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

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:

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
}

chevron_right


Output:

32 16 8 4 2 1

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

competitive-programming-img




My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.