Check if a number can be represented as sum of non zero powers of 2

Given an integer N, the task is to check whether N can be represented as the sum of powers of 2 where all the powers are > 0 i.e. 2⁰ cannot contribute to the sum.
Examples:

Input: N = 10 
Output: 1 
2³ + 2¹ = 10
Input: N = 9 
Output: 0 

Approach: There are two cases: 

1. When N is even then it can always be represented as the sum of powers of 2 when power > 0.
2. When N is odd then it can never be represented as the sum of powers of 2 if 2⁰ is not included in the sum.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function that return true if n
// can be represented as the sum
// of powers of 2 without using 2^0
bool isSumOfPowersOfTwo(int n)
{
    if (n % 2 == 1)
        return false;
    else
        return true;
}

// Driver code
int main()
{
    int n = 10;
    if (isSumOfPowersOfTwo(n))
        cout << "Yes";
    else
        cout << "No";

    return 0;
}

Java

// Java implementation of the approach
class GFG {

    // Function that return true if n
    // can be represented as the sum
    // of powers of 2 without using 2^0
    static boolean isSumOfPowersOfTwo(int n)
    {
        if (n % 2 == 1)
            return false;
        else
            return true;
    }

    // Driver code
    public static void main(String args[])
    {
        int n = 10;
        if (isSumOfPowersOfTwo(n))
            System.out.print("Yes");
        else
            System.out.print("No");
    }
}

Python3

# Python3 implementation of the approach

# Function that return true if n
# can be represented as the sum
# of powers of 2 without using 2^0
def isSumOfPowersOfTwo(n):
    if n % 2 == 1:
        return False
    else:
        return True

# Driver code
n = 10
if isSumOfPowersOfTwo(n):
    print("Yes")
else:
    print("No")

# This code is contributed
# by Shrikant13

C#

// C# implementation of the approach
using System;

class GFG {

    // Function that return true if n
    // can be represented as the sum
    // of powers of 2 without using 2^0
    static bool isSumOfPowersOfTwo(int n)
    {
        if (n % 2 == 1)
            return false;
        else
            return true;
    }

    // Driver code
    public static void Main()
    {
        int n = 10;
        if (isSumOfPowersOfTwo(n))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}

PHP

<?php
// PHP implementation of the approach

// Function that return true if n
// can be represented as the sum
// of powers of 2 without using 2^0
function isSumOfPowersOfTwo($n)
{
    if ($n % 2 == 1)
        return false;
    else
        return true;
}

// Driver code
$n = 10;
if (isSumOfPowersOfTwo($n))
    echo("Yes");
else
    echo("No");

// This code is contributed
// by Code_Mech
?>

Javascript

<script>
// Javascript implementation of the approach

// Function that return true if n
// can be represented as the sum
// of powers of 2 without using 2^0
function isSumOfPowersOfTwo(n)
{
    if (n % 2 == 1)
        return false;
    else
        return true;
}

// Driver code
var n = 10;
if (isSumOfPowersOfTwo(n))
    document.write("Yes");
else
    document.write("No");

// This code is contributed by noob2000.
</script>

Output:

Yes

Time Complexity: O(1)

Auxiliary Space: O(1)

Approach 2:

Here’s another approach to solve the problem:

• Start with the given number n.
• Initialize a variable i to 1.
• While i is less than n, do the following steps:
• a. If i is a power of 2, compute n-i.
• b. If n-i is also a power of 2, return true.
• c. Increment i by 1.
• If none of the pairs (i, n-i) are both powers of 2, return false.

Here’s the code for the above approach:

C++

#include <iostream>
#include <cmath>
using namespace std;

bool isPowerOfTwo(int n) {
    if (n == 0) {
        return false;
    }
    return (ceil(log2(n)) == floor(log2(n)));
}

bool canSumToPowerOfTwo(int n) {
    for (int i = 1; i < n; i++) {
        if (isPowerOfTwo(i) && isPowerOfTwo(n-i)) {
            return true;
        }
    }
    return false;
}

int main() {
    int n = 10;
    if (canSumToPowerOfTwo(n)) {
        cout << "Yes" << endl;
    }
    else {
        cout << "No" << endl;
    }
    return 0;
}

Javascript

// This function determines if a given integer n is a power of two
function isPowerOfTwo(n) {
    if (n == 0) {
        return false;
    }
    return (Math.ceil(Math.log2(n)) == Math.floor(Math.log2(n)));
}

// This function determines if it is possible to represent a given integer n
// as the sum of two distinct powers of two
function canSumToPowerOfTwo(n) {

    // Loop through all possible pairs of powers of two that sum to n
    for (let i = 1; i < n; i++) {
        if (isPowerOfTwo(i) && isPowerOfTwo(n-i)) {
            return true;
        }
    }

    // If no such pair exists, return false
    return false;
}

// Test the function with some sample input
let n = 10;
if (canSumToPowerOfTwo(n)) {
    console.log("Yes");
}
else {
    console.log("No");
}

Output: 

Yes

Time Complexity: O(1)

Auxiliary Space: O(1)