Highest power of 2 less than or equal to given number

Given a number n, find the highest power of 2 that is smaller than or equal to n.

Examples :

Input : n = 10
Output : 8

Input : n = 19
Output : 16

Input : n = 32
Output : 32

A simple solution is to start checking from n and keep decrementing until we find a power of 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>

Output

8

Time complexity : O(n). In worst case, the loop runs floor(n/2) times. The worst case happens when n is of the form 2^x – 1.

Auxiliary Space : O(1) since only constant space is used for variables

An efficient solution is to use bitwise left shift operator to find all powers of 2 starting from 1. For every power check if it is smaller than or equal to n or not. Below is the implementation of the 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>

Output

8

Time Complexity: O(32)

Auxiliary Space: O(1)

A Solution using Log(n)
Thanks to Anshuman Jha for suggesting this solution. 

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>

Output

8

Time Complexity: O(logn)

Auxiliary Space: O(1)

Solution using bitmasks :

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>

Output

8

Time Complexity: O(1)

Auxiliary Space: O(1) since only constant space is used for variables

A solution using MSB

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

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

Java

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

class GFG
{
  static int highestPowerof2(int N)
  {

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

    // else set only the most significant bit
    return (1 << (Integer.toBinaryString(N).length() - 1));
  }

  // Driver Code
  public static void main(String[] args)
  {
    int n = 5;
    System.out.println(highestPowerof2(n));
  }
}

// This code is contributed by phasing17

Python3

# Python3 program to find
# smallest power of 2
# smaller than or equal to n

def highestPowerof2(N):

    # if N is a power of two simply return it
    if (not (N & (N - 1))):
        return N;

    # else set only the most significant bit
    return 0x8000000000000000 >>  (64 - N.bit_length())


# Driver Code
n = 5;
print(highestPowerof2(n))

# This code is contributed by phasing17

C#

// C# program to find
// smallest power of 2
// smaller than or equal to n
using System;
using System.Collections.Generic;

class GFG
{
  static int highestPowerof2(int N)
  {
    // if N is a power of two simply return it
    if ((N & (N - 1)) == 0)
      return N;

    // else set only the most significant bit
    return (1 << ((Convert.ToString(N, 2).Length) - 1));
  }

  // Driver Code
  public static void Main(string[] args)
  {
    int n = 5;
    Console.WriteLine(highestPowerof2(n));
  }
}

// This code is contributed by phasing17

Javascript

// JS program to find
// smallest power of 2
// smaller than or equal to n

function highestPowerof2(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 1 << ((N.toString(2)).length) - 1;
}

// Driver Code
let n = 5;
console.log(highestPowerof2(n));

// This code is contributed by phasing17

Output

4

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

Application Problem:
Some people are standing in a queue. A selection process follows a rule where people standing on even positions are selected. Of the selected people a queue is formed and again out of these only people on even position are selected. This continues until we are left with one person. Find out the position of that person in the original queue. 
Print the position(original queue) of that person who is left. 

Examples :

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

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