<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:
        dp = defaultdict(int)
        dp = [[-1] * 2 for _ in range(n + 1)]
        def dfs(i, p):
            if i == 0:
                return False if p == 0 else True
            if dp[i][p] != -1:
                return dp[i][p]
            k = 1
            win = True if p == 1 else False
            while i - k**2 >= 0:
                if p == 0:
                    win = win | dfs(i - k**2, 1 - p)
                    if win == True:
                        break
                else:
                    win = win & dfs(i - k**2, 1 - p)
                    if win == False:
                        break
                k += 1
            dp[i][p] = win
            return win
        return dfs(n, 0)
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