Alice and Bob take turns playing a game, with Alice starting first.

Initially, there are `n`

stones in a pile. On each player’s turn, that player makes a *move* consisting of removing **any** non-zero **square number** of stones in the pile.

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

Given a positive integer `n`

. Return `True`

if and only if Alice wins the game otherwise return `False`

, assuming both players play optimally.

**Example 1:**

Input:n = 1Output:trueExplanation:Alice can remove 1 stone winning the game because Bob doesn't have any moves.

**Example 2:**

Input:n = 2Output:falseExplanation:Alice can only remove 1 stone, after that Bob removes the last one winning the game (2 -> 1 -> 0).

**Example 3:**

Input:n = 4Output:trueExplanation:n is already a perfect square, Alice can win with one move, removing 4 stones (4 -> 0).

**Example 4:**

Input:n = 7Output:falseExplanation:Alice can't win the game if Bob plays optimally. If Alice starts removing 4 stones, Bob will remove 1 stone then Alice should remove only 1 stone and finally Bob removes the last one (7 -> 3 -> 2 -> 1 -> 0). If Alice starts removing 1 stone, Bob will remove 4 stones then Alice only can remove 1 stone and finally Bob removes the last one (7 -> 6 -> 2 -> 1 -> 0).

**Example 5:**

Input:n = 17Output:falseExplanation:Alice can't win the game if Bob plays optimally.

**Constraints:**

`1 <= n <= 10^5`

**Solution: Recursion w/ Memoization / DP**

Let win(n) denotes whether the current play will win or not.

Try all possible square numbers and see whether the other player will lose or not.

win(n) = any(win(n – i*i) == False) ? True : False

base case: win(0) = False

Time complexity: O(nsqrt(n))

Space complexity: O(n)

## C++

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class Solution { public: bool winnerSquareGame(int n) { vector<int> cache(n + 1, 0); // 0:Unknown, 1:Win, -1:Lose function<int(int)> win = [&](int n) -> int { if (n == 0) return -1; if (cache[n]) return cache[n]; for (int i = sqrt(n); i >= 1; --i) if (win(n - i * i) < 0) return cache[n] = 1; return cache[n] = -1; }; return win(n) > 0; } }; |

## Java

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class Solution { private int[] cache; public boolean winnerSquareGame(int n) { this.cache = new int[n + 1]; return this.win(n) > 0; } private int win(int n) { if (n == 0) return -1; if (this.cache[n] != 0) return this.cache[n]; for (int i = (int)Math.sqrt(n); i >= 1; --i) if (win(n - i * i) < 0) return this.cache[n] = 1; return this.cache[n] = -1; } } |

## Python3

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class Solution: def winnerSquareGame(self, n: int) -> bool: dp = [None] * (n + 1) dp[0] = False for i in range(0, n): if dp[i]: continue for j in range(1, n + 1): if i + j * j > n: break dp[i + j * j] = True return dp[n] |