Skip to content
Related Articles
Open in App
Not now

Related Articles

Find all powers of 2 less than or equal to a given number

Improve Article
Save Article
  • Difficulty Level : Easy
  • Last Updated : 29 Sep, 2022
Improve Article
Save Article

Given a positive number N, the task is to find out all the perfect powers of two which are less than or equal to the given number N. Examples: 

Input: N = 63 Output: 32 16 8 4 2 1 Explanation: There are total of 6 powers of 2, which are less than or equal to the given number N. Input: N = 193 Output: 128 64 32 16 8 4 2 1 Explaination: There are total of 8 powers of 2, which are less than or equal to the given number N.

Naive Approach: The idea is to traverse each number from N to 1 and check if it is a perfect power of 2 or not. If yes, then print that number. Another Approach: The idea is to find all powers of 2 and simply print the powers that are lesser than or equal to N. Another Approach: The idea is based on the concept that all powers of 2 has all bits set, in its binary form. Bitset function is used in this approach solve the above problem. Below are the steps:

  1. Find the largest power of 2(say temp) which is used to evaluate the number less than or equal to N.
  2. Initialise an bitset array arr[] of maximum size 64, to store the binary representation of the given number N.
  3. Reset all the bits in the bitset array using reset() function.
  4. Iterate a loop from total to 0, and sequentially make each bit 1, and find the value of that binary expression and then reset the bit.

Below is the implementation of the above approach: 

CPP




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
const int MAX = 64;
 
// Function to return max exponent of
// 2 which evaluates a number less
// than or equal to N
int max_exponent(int n)
{
    return (int)(log2(n));
}
 
// Function to print all the powers
// of 2 less than or equal to N
void all_powers(int N)
{
    bitset<64> arr(N);
 
    // Reset all the bits
    arr.reset();
 
    int total = max_exponent(N);
 
    // Iterate from total to 0
    for (int i = total; i >= 0; i--) {
 
        // Reset the next bit
        arr.reset(i + 1);
 
        // Set the current bit
        arr.set(i);
 
        // Value of the binary expression
        cout << arr.to_ulong() << " ";
    }
}
 
// Driver Code
int main()
{
    // Given Number
    int N = 63;
 
    // Function Call
    all_powers(N);
    return 0;
}

Java




// Java program for the above approach
 
import java.util.*;
 
class GFG {
 
  // Function to return max exponent of
  // 2 which evaluates a number less
  // than or equal to N
  static int max_exponent(int n)
  {
    return (int)(Math.log(n) / Math.log(2));
  }
 
  // Function to print all the powers
  // of 2 less than or equal to N
  static void all_powers(int N)
  {
    int total = max_exponent(N);
 
    // Reset all the bits
    char[] arr = new char[total + 2];
    for (int i = 0; i <= total + 1; i++)
      arr[i] = '0';
 
    // Iterate from total to 0
    for (int i = 1; i < total + 2; i++) {
 
      // Reset the next bit
      arr[i - 1] = '0';
 
      // Set the current bit
      arr[i] = '1';
 
      // Value of the binary expression
      System.out.print(
        Integer.parseInt(new String(arr), 2) + " ");
    }
    System.out.print("\n");
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    // Given Number
    int N = 63;
 
    // Function Call
    all_powers(N);
  }
}
 
// This code is contributed by phasing17.

Python3




# Python3 program for the above approach
from math import log
MAX = 64;
 
# Function to return max exponent of
# 2 which evaluates a number less
# than or equal to N
def max_exponent(n):
 
    return  int(log(n,2))
 
# Function to print all the powers
# of 2 less than or equal to N
def all_powers(N):
     
    total = max_exponent(N);
     
    # Reset all the bits
    arr = ['0' for _ in range(total + 2)]
     
    # Iterate from total to 0
    for i in range(1, total + 2):
 
        # Reset the next bit
        arr[i - 1] = '0';
 
        # Set the current bit
        arr[i] = '1';
 
        # Value of the binary expression
        print( int("".join(arr), 2), end = " ")
         
    print()
     
# Driver Code
 
# Given Number
N = 63;
 
# Function Call
all_powers(N)
 
# This code is contributed by phasing17.

C#




// C# program for the above approach
 
using System;
using System.Collections.Generic;
 
class GFG
{
 
    // Function to return max exponent of
    // 2 which evaluates a number less
    // than or equal to N
    static int max_exponent(int n)
    {
        return  (int)(Math.Log(n) / Math.Log(2));
    }
     
    // Function to print all the powers
    // of 2 less than or equal to N
    static void all_powers(int N)
    {
        int total = max_exponent(N);
         
        // Reset all the bits
        char[] arr = new char[total + 2];
        for (int i = 0; i <= total + 1; i++)
            arr[i] = '0';
 
        // Iterate from total to 0
        for (int i = 1; i < total + 2; i++)
        {
     
            // Reset the next bit
            arr[i - 1] = '0';
     
            // Set the current bit
            arr[i] = '1';
     
            // Value of the binary expression
            Console.Write(Convert.ToInt32(new string(arr), 2) + " ");
        }
        Console.Write("\n");
    }
         
    // Driver Code
    public static void Main(string[] args)
    {
        // Given Number
        int N = 63;
         
        // Function Call
        all_powers(N);
    }
}
 
 
// This code is contributed by phasing17.

Javascript




// JS program for the above approach
 
let MAX = 64;
 
// Function to return max exponent of
// 2 which evaluates a number less
// than or equal to N
function max_exponent(n)
{
    return  Math.floor(Math.log(n) / Math.log(2))
}
 
// Function to print all the powers
// of 2 less than or equal to N
function all_powers(N)
{
     
    let total = max_exponent(N);
     
    let arr = new Array(total + 1);
 
    // Reset all the bits
    arr.fill(0)
 
    // Iterate from total to 0
    for (var i = total; i >= 0; i--) {
 
        // Reset the next bit
        arr[i + 1] = 0;
 
        // Set the current bit
        arr[i] = 1;
 
        // Value of the binary expression
        process.stdout.write( parseInt(arr.join(""), 2)/2 + " ");
    }
}
 
// Driver Code
 
// Given Number
let N = 63;
 
// Function Call
all_powers(N);
 
// This code is contributed by phasing17.

Output:

32 16 8 4 2 1

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


My Personal Notes arrow_drop_up
Related Articles

Start Your Coding Journey Now!