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3021. Alice and Bob Playing Flower Game

Description

Alice and Bob are playing a turn-based game on a circular field surrounded by flowers. The circle represents the field, and there are x flowers in the clockwise direction between Alice and Bob, and y flowers in the anti-clockwise direction between them.

The game proceeds as follows:

  1. Alice takes the first turn.
  2. In each turn, a player must choose either the clockwise or anti-clockwise direction and pick one flower from that side.
  3. At the end of the turn, if there are no flowers left at all, the current player captures their opponent and wins the game.

Given two integers, n and m, the task is to compute the number of possible pairs (x, y) that satisfy the conditions:

  • Alice must win the game according to the described rules.
  • The number of flowers x in the clockwise direction must be in the range [1,n].
  • The number of flowers y in the anti-clockwise direction must be in the range [1,m].

Return the number of possible pairs (x, y) that satisfy the conditions mentioned in the statement.

 

Example 1:

Input: n = 3, m = 2
Output: 3
Explanation: The following pairs satisfy conditions described in the statement: (1,2), (3,2), (2,1).

Example 2:

Input: n = 1, m = 1
Output: 0
Explanation: No pairs satisfy the conditions described in the statement.

 

Constraints:

  • 1 <= n, m <= 105

Solutions

Solution 1: Mathematics

According to the problem description, in each move, the player will choose to move in a clockwise or counterclockwise direction and then pick a flower. Since Alice moves first, when \(x + y\) is odd, Alice will definitely win the game.

Therefore, the number of flowers \(x\) and \(y\) meet the following conditions:

  1. \(x + y\) is odd;
  2. \(1 \le x \le n\);
  3. \(1 \le y \le m\).

If \(x\) is odd, \(y\) must be even. At this time, the number of values of \(x\) is \(\lceil \frac{n}{2} \rceil\), the number of values of \(y\) is \(\lfloor \frac{m}{2} \rfloor\), so the number of pairs that meet the conditions is \(\lceil \frac{n}{2} \rceil \times \lfloor \frac{m}{2} \rfloor\).

If \(x\) is even, \(y\) must be odd. At this time, the number of values of \(x\) is \(\lfloor \frac{n}{2} \rfloor\), the number of values of \(y\) is \(\lceil \frac{m}{2} \rceil\), so the number of pairs that meet the conditions is \(\lfloor \frac{n}{2} \rfloor \times \lceil \frac{m}{2} \rceil\).

Therefore, the number of pairs that meet the conditions is \(\lceil \frac{n}{2} \rceil \times \lfloor \frac{m}{2} \rfloor + \lfloor \frac{n}{2} \rfloor \times \lceil \frac{m}{2} \rceil\), which is \(\lfloor \frac{n + 1}{2} \rfloor \times \lfloor \frac{m}{2} \rfloor + \lfloor \frac{n}{2} \rfloor \times \lfloor \frac{m + 1}{2} \rfloor\).

The time complexity is \(O(1)\), and the space complexity is \(O(1)\).

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class Solution:
    def flowerGame(self, n: int, m: int) -> int:
        a1 = (n + 1) // 2
        b1 = (m + 1) // 2
        a2 = n // 2
        b2 = m // 2
        return a1 * b2 + a2 * b1

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class Solution {
    public long flowerGame(int n, int m) {
        long a1 = (n + 1) / 2;
        long b1 = (m + 1) / 2;
        long a2 = n / 2;
        long b2 = m / 2;
        return a1 * b2 + a2 * b1;
    }
}

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class Solution {
public:
    long long flowerGame(int n, int m) {
        long long a1 = (n + 1) / 2;
        long long b1 = (m + 1) / 2;
        long long a2 = n / 2;
        long long b2 = m / 2;
        return a1 * b2 + a2 * b1;
    }
};

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func flowerGame(n int, m int) int64 {
    a1, b1 := (n+1)/2, (m+1)/2
    a2, b2 := n/2, m/2
    return int64(a1*b2 + a2*b1)
}

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function flowerGame(n: number, m: number): number {
    const [a1, b1] = [(n + 1) >> 1, (m + 1) >> 1];
    const [a2, b2] = [n >> 1, m >> 1];
    return a1 * b2 + a2 * b1;
}

Solution 2: Mathematics (Optimized)

The result obtained from Solution 1 is \(\lfloor \frac{n + 1}{2} \rfloor \times \lfloor \frac{m}{2} \rfloor + \lfloor \frac{n}{2} \rfloor \times \lfloor \frac{m + 1}{2} \rfloor\).

If both \(n\) and \(m\) are odd, then the result is \(\frac{n + 1}{2} \times \frac{m - 1}{2} + \frac{n - 1}{2} \times \frac{m + 1}{2}\), which is \(\frac{n \times m - 1}{2}\).

If both \(n\) and \(m\) are even, then the result is \(\frac{n}{2} \times \frac{m}{2} + \frac{n}{2} \times \frac{m}{2}\), which is \(\frac{n \times m}{2}\).

If \(n\) is odd and \(m\) is even, then the result is \(\frac{n + 1}{2} \times \frac{m}{2} + \frac{n - 1}{2} \times \frac{m}{2}\), which is \(\frac{n \times m}{2}\).

If \(n\) is even and \(m\) is odd, then the result is \(\frac{n}{2} \times \frac{m - 1}{2} + \frac{n}{2} \times \frac{m + 1}{2}\), which is \(\frac{n \times m}{2}\).

The above four cases can be combined into \(\lfloor \frac{n \times m}{2} \rfloor\).

The time complexity is \(O(1)\), and the space complexity is \(O(1)\).

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class Solution:
    def flowerGame(self, n: int, m: int) -> int:
        return (n * m) // 2

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class Solution {
    public long flowerGame(int n, int m) {
        return ((long) n * m) / 2;
    }
}

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class Solution {
public:
    long long flowerGame(int n, int m) {
        return ((long long) n * m) / 2;
    }
};

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func flowerGame(n int, m int) int64 {
    return int64((n * m) / 2)
}

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function flowerGame(n: number, m: number): number {
    return Number(((BigInt(n) * BigInt(m)) / 2n) | 0n);
}

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