Count ways to represent N as sum of powers of 2

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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:

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

Another Efficient Approach ( using DP) : First we iterate all values of power of 2<= N , starting from 1. Then , we will find value of big problem using value of small sub-problems .

Below is the implementation of the above approach:

C++

// C++ program for above implementation
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to count the number of ways to represent 
// n as the power of 2
int numberOfWays(int n)
{
   // Initialize an array dp with size n+1
   int dp[n+1] = {0};
   
   // Base case- there is only 1 way to represent 0 
   // as a sum of powers of 2
   dp[0] = 1;
 
   // Iterate all powers of 2 starting from 1
   for (int i = 1; i <= n; i = i*2) 
   {
        // Iterate through all numbers from 1 to n
       for (int j = i; j <= n; j++) 
       {
        // Using sub-problems that is already calculated to find the
        // number of ways to represent j as a sum of powers of 2
           dp[j] += dp[j-i];
       }
   }
 
   // Return the number of ways to represent 
   // n as a sum of powers of 2
   return dp[n];
}
 
//Drive code
int main() {
   int n=4;
   
   //Function call
   cout << "Number of ways: "<< numberOfWays(n) << endl;
   return 0;
}
 
// This code is contributed by nikhilsainiofficial546

Java

// Java program for above implementation
import java.util.*;
 
public class Main {
 
    // Function to count the number of ways to represent
    // n as the power of 2
    static int numberOfWays(int n)
    {
        // Initialize an array dp with size n+1
        int dp[] = new int[n + 1];
 
        // Base case- there is only 1 way to represent 0
        // as a sum of powers of 2
        dp[0] = 1;
 
        // Iterate all powers of 2 starting from 1
        for (int i = 1; i <= n; i = i * 2) {
            // Iterate through all numbers from 1 to n
            for (int j = i; j <= n; j++) {
                // Using sub-problems that is already
                // calculated to find the number of ways to
                // represent j as a sum of powers of 2
                dp[j] += dp[j - i];
            }
        }
 
        // Return the number of ways to represent
        // n as a sum of powers of 2
        return dp[n];
    }
 
    // Drive code
    public static void main(String[] args)
    {
        int n = 4;
 
        // Function call
        System.out.println("Number of ways: "
                           + numberOfWays(n));
    }
}

Python3

# Python3 program for above implementation
 
# Function to count the number of ways to represent 
# n as the power of 2
def numberOfWays(n):
    # Initialize an array dp with size n+1
    dp = [0 for i in range(n+1)]
   
    # Base case- there is only 1 way to represent 0 
    # as a sum of powers of 2
    dp[0] = 1
 
    # Iterate all powers of 2 starting from 1
    i = 1
    while i <= n:
       
        # Iterate through all numbers from 1 to n
        j = i
        while j <= n:
           
            # Using sub-problems that is already calculated to find the
            # number of ways to represent j as a sum of powers of 2
            dp[j] += dp[j-i]
            j += 1
        i *= 2
 
    # Return the number of ways to represent 
    # n as a sum of powers of 2
    return dp[n]
 
# Drive code
if __name__ == '__main__':
    n = 4
   
    # Function call
    print("Number of ways:", numberOfWays(n))
 
# This code is contributed by nikhilsainiofficial546

C#

// C# program for the above approach
using System;
 
public class GFG
{
    // Function to count the number of ways to represent
    // n as the power of 2
    static int NumberOfWays(int n)
    {
        // Initialize an array dp with size n+1
        int[] dp = new int[n + 1];
            // Base case- there is only 1 way to represent 0
        // as a sum of powers of 2
        dp[0] = 1;
     
        // Iterate all powers of 2 starting from 1
        for (int i = 1; i <= n; i = i * 2)
        {
            // Iterate through all numbers from 1 to n
            for (int j = i; j <= n; j++)
            {
                // Using sub-problems that is already
                // calculated to find the number of ways to
                // represent j as a sum of powers of 2
                dp[j] += dp[j - i];
            }
        }
     
        // Return the number of ways to represent
        // n as a sum of powers of 2
        return dp[n];
    }
 
    // Drive code
    public static void Main(string[] args)
    {
        int n = 4;
     
        // Function call
        Console.WriteLine("Number of ways: " + NumberOfWays(n));
    }
}
 
// This code is contributed by sdeadityasharma

Javascript

// Function to count the number of ways to represent 
// n as the power of 2
function numberOfWays(n) {
    // Initialize an array dp with size n+1
    let dp = new Array(n+1).fill(0);
   
    // Base case- there is only 1 way to represent 0 
    // as a sum of powers of 2
    dp[0] = 1;
 
    // Iterate all powers of 2 starting from 1
    let i = 1;
    while (i <= n) {
       
        // Iterate through all numbers from 1 to n
        let j = i;
        while (j <= n) {
           
            // Using sub-problems that is already calculated to find the
            // number of ways to represent j as a sum of powers of 2
            dp[j] += dp[j-i];
            j += 1;
        }
        i *= 2;
    }
 
    // Return the number of ways to represent 
    // n as a sum of powers of 2
    return dp[n];
}
 
// Driver code
let n = 4;
   
// Function call
console.log("Number of ways:", numberOfWays(n));

Output

Number of ways: 4

Time Complexity: O(N*logN), logN time to iterate all powers of 2 that is <=N
Auxiliary Space: O(N)

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