Count ways to represent N as sum of powers of 2

Given an integer N, the task is to count the number of ways to represent N as the sum of powers of 2.

Examples:

Input: N = 4
Output: 4
Explanation: All possible ways to obtains sum N using powers of 2 are  {4, 2+2, 1+1+1+1, 2+1+1}.

Input: N = 5
Output: 4
Explanation: All possible ways to obtains sum N using powers of 2 are  {4 + 1, 2+2 + 1, 1+1+1+1 + 1, 2+1+1 + 1}

Naive Approach: The simplest approach to solve the problem is to generate all powers of 2 whose values are less than N and print all combinations to represent the sum N.

Efficient Approach: To optimize the above approach, the idea is to use recursion. Define a function f(N, K) which represents the number of ways to express N as a sum of powers of 2 with all the numbers having power less than or equal to k where K ( = log₂(N)) is the maximum power of 2 which satisfies 2^K ≤ N.

If (power(2, K) ≤ N) :
f(N, K) = f(N – power(2, K), K) + f(N, K – 1) //to check if power(2, k) can be one of the number.    
Otherwise:
f(N, K)=f(N, K – 1)
Base cases :

• If (N = 0) f(N, K)=1 (Only 1 possible way exists to represent N)
• If (k==0) f(N, K)=1 (Only 1 possible way exists to represent N by taking 2⁰)

Below is the implementation of the above approach:

C++

// C++ program for above implementation
#include <bits/stdc++.h>
using namespace std;
int numberOfWays(int n, int k)
{

    // Base Cases
    if (n == 0)
        return 1;

    if (k == 0)
        return 1;

    // Check if 2^k can be used as
    // one of the numbers or not
    if (n >= pow(2, k)) {
        int curr_val = pow(2, k);
        return numberOfWays(n - curr_val, k)
               + numberOfWays(n, k - 1);
    }
    // Otherwise
    else

        // Count number of  ways to
        // N using 2 ^ k - 1
        return numberOfWays(n, k - 1);
}

// Driver Code
int main()
{
    int n = 4;
    int k = log2(n);

    cout << numberOfWays(n, k) << endl;
}

Java

// Java program to implement
// the above approach
import java.util.*;
class GFG
{
static int numberOfWays(int n, int k)
{

    // Base Cases
    if (n == 0)
        return 1;
    if (k == 0)
        return 1;

    // Check if 2^k can be used as
    // one of the numbers or not
    if (n >= (int)Math.pow(2, k))
    {
        int curr_val = (int)Math.pow(2, k);
        return numberOfWays(n - curr_val, k)
               + numberOfWays(n, k - 1);
    }

    // Otherwise
    else

        // Count number of  ways to
        // N using 2 ^ k - 1
        return numberOfWays(n, k - 1);
}

// Driver code
public static void main(String[] args)
{
    int n = 4;
    int k = (int)(Math.log(n) / Math.log(2));
     System.out.println(numberOfWays(n, k));
}
}

// This code is contributed by susmitakundugoaldanga.

Python3

# Python3 program for above implementation
from math import log2
def numberOfWays(n, k):

    # Base Cases
    if (n == 0):
        return 1
    if (k == 0):
        return 1

    # Check if 2^k can be used as
    # one of the numbers or not
    if (n >= pow(2, k)):
        curr_val = pow(2, k)
        return numberOfWays(n - curr_val, k) + numberOfWays(n, k - 1)

    # Otherwise
    else:

        # Count number of  ways to
        # N using 2 ^ k - 1
        return numberOfWays(n, k - 1)

# Driver Code
if __name__ == '__main__':
    n = 4
    k = log2(n)

    print(numberOfWays(n, k))

# This code is contributed by mohit kumar 29

C#

// C# program to implement
// the above approach
using System;
class GFG
{
static int numberOfWays(int n, int k)
{

    // Base Cases
    if (n == 0)
        return 1;
    if (k == 0)
        return 1;

    // Check if 2^k can be used as
    // one of the numbers or not
    if (n >= (int)Math.Pow(2, k))
    {
        int curr_val = (int)Math.Pow(2, k);
        return numberOfWays(n - curr_val, k)
               + numberOfWays(n, k - 1);
    }

    // Otherwise
    else

        // Count number of  ways to
        // N using 2 ^ k - 1
        return numberOfWays(n, k - 1);
}

// Driver code
public static void Main(String[] args)
{
    int n = 4;
    int k = (int)(Math.Log(n) / Math.Log(2));
     Console.WriteLine(numberOfWays(n, k));
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>

// JavaScript program for above implementation

function numberOfWays(n, k)
{

    // Base Cases
    if (n == 0)
        return 1;

    if (k == 0)
        return 1;

    // Check if 2^k can be used as
    // one of the numbers or not
    if (n >= Math.pow(2, k)) {
        let curr_val = Math.pow(2, k);
        return numberOfWays(n - curr_val, k)
            + numberOfWays(n, k - 1);
    }
    // Otherwise
    else

        // Count number of ways to
        // N using 2 ^ k - 1
        return numberOfWays(n, k - 1);
}

// Driver Code

    let n = 4;
    let k = Math.log2(n);

    document.write(numberOfWays(n, k) + "<br>");

// This code is contributed by Mayank Tyagi

</script>

Output:

4

Time Complexity: O((logN+K)^K ), where K is log₂(N)
Auxiliary Space: O(1)