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:


Solution:

class Solution {
    Map<Integer, Boolean> map = new HashMap();
    
    public boolean winnerSquareGame(int n) {
        if (n == 1) return true;
        if (map.containsKey(n)) return map.get(n);
        for (int i = 1; i <= Math.sqrt(n); i ++) {
            if (!winnerSquareGame(n - i * i)) {
                map.put(n, true);
                return true;
            }
        }
        map.put(n, false);
        return false;
    }
}

class Solution {    
    public boolean winnerSquareGame(int n) {
        boolean[] dp = new boolean[n + 1];
        for (int i = 1; i <= n; i ++) {
            for (int k = 1; k * k <= i; k ++) {
                if (!dp[i - k * k]) {
                    dp[i] = true;
                    break;
                }
            }
        }
        return dp[n];
    }
}