<p>Given an integer <code>n</code>, return <code>true</code> <em>if it is possible to represent </em><code>n</code><em> as the sum of distinct powers of three.</em> Otherwise, return <code>false</code>.</p>

<p>An integer <code>y</code> is a power of three if there exists an integer <code>x</code> such that <code>y == 3<sup>x</sup></code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 12
<strong>Output:</strong> true
<strong>Explanation:</strong> 12 = 3<sup>1</sup> + 3<sup>2</sup>
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 91
<strong>Output:</strong> true
<strong>Explanation:</strong> 91 = 3<sup>0</sup> + 3<sup>2</sup> + 3<sup>4</sup>
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> n = 21
<strong>Output:</strong> false
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 10<sup>7</sup></code></li>
</ul>

class Solution:
    def checkPowersOfThree(self, n: int) -> bool:
        #can find all powers of 3 less than n
        #see if any subset will add
            #recursive???
        def go(pows, s):
            # print(pows, s)
            if s == n:
                ans.append(True)
            if s > n:
                return False
            else:
                for i in range(len(pows)):
                    return go(pows[:i] + pows[i+1:], s + pows[i]) or go(pows[:i] + pows[i+1:], s)

        pows = [1]
        while pows[-1] * 3 <= n:
            pows.append(pows[-1] * 3)
        print(pows)

        if sum(pows) < n:
            return False
        
        ans = []
        go(pows, 0)
        print(ans)
        return len(ans) > 0

        
                



        
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