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1510. Stone Game IV

SourceBiweekly Contest 30 Q4DifficultyHardRating1786

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: Memoization

Thinking

Players alternately remove a square number of stones from a pile of \(n\); the one who takes the last stone wins. \(n\le 10^5\), so expanding the full game tree would recompute the same remainder many times.

A position depends only on the remaining count: it is winning if some move leaves the opponent in a losing position. Memoized \(\textit{dfs}(i)\) answers whether the player to move wins with \(i\) stones, trying every \(j^2\le i\). Each state has \(O(\sqrt{i})\) transitions.

We design a function \(\textit{dfs}(i)\), which represents whether the current player can win the game when there are \(i\) stones in the pile. If the current player can win, it returns \(\textit{true}\); otherwise, it returns \(\textit{false}\). The answer is \(\textit{dfs}(n)\).

The calculation process of the function \(\textit{dfs}(i)\) is as follows:

  • If \(i \leq 0\), it means the current player cannot make any move, so the current player loses the game, return \(\textit{false}\);
  • Otherwise, enumerate the number of stones \(j\) that the current player can take away, where \(j\) is a square number. If the other player cannot win the game after the current player takes away \(j\) stones, then the current player wins the game, return \(\textit{true}\). If all \(j\) enumerated cannot satisfy the above condition, then the current player loses the game, return \(\textit{false}\).

To avoid repeated calculations, we can use memoization, i.e., use an array \(f\) to record the calculation results of the function \(\textit{dfs}(i)\).

The time complexity is \(O(n \sqrt{n})\), and the space complexity is \(O(n)\), where \(n\) is the number of stones in the pile.

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@cache
def dfs(i: int) -> bool:
    if i <= 0:
        return False
    k = isqrt(i)
    return any(not dfs(i - j * j) for j in range(1, k + 1))


class Solution:
    def winnerSquareGame(self, n: int) -> bool:
        return dfs(n)

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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];
        }
        int k = (int) Math.sqrt(i);
        for (int j = 1; j <= k; j++) {
            if (!dfs(i - j * j)) {
                return f[i] = true;
            }
        }
        return f[i] = false;
    }
}

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class Solution {
public:
    bool winnerSquareGame(int n) {
        vector<int> f(n + 1, -1);

        auto dfs = [&](this auto&& dfs, int i) -> bool {
            if (i <= 0) {
                return false;
            }
            if (f[i] != -1) {
                return f[i];
            }

            int k = sqrt(i);
            for (int j = 1; j <= k; j++) {
                if (!dfs(i - j * j)) {
                    return f[i] = true;
                }
            }

            return f[i] = false;
        };

        return dfs(n);
    }
};

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func winnerSquareGame(n int) bool {
    f := make([]int8, 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
        }
        k := int(math.Sqrt(float64(i)))
        for j := 1; j <= k; j++ {
            if !dfs(i - j*j) {
                f[i] = 1
                return true
            }
        }
        f[i] = -1
        return false
    }

    return dfs(n)
}

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function winnerSquareGame(n: number): boolean {
    const f = new Array<number>(n + 1).fill(-1);

    const dfs = (i: number): boolean => {
        if (i <= 0) {
            return false;
        }
        if (f[i] !== -1) {
            return f[i] === 1;
        }

        const k = Math.floor(Math.sqrt(i));
        for (let j = 1; j <= k; j++) {
            if (!dfs(i - j * j)) {
                f[i] = 1;
                return true;
            }
        }

        f[i] = 0;
        return false;
    };

    return dfs(n);
}

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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: Dynamic Programming

Thinking

Solution 1 already fills this win/loss table from fewer stones to more. The \(f[i]\) below is that same table.

Solution 1 already fills this table from fewer stones to more. Define an array \(f\), where \(f[i]\) represents whether the current player can win the game when there are \(i\) stones in the pile. If the current player can win, then \(f[i]\) is \(\textit{true}\), otherwise it is \(\textit{false}\). The answer is \(f[n]\).

We enumerate \(i\) in the range \([1,..n]\), and enumerate \(j\) in the range \([1,..i]\), where \(j\) is a square number. If the other player cannot win the game after the current player takes away \(j\) stones, then the current player wins the game, i.e., \(f[i] = \textit{true}\). If all \(j\) enumerated cannot satisfy the above condition, then the current player loses the game, i.e., \(f[i] = \textit{false}\). Therefore, we can get the state transition equation:

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

Finally, we return \(f[n]\).

The time complexity is \(O(n \sqrt{n})\), and the space complexity is \(O(n)\), where \(n\) is the number of stones in the pile.

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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]

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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];
    }
}

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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];
    }
};

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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]
}

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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];
}

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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]
    }
}

Solution 3: Dynamic Programming

Thinking

Expanding the whole game tree is exponential. There can be \(10^5\) stones, and the first move always removes one stone, so the search has depth \(n\). Whether a position is winning depends only on the remaining count, and only on smaller counts. Let \(f[i]\) record a win with \(i\) stones, with \(f[0]\) losing, and scan \(i\) upward. A square \(j^2\) that leaves a losing position makes \(f[i]\) winning.

Let \(f[i]\) be whether the player to move wins with \(i\) stones. \(f[0]\) is \(\textit{false}\), since no move remains. For \(i\) from \(1\) to \(n\), enumerate squares \(j^2 \le i\). If \(f[i-j^2]\) is \(\textit{false}\), the opponent loses from there, so \(f[i]\) is \(\textit{true}\) and the remaining squares can be skipped. If every square leaves a winning position for the opponent, \(f[i]\) stays \(\textit{false}\). The answer is \(f[n]\).

The time complexity is \(O(n \sqrt{n})\), and the space complexity is \(O(n)\), where \(n\) is the number of stones in the pile.

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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]

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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];
    }
}

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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];
    }
};

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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]
}

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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];
}

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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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