<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:
        '''
        * game starts with alice and n stones in a pile.. 
        * a player remove square number of stones in each turn.. if they can't find a square number of stones to remove.. they lose.. 
        * if both alice and bob plays optimally, return true if alice can win the game.. 
        ====
        * I can simulate the game --> but 1 <= n <= 10**5; might not get accepted.. 
        * what would optimally mean here? => on each turn they should take the biggest square number of stones possible out.. that way next player will be left with nothing (or at least left with non square stones..) => actually recursive solution works! (strightforward gameplay..)
        '''
        @cache
        def helper(n): 
            if n == 0: return False # no stones left to play.. 

            for square_num in range(1, int(math.sqrt(n)) + 1)[::-1]: # all the square number that can be made with this n stones in reverse.. 
                if not helper(n - (square_num * square_num)): # alice take out(square_num * square_num).. recursively simulate and check if there is at least one scenario in which bob loses.. 
                    return True # alice wins!
            return False

        return helper(n)
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