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

Input:N= 5Output:4Explanation: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** ( = **log _{2}(N)**) is the maximum power of 2 which satisfies

**2**.

^{K }â‰¤^{ }N

If (power(2, K) â‰¤ 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 2^{0})

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 log_{2}(N)**Auxiliary Space: **O(1)