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

Example 2:

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

Example 3:

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

Example 4:

Input: n = 7
Output: false
Explanation: 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 = 17
Output: false
Explanation: Alice can't win the game if Bob plays optimally.

Constraints:

1 <= n <= 10^5

Solution

Time complexity : O(n*sqrt(n))
Space complexity : O(n)

class Solution {
public:
    bool winnerSquareGame(int n) {
        vector<int> mem(n+1, 0);
        function<int(int)> win = [&](int n) -> int {
            if (0 == n) return -1;
            if (mem[n]) return mem[n];
            for (int take=sqrt(n); 1<=take; --take) {
                if (win(n-take*take) < 0) return mem[n] = 1;
            }
            return mem[n] = -1;
        };
        return win(n) > 0;
    }
};

若alice想要贏得遊戲，就思考如何做選擇讓Bob輸。

以目前的石頭數，recursive考量可能的選擇，找出會讓Bob輸的去執行。
若都找不到會讓Bob輸的路，就是Alice輸。

用memoization記憶已做過的結果，避免重覆計算，增加效能。