<p>Alice and Bob take turns playing a game, with Alice starting first.</p>

<p>Initially, there are <code>n</code> stones in a pile. On each player&#39;s turn, that player makes a <em>move</em> consisting of removing <strong>any</strong> non-zero <strong>square number</strong> of stones in the pile.</p>

<p>Also, if a player cannot make a move, he/she loses the game.</p>

<p>Given a positive integer <code>n</code>, return <code>true</code> if and only if Alice wins the game otherwise return <code>false</code>, assuming both players play optimally.</p>

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

<pre>
<strong>Input:</strong> n = 1
<strong>Output:</strong> true
<strong>Explanation: </strong>Alice can remove 1 stone winning the game because Bob doesn&#39;t have any moves.</pre>

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

<pre>
<strong>Input:</strong> n = 2
<strong>Output:</strong> false
<strong>Explanation: </strong>Alice can only remove 1 stone, after that Bob removes the last one winning the game (2 -&gt; 1 -&gt; 0).
</pre>

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

<pre>
<strong>Input:</strong> n = 4
<strong>Output:</strong> true
<strong>Explanation:</strong> n is already a perfect square, Alice can win with one move, removing 4 stones (4 -&gt; 0).
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution:
    def winnerSquareGame(self, n: int) -> bool:

        k = math.ceil(sqrt(n))
        squares = [x*x for x in range(1, k+1)]
  
        @lru_cache(None)
        def dp(remaining): 
            if remaining == 0:
                return False
            
            for x in squares:
                if x > remaining:
                    return False 
                if not dp(remaining - x):
                    return True
        
        return dp(n)
    
            

        

                    


            






                
            

        
        


        
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