# 1510-stone-game-iv


Try it on <a href='https://leetcode.com/problems/1510-stone-game-iv'>leetcode</a>

## Description
<div class="description">
<div><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'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>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't have any moves.</pre>

<p><strong>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>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>
</div>
</div>

## Solution(Python)
```Python
class Solution:
    #     @cache
    #     def winnerSquareGame(self, n: int) -> bool:
    #         if n == 0:
    #             return False

    #         for i in range(1,int(pow(n,0.5)+1)):
    #             if not self.winnerSquareGame(n-i*i):
    #                 return True
    #         return False
    # Time Complexity : N^(1+1/2)
    # Space Complexity: N
    def winnerSquareGame(self, n: int) -> bool:
        dp = [False] * (n + 1)

        for i in range(1, n + 1):
            for j in range(1, int(pow(i, 0.5) + 1)):
                if not dp[i - j * j]:
                    dp[i] = True
                    break
        return dp[n]

```