Stone Game IV

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

class Solution {
    Boolean[][] dp;
    // O(N*Root(N))
    public boolean winnerSquareGame(int n) {
        dp = new Boolean[n + 1][2];
        return helper(n, 0);
    }

    public boolean helper(int n, int player) {
        if (n == 0)
            return false;
        if (dp[n][player] != null)
            return dp[n][player];
        boolean ans = false;
        for (int i = 1; i * i <= n; i++) {
            ans = ans || !(helper(n - i * i, 1 - player));
            if (ans)
                break;
        }
        dp[n][player] = ans;
        return ans;
    }
}