<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:

        res = []
        self.res = False

        for i in range(0, 20):
            res.append(3**i)

        def dfs(ind, summ):
            if summ == 0:
                self.res = True
                return 
            if summ < 0 or ind >= len(res) or res[ind] > summ:
                return 

            for i in range(ind, len(res)):
                dfs(i + 1, summ - res[i])

        print('res', res)
        
        dfs(0, n)
        return self.res
        
        
        
16500