Powers of 2 to required sum using Bit Masking

Given an integer N, the task is to find the numbers which when added after being raised to the Power of 2 gives the integer N.
Examples:

Input: N = 12345 
Output: 0, 3, 4, 5, 12, 13 
Explanation:
12345 = 2^0 + 2^3 + 2^4 + 2^5 + 2^12 + 2^13
Input: N = 10000 
Output: 4, 8, 9, 10, 13 
Explanation:
10000 = 2^4 + 2^8 + 2^9 + 2^10 + 2^13 

Approach:

• Since every number can be expressed as sum of powers of 2, the task is to print every i when i^th bit is set in the binary representation of N.

Illustration:
(29)₁₀ = (11101)₂
Thus, in 29, {0, 2, 3, 4} are the indices of the set bits. 
2⁰ + 2² + 2³ + 2⁴
= 1 + 4 + 8 + 16 
= 29 

• In order to check if i^th bit is set, we only need to check: 
  

if( N & (1 << i) ) == 1 
Illustration:
(29)₁₀ = (11101)₂
0^th bit: 11101 & 1 = 1 
1^st bit: 11101 & 10 = 0 
2^nd bit: 11101 & 100 = 1 
3^rd bit: 11101 & 1000 = 1 
4^th bit: 11101 & 10000 = 1 

Below is the implementation of the above approach.

C++

// C++ Program to split N
// into numbers which when
// added after being raised
// to the power of 2 gives N

#include <bits/stdc++.h>
using namespace std;

// Function to print the numbers
// which raised to power of 2
// adds up to N
void PowerOfTwo(long long N)
{
    for (int i = 0; i < 64; i++) {
        long long x = 1;

        // Checking if i-th bit is
        // set in N or not by
        // shifting 1 i bits to
        // the left and performing
        // AND with N
        if (N & (x << i))
            cout << i << " ";
    }
}

// Driver Code
int main()
{

    long long N = 12345;
    PowerOfTwo(N);

    return 0;
}

Java

// Java Program to split N
// into numbers which when
// added after being raised
// to the power of 2 gives N
class GFG{

// Function to print the numbers
// which raised to power of 2
// adds up to N
static void PowerOfTwo(long N)
{
    for (int i = 0; i < 64; i++)
    {
        long x = 1;

        // Checking if i-th bit is
        // set in N or not by
        // shifting 1 i bits to
        // the left and performing
        // AND with N
        if ((N & (x << i)) > 0)
            System.out.print(i + " ");
    }
}

// Driver Code
public static void main(String[] args)
{
    long N = 12345;
    PowerOfTwo(N);
}
}

// This code is contributed by Amit Katiyar

Python3

# Python3 program to split N
# into numbers which when
# added after being raised
# to the power of 2 gives N

# Function to print the numbers
# which raised to power of 2
# adds up to N
def PowerOfTwo(N):

    for i in range(0, 64):
        x = 1

        # Checking if i-th bit is
        # set in N or not by
        # shifting 1 i bits to
        # the left and performing
        # AND with N
        if (N & (x << i)) > 0:
            print(i, end = " ")

# Driver Code
if __name__ == "__main__":

    # Given number
    N = 12345;

    # Function call
    PowerOfTwo(N)

# This code is contributed by rock_cool

C#

// C# Program to split N
// into numbers which when
// added after being raised
// to the power of 2 gives N
using System;
class GFG{

// Function to print the numbers
// which raised to power of 2
// adds up to N
static void PowerOfTwo(long N)
{
    for (int i = 0; i < 64; i++)
    {
        long x = 1;

        // Checking if i-th bit is
        // set in N or not by
        // shifting 1 i bits to
        // the left and performing
        // AND with N
        if ((N & (x << i)) > 0)
            Console.Write(i + " ");
    }
}

// Driver Code
public static void Main()
{
    long N = 12345;
    PowerOfTwo(N);
}
}

// This code is contributed by Nidhi_biet

Javascript

<script>

// Javascript Program to split N
// into numbers which when
// added after being raised
// to the power of 2 gives N

// Function to print the numbers
// which raised to power of 2
// adds up to N
function PowerOfTwo(N)
{
    for (var i = 0; i < 64; i++) {
        var x = 1;

        // Checking if i-th bit is
        // set in N or not by
        // shifting 1 i bits to
        // the left and performing
        // AND with N
        if ((N & (x * Math.pow(2,i))) >0)
            document.write( i + " ");
    }
}

// Driver Code
var N = 12345;
PowerOfTwo(N);

// This code is contributed by importantly.
</script>

Output:

0 3 4 5 12 13

Time Complexity: O(log N) 
Space Complexity: O(1)
For an alternative approach, refer to Powers of 2 to required sum