

unsigned int nextPowerOf2(unsigned int n)
{
unsigned count = 0;

// First n in the below condition
// is for the case where n is 0
if (n && !(n & (n - 1)))
    return n;

while( n != 0)
{
    n >>= 1;
    count += 1;
}

return 1 << count;
}

// Driver Code
int main()
{
unsigned int n = 0;
printf("%d", nextPowerOf2(n));
return 0;
}

Java

// Java program to find
// smallest power of 2
// greater than or equal to n
import java.io.*;

class GFG
{
    static int nextPowerOf2(int n)
    {
        int count = 0;

        // First n in the below
        // condition is for the
        // case where n is 0
        if (n > 0 && (n & (n - 1)) == 0)
            return n;

        while(n != 0)
        {
            n >>= 1;
            count += 1;
        }

        return 1 << count;
    }

    // Driver Code
    public static void main(String args[])
    {
        int n = 0;
        System.out.println(nextPowerOf2(n));
    }
}

// This article is contributed
// by Anshika Goyal.

Python3

def nextPowerOf2(n):
    count = 0

    # First n in the below
    # condition is for the
    # case where n is 0
    if (n and not(n & (n - 1))):
        return n

    while( n != 0):
        n >>= 1
        count += 1

    return 1 << count


# Driver Code
n = 0
print(nextPowerOf2(n))
# This code is contributed
# by Smitha Dinesh Semwal

C#

// C# program to find smallest
// power of 2 greater than
// or equal to n
using System;

class GFG
{
    static int nextPowerOf2(int n)
    {
        int count = 0;

        // First n in the below
        // condition is for the
        // case where n is 0
        if (n > 0 && (n & (n - 1)) == 0)
            return n;

        while(n != 0)
        {
            n >>= 1;
            count += 1;
        }

        return 1 << count;
    }

    // Driver Code
    public static void Main()
    {
        int n = 0;
        Console.WriteLine(nextPowerOf2(n));
    }
}

// This code is contributed by anuj_67.

PHP

<?php
// PHP program to find smallest
// power of 2 greater than or
// equal to n

function nextPowerOf2($n)
{
$count = 0;

// First n in the below condition
// is for the case where n is 0
if ($n && !($n&($n - 1)))
    return $n;

while($n != 0)
{
    $n >>= 1;
    $count += 1;
}

return 1 << $count;
}

// Driver Code
$n = 0;
echo (nextPowerOf2($n));

// This code is contributed by vt_m
?>

Javascript

<script>

// JavaScript program to find
// smallest power of 2
// greater than or equal to n

function nextPowerOf2(n)
{
    var count = 0;

    // First n in the below condition
    // is for the case where n is 0
    if (n && !(n & (n - 1)))
        return n;

    while( n != 0)
    {
        n >>= 1;
        count += 1;
    }

    return 1 << count;
}

// Driver Code
var n = 0;
document.write(nextPowerOf2(n));

</script>

Output

1

Time Complexity: O(log n)
Auxiliary Space: O(1)

Method 3(Shift result one by one)
Thanks to coderyogi for suggesting this method . This method is a variation of method 2 where instead of getting count, we shift the result one by one in a loop. 

C++

// C++ program to find smallest
// power of 2 greater than or
// equal to n
#include<bits/stdc++.h>
using namespace std;
unsigned int nextPowerOf2(unsigned int n)
{
    unsigned int p = 1;
    if (n && !(n & (n - 1)))
        return n;

    while (p < n)
        p <<= 1;

    return p;
}

// Driver Code
int main()
{
    unsigned int n = 5;
    cout << nextPowerOf2(n);
    return 0;
}

// This code is contributed by rathbhupendra

C

#include<stdio.h>
unsigned int nextPowerOf2(unsigned int n)
{
    unsigned int p = 1;
    if (n && !(n & (n - 1)))
        return n;

    while (p < n)
        p <<= 1;

    return p;
}

// Driver Code
int main()
{
unsigned int n = 5;
printf("%d", nextPowerOf2(n));
return 0;
}

Java

// Java program to find smallest
// power of 2 greater than or
// equal to n
import java.io.*;

class GFG
{
    static int nextPowerOf2(int n)
    {
        int p = 1;
        if (n > 0 && (n & (n - 1)) == 0)
            return n;

        while (p < n)
            p <<= 1;

        return p;
    }

    // Driver Code
    public static void main(String args[])
    {
        int n = 5;
        System.out.println(nextPowerOf2(n));
    }
}

// This article is contributed
// by Anshika Goyal.

Python3

def nextPowerOf2(n):

    p = 1
    if (n and not(n & (n - 1))):
        return n

    while (p < n) :
        p <<= 1

    return p;


# Driver Code
n = 5
print(nextPowerOf2(n));

# This code is contributed by
# Smitha Dinesh Semwal

C#

// C# program to find smallest
// power of 2 greater than or
// equal to n
using System;

class GFG
{

    static int nextPowerOf2(int n)
    {
        int p = 1;
        if (n > 0 && (n & (n - 1)) == 0)
            return n;

        while (p < n)
            p <<= 1;

        return p;
    }

    // Driver Code
    public static void Main()
    {
        int n = 5;
        Console.Write(nextPowerOf2(n));
    }
}

// This code is contributed by Smitha.

PHP

<?php

function nextPowerOf2($n)
{
    $p = 1;
    if ($n && !($n & ($n - 1)))
        return $n;

    while ($p < $n)
        $p <<= 1;

    return $p;
}

// Driver Code
$n = 5;
echo ( nextPowerOf2($n));

// This code is contributed by vt_m.
?>

Javascript

<script>
// Program to find smallest
// power of 2 greater than or
// equal to n

function  nextPowerOf2( n)
{
    p = 1;
    if (n && !(n & (n - 1)))
        return n;

    while (p < n)
        p <<= 1;

    return p;
}

// Driver Code

     n = 5;
    document.write (nextPowerOf2(n));

    //This code is contributed by simranarora5sos
</script>

Output

8

Time Complexity: O(log(n)) 
Auxiliary Space: O(1)

Method 4(Customized and Fast) 

    1. Subtract n by 1
       n = n -1

    2. Set all bits after the leftmost set bit.

    /* Below solution works only if integer is 32 bits */
                n = n | (n >> 1);
                n = n | (n >> 2);
                n = n | (n >> 4);
                n = n | (n >> 8);
                n = n | (n >> 16);
    3. Return n + 1

Example :

Steps 1 & 3 of above algorithm are to handle cases 
of power of 2 numbers e.g., 1, 2, 4, 8, 16,

    Let us try for 17(10001)
    step 1
       n = n - 1 = 16 (10000)  
    step 2
       n = n | n >> 1
       n = 10000 | 01000
       n = 11000
       n = n | n >> 2
       n = 11000 | 00110
       n = 11110
       n = n | n >> 4
       n = 11110 | 00001
       n = 11111
       n = n | n >> 8
       n = 11111 | 00000
       n = 11111
       n = n | n >> 16
       n = 11110 | 00000
       n = 11111    

    step 3: Return n+1
     We get n + 1 as 100000 (32)

Program: 

C++

// C++ program to
// Finds next power of two
// for n. If n itself is a
// power of two then returns n
#include <bits/stdc++.h>
using namespace std;

unsigned int nextPowerOf2(unsigned int n)
{
    n--;
    n |= n >> 1;
    n |= n >> 2;
    n |= n >> 4;
    n |= n >> 8;
    n |= n >> 16;
    n++;
    return n;
}

// Driver Code
int main()
{
    unsigned int n = 5;
    cout << nextPowerOf2(n);
    return 0;
}

// This code is contributed by SHUBHAMSINGH10

C

#include <stdio.h>
// Finds next power of two
// for n. If n itself is a
// power of two then returns n
unsigned int nextPowerOf2(unsigned int n)
{
    n--;
    n |= n >> 1;
    n |= n >> 2;
    n |= n >> 4;
    n |= n >> 8;
    n |= n >> 16;
    n++;
    return n;
}

// Driver Code
int main()
{
    unsigned int n = 5;
    printf("%d", nextPowerOf2(n));
    return 0;
}

Java

// Java program to find smallest
// power of 2 greater than or
// equal to n
import java.io.*;

class GFG
{
    // Finds next power of two
    // for n. If n itself is a
    // power of two then returns n
    static int nextPowerOf2(int n)
    {
        n--;
        n |= n >> 1;
        n |= n >> 2;
        n |= n >> 4;
        n |= n >> 8;
        n |= n >> 16;
        n++;

        return n;
    }

    // Driver Code
    public static void main(String args[])
    {
        int n = 5;
        System.out.println(nextPowerOf2(n));
    }
}

// This article is contributed
// by Anshika Goyal.

Python3

# Finds next power of two
# for n. If n itself is a
# power of two then returns n
def nextPowerOf2(n):

    n -= 1
    n |= n >> 1
    n |= n >> 2
    n |= n >> 4
    n |= n >> 8
    n |= n >> 16
    n += 1
    return n

# Driver program to test
# above function
n = 5
print(nextPowerOf2(n))

# This code is contributed
# by Smitha

C#

// C# program to find smallest
// power of 2 greater than or
// equal to n
using System;

class GFG
{

    // Finds next power of two
    // for n. If n itself is a
    // power of two then returns n
    static int nextPowerOf2(int n)
    {
        n--;
        n |= n >> 1;
        n |= n >> 2;
        n |= n >> 4;
        n |= n >> 8;
        n |= n >> 16;
        n++;

        return n;
    }

    // Driver Code
    public static void Main()
    {
        int n = 5;
        Console.WriteLine(nextPowerOf2(n));
    }
}

// This code is contributed by anuj_67.

PHP

<?php
// PHP program to find smallest
// power of 2 greater than or
// equal to n

// Finds next power of
// two for n. If n itself
// is a power of two then
// returns n
function nextPowerOf2($n)
{
    $n--;
    $n |= $n >> 1;
    $n |= $n >> 2;
    $n |= $n >> 4;
    $n |= $n >> 8;
    $n |= $n >> 16;
    $n++;
    return $n;
}

    // Driver Code
    $n = 5;
    echo nextPowerOf2($n);

// This code contributed by Ajit
?>

Javascript

<script>

// Javascript program to find smallest
// power of 2 greater than or
// equal to n

// Finds next power of
// two for n. If n itself
// is a power of two then
// returns n
function nextPowerOf2(n)
{
    n -= 1
    n |= n >> 1
    n |= n >> 2
    n |= n >> 4
    n |= n >> 8
    n |= n >> 16
    n += 1
    return n
}

// Driver Code
n = 5;
document.write (nextPowerOf2(n));

// This code is contributed by bunnyram19

</script>

Output

8

Time Complexity : O(log(n)) 
Auxiliary Space: O(1)

Efficient approach:
If the given number is the power of two then it is the required number otherwise set only the left bit of most significant bit which gives us the required number.

C++

// C++ program to find
// smallest power of 2
// greater than or equal to n
#include <iostream>
using namespace std;

long long nextPowerOf2(long long N)
{
    // if N is a power of two simply return it
    if (!(N & (N - 1)))
        return N;
    // else set only the left bit of most significant bit
    return 0x8000000000000000 >> (__builtin_clzll(N) - 1);
}

// Driver Code
int main()
{
    long long n = 5;
    cout << nextPowerOf2(n);
    return 0;
}

// This code is contributed by Kasina Dheeraj.

C

// C program to find
// smallest power of 2
// greater than or equal to n
#include <stdio.h>

long long nextPowerOf2(long long N)
{
    // if N is a power of two simply return it
    if (!(N & (N - 1)))
        return N;
    // else set only the left bit of most significant bit
    return 0x8000000000000000 >> (__builtin_clzll(N) - 1);
}

// Driver Code
int main()
{
    long long n = 5;
    printf("%lld", nextPowerOf2(n));
    return 0;
}

// This code is contributed by Kasina Dheeraj.

Java

// Java program to find
// smallest power of 2
// greater than or equal to n
import java.io.*;

class GFG {
    static long nextPowerOf2(long N)
    {
        // if N is a power of two simply return it
        if ((N & (N - 1)) == 0)
            return N;
        // else set only the left bit of most significant
        // bit as in Java right shift is filled with most
        // significant bit we consider
        return 0x4000000000000000L
            >> (Long.numberOfLeadingZeros(N) - 2);
    }

    // Driver Code
    public static void main(String args[])
    {
        long n = 5;
        System.out.println(nextPowerOf2(n));
    }
}

// This code is contributed
// by Kasina Dheeraj.

Python3

# Python program to find
# smallest power of 2
# greater than or equal to n
#include <iostream>
def nextPowerOf2(N):
    # if N is a power of two simply return it
    if not (N & (N - 1)):
        return N
    # else set only the left bit of most significant bit
    return  int("1" + (len(bin(N)) - 2) * "0", 2)

# Driver Code
n = 5
print(nextPowerOf2(n))

# this code is contributed by phasing17

C#

// C# program to find
// smallest power of 2
// greater than or equal to n
using System;
using System.Linq;

class GFG {

  static int nextPowerOf2(int N)
  {

    // if N is a power of two simply return it
    if ((N & (N - 1)) == 0)
      return N;

    // else set only the left bit of most significant
    // bit
    return Convert.ToInt32(
      "1"
      + new string('0',
                   Convert.ToString(N, 2).Length),
      2);
  }

  // Driver Code
  public static void Main(string[] args)
  {
    int n = 5;

    // Function call
    Console.WriteLine(nextPowerOf2(n));
  }
}

// This code is contributed
// by phasing17

Javascript

// JavaScript program to find
// smallest power of 2
// greater than or equal to n
function nextPowerOf2(N)
{
    // if N is a power of two simply return it
    if (!(N & (N - 1)))
        return N;

    // else set only the left bit of most significant bit
    return  parseInt("1" + "0".repeat(N.toString(2).length), 2);
}

// Driver Code
let n = 5;
console.log(nextPowerOf2(n));

// this code is contributed by phasing17

Output

8

Time Complexity : O(1) as counting leading zeroes can cause at most O(64) time complexity.
Auxiliary Space: O(1)

References :
http://en.wikipedia.org/wiki/Power_of_2