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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

Solution 1: Depth-First Search + Memoization

We design a function $dfs(i)$ to indicate whether the current player can win the game when there are $i$ stones in the current pile of stones. Returns $true$ if the current player can win the game, otherwise returns $false$. Then the answer is $dfs(n)$.

The calculation process of function $dfs(i)$ is as follows:

  • If $i \leq 0$, it means that the current player cannot perform any operation, so the current player loses the game and returns $false$;
  • Otherwise, enumerate the number of stones $j$ that the current player can take, where $j$ is a square number. If another player cannot win the game after the current player takes $j$ stones, the current player wins the game and $true$ is returned. If all $j$ are enumerated and none of the above conditions are met, the current player loses the game and $false$ is returned.

In order to avoid repeated calculations, we can use memoized search, that is, use the array $f$ to record the calculation results of the function $dfs(i)$.

Time complexity $O(n \times \sqrt{n})$, space complexity $O(n)$. Among them, $n$ is the number of stones in the pile of stones.

Solution 2

We can also use dynamic programming to solve this problem.

Define the array $f$, where $f[i]$ indicates whether the current player can win the game when there are $i$ stones in the current pile of stones. $f[i]$ is $true$ if the current player can win the game, $false$ otherwise. Then the answer is $f[n]$.

We enumerate $i$ in the range of $[1,..n]$ and enumerate $j$ in the range of $[1,..i]$, where $j$ is a square number. If another player cannot win the game after the current player takes $j$ stones, the current player wins the game, that is $f[i] = true$. If all $j$ are enumerated and none of the above conditions are met, the current player loses the game, which is $f[i] = false$. Therefore we can get the state transition equation:

\[f[i]= \begin{cases} true, & \textit{if } \exists j \in [1,..i], j^2 \leq i \textit{ and } f[i-j^2] = false\\ false, & \textit{otherwise} \end{cases}\]

Finally, we return $f[n]$.

Time complexity $O(n \times \sqrt{n})$, space complexity $O(n)$. Among them, $n$ is the number of stones in the pile of stones.

  • 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;
    }
}


// Solution 2
class Solution {
    public boolean winnerSquareGame(int n) {
        boolean[] f = new boolean[n + 1];
        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= i / j; ++j) {
                if (!f[i - j * j]) {
                    f[i] = true;
                    break;
                }
            }
        }
        return f[n];
    }
}


  • 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);
    }
};


// Solution 2
class Solution {
public:
    bool winnerSquareGame(int n) {
        bool f[n + 1];
        memset(f, false, sizeof(f));
        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= i / j; ++j) {
                if (!f[i - j * j]) {
                    f[i] = true;
                    break;
                }
            }
        }
        return f[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)


# Solution 2
class Solution:
    def winnerSquareGame(self, n: int) -> bool:
        f = [False] * (n + 1)
        for i in range(1, n + 1):
            j = 1
            while j <= i // j:
                if not f[i - j * j]:
                    f[i] = True
                    break
                j += 1
        return f[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)
}


// Solution 2
func winnerSquareGame(n int) bool {
	f := make([]bool, n+1)
	for i := 1; i <= n; i++ {
		for j := 1; j <= i/j; j++ {
			if !f[i-j*j] {
				f[i] = true
				break
			}
		}
	}
	return f[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);
}


// Solution 2
function winnerSquareGame(n: number): boolean {
    const f: boolean[] = new Array(n + 1).fill(false);
    for (let i = 1; i <= n; ++i) {
        for (let j = 1; j * j <= i; ++j) {
            if (!f[i - j * j]) {
                f[i] = true;
                break;
            }
        }
    }
    return f[n];
}


  • impl Solution {
    pub fn winner_square_game(n: i32) -> bool {
        let mut f = vec![-1; (n + 1) as usize];

        fn dfs(i: i32, f: &mut Vec<i8>) -> bool {
            if i <= 0 {
                return false;
            }

            let idx = i as usize;
            if f[idx] != -1 {
                return f[idx] == 1;
            }

            let k = (i as f64).sqrt() as i32;
            for j in 1..=k {
                if !dfs(i - j * j, f) {
                    f[idx] = 1;
                    return true;
                }
            }

            f[idx] = 0;
            false
        }

        dfs(n, &mut f)
    }
}


// Solution 2
impl Solution {
    pub fn winner_square_game(n: i32) -> bool {
        let n = n as usize;
        let mut f = vec![false; n + 1];

        for i in 1..=n {
            let mut j = 1;
            while j <= i / j {
                if !f[i - j * j] {
                    f[i] = true;
                    break;
                }
                j += 1;
            }
        }

        f[n]
    }
}

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