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Powers of 2 to required sum using Bit Masking

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  • Difficulty Level : Easy
  • Last Updated : 03 Jun, 2021

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 ith 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 ith bit is set, we only need to check: 
     

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


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