<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:
        """
        we can use backtracking sort of approach
        State that define the game at any instant is
        (n, chance) -> (number of stones remaining, chance of whom to pick)

        in Below code:
            if chance = 1 that means it's Alice's chance to pick
            else it's Bob's Chance to pick
        if fun returns True that means Alice wins else Bob Wins
        """
        lookup = {}
        def fun(n, chance):
            if (n, chance) in lookup:
                return lookup[(n, chance)]
            if n <= 0:
                if chance == 1:
                    return False
                return True

            for i in range(1, floor(sqrt(n))+1):
                #print(f'i={i}, n={n}, chance={chance}')
                if chance == 1:
                    val = False
                    # if fun() in Bob's chance returns True, that means Alice wins
                    if fun(n - (i*i), chance^1):
                        val = True
                        break
                else:
                    val = True
                    # if fun() in Alice's chance returns False, that means Bob wins
                    if not fun(n - (i*i), chance^1):
                        val = False
                        break
            lookup[(n, chance)] = val
            return val
        return fun(n, 1)
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