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1510. Stone Game IV

Description

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).

 

Constraints:

  • 1 <= n <= 105

Solutions

  • class Solution {
    private Boolean[] f;

    public boolean winnerSquareGame(int n) {
        f = new Boolean[n + 1];
        return dfs(n);
    }

    private boolean dfs(int i) {
        if (i <= 0) {
            return false;
        }
        if (f[i] != null) {
            return f[i];
        }
        for (int j = 1; j <= i / j; ++j) {
            if (!dfs(i - j * j)) {
                return f[i] = true;
            }
        }
        return f[i] = false;
    }
}

  • class Solution {
public:
    bool winnerSquareGame(int n) {
        int f[n + 1];
        memset(f, 0, sizeof(f));
        function<bool(int)> dfs = [&](int i) -> bool {
            if (i <= 0) {
                return false;
            }
            if (f[i] != 0) {
                return f[i] == 1;
            }
            for (int j = 1; j <= i / j; ++j) {
                if (!dfs(i - j * j)) {
                    f[i] = 1;
                    return true;
                }
            }
            f[i] = -1;
            return false;
        };
        return dfs(n);
    }
};

  • class Solution:
    def winnerSquareGame(self, n: int) -> bool:
        @cache
        def dfs(i: int) -> bool:
            if i == 0:
                return False
            j = 1
            while j * j <= i:
                if not dfs(i - j * j):
                    return True
                j += 1
            return False

        return dfs(n)


  • func winnerSquareGame(n int) bool {
	f := make([]int, n+1)
	var dfs func(int) bool
	dfs = func(i int) bool {
		if i <= 0 {
			return false
		}
		if f[i] != 0 {
			return f[i] == 1
		}
		for j := 1; j <= i/j; j++ {
			if !dfs(i - j*j) {
				f[i] = 1
				return true
			}
		}
		f[i] = -1
		return false
	}
	return dfs(n)
}

  • function winnerSquareGame(n: number): boolean {
    const f: number[] = new Array(n + 1).fill(0);
    const dfs = (i: number): boolean => {
        if (i <= 0) {
            return false;
        }
        if (f[i] !== 0) {
            return f[i] === 1;
        }
        for (let j = 1; j * j <= i; ++j) {
            if (!dfs(i - j * j)) {
                f[i] = 1;
                return true;
            }
        }
        f[i] = -1;
        return false;
    };
    return dfs(n);
}

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