1900. The Earliest and Latest Rounds Where Players Compete

Description

There is a tournament where n players are participating. The players are standing in a single row and are numbered from 1 to n based on their initial standing position (player 1 is the first player in the row, player 2 is the second player in the row, etc.).

The tournament consists of multiple rounds (starting from round number 1). In each round, the i^th player from the front of the row competes against the i^th player from the end of the row, and the winner advances to the next round. When the number of players is odd for the current round, the player in the middle automatically advances to the next round.

For example, if the row consists of players 1, 2, 4, 6, 7
- Player 1 competes against player 7.
- Player 2 competes against player 6.
- Player 4 automatically advances to the next round.

After each round is over, the winners are lined back up in the row based on the original ordering assigned to them initially (ascending order).

The players numbered firstPlayer and secondPlayer are the best in the tournament. They can win against any other player before they compete against each other. If any two other players compete against each other, either of them might win, and thus you may choose the outcome of this round.

Given the integers n, firstPlayer, and secondPlayer, return an integer array containing two values, the earliest possible round number and the latest possible round number in which these two players will compete against each other, respectively.

Example 1:

Input: n = 11, firstPlayer = 2, secondPlayer = 4
Output: [3,4]
Explanation:
One possible scenario which leads to the earliest round number:
First round: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11
Second round: 2, 3, 4, 5, 6, 11
Third round: 2, 3, 4
One possible scenario which leads to the latest round number:
First round: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11
Second round: 1, 2, 3, 4, 5, 6
Third round: 1, 2, 4
Fourth round: 2, 4

Example 2:

Input: n = 5, firstPlayer = 1, secondPlayer = 5
Output: [1,1]
Explanation: The players numbered 1 and 5 compete in the first round.
There is no way to make them compete in any other round.

Constraints:

2 <= n <= 28
1 <= firstPlayer < secondPlayer <= n

Solutions

class Solution:
    def earliestAndLatest(
        self, n: int, firstPlayer: int, secondPlayer: int
    ) -> List[int]:
        # dp[i][j][k] := (earliest, latest) pair w/ firstPlayer is i-th player from
        # Front, secondPlayer is j-th player from end, and there're k people
        @functools.lru_cache(None)
        def dp(l: int, r: int, k: int) -> List[int]:
            if l == r:
                return [1, 1]
            if l > r:
                return dp(r, l, k)

            a = math.inf
            b = -math.inf

            # Enumerate all possible positions
            for i in range(1, l + 1):
                for j in range(l - i + 1, r - i + 1):
                    if not l + r - k // 2 <= i + j <= (k + 1) // 2:
                        continue
                    x, y = dp(i, j, (k + 1) // 2)
                    a = min(a, x + 1)
                    b = max(b, y + 1)

            return [a, b]

        return dp(firstPlayer, n - secondPlayer + 1, n)

class Solution {
    static int[][][] f = new int[30][30][31];

    public int[] earliestAndLatest(int n, int firstPlayer, int secondPlayer) {
        return dfs(firstPlayer - 1, secondPlayer - 1, n);
    }

    private int[] dfs(int l, int r, int n) {
        if (f[l][r][n] != 0) {
            return decode(f[l][r][n]);
        }
        if (l + r == n - 1) {
            f[l][r][n] = encode(1, 1);
            return new int[] {1, 1};
        }
        int min = Integer.MAX_VALUE, max = Integer.MIN_VALUE;
        int m = n >> 1;
        for (int i = 0; i < (1 << m); i++) {
            boolean[] win = new boolean[n];
            for (int j = 0; j < m; j++) {
                if (((i >> j) & 1) == 1) {
                    win[j] = true;
                } else {
                    win[n - 1 - j] = true;
                }
            }
            if ((n & 1) == 1) {
                win[m] = true;
            }
            win[n - 1 - l] = win[n - 1 - r] = false;
            win[l] = win[r] = true;
            int a = 0, b = 0, c = 0;
            for (int j = 0; j < n; j++) {
                if (j == l) {
                    a = c;
                }
                if (j == r) {
                    b = c;
                }
                if (win[j]) {
                    c++;
                }
            }
            int[] t = dfs(a, b, c);
            min = Math.min(min, t[0] + 1);
            max = Math.max(max, t[1] + 1);
        }
        f[l][r][n] = encode(min, max);
        return new int[] {min, max};
    }

    private int encode(int x, int y) {
        return (x << 8) | y;
    }

    private int[] decode(int val) {
        return new int[] {val >> 8, val & 255};
    }
}

int f[30][30][31];
class Solution {
public:
    vector<int> earliestAndLatest(int n, int firstPlayer, int secondPlayer) {
        return dfs(firstPlayer - 1, secondPlayer - 1, n);
    }

private:
    vector<int> dfs(int l, int r, int n) {
        if (f[l][r][n] != 0) {
            return decode(f[l][r][n]);
        }
        if (l + r == n - 1) {
            f[l][r][n] = encode(1, 1);
            return {1, 1};
        }

        int min = INT_MAX, max = INT_MIN;
        int m = n >> 1;

        for (int i = 0; i < (1 << m); i++) {
            vector<bool> win(n, false);
            for (int j = 0; j < m; j++) {
                if ((i >> j) & 1) {
                    win[j] = true;
                } else {
                    win[n - 1 - j] = true;
                }
            }
            if (n & 1) {
                win[m] = true;
            }

            win[n - 1 - l] = false;
            win[n - 1 - r] = false;
            win[l] = true;
            win[r] = true;

            int a = 0, b = 0, c = 0;
            for (int j = 0; j < n; j++) {
                if (j == l) a = c;
                if (j == r) b = c;
                if (win[j]) c++;
            }

            vector<int> t = dfs(a, b, c);
            min = std::min(min, t[0] + 1);
            max = std::max(max, t[1] + 1);
        }

        f[l][r][n] = encode(min, max);
        return {min, max};
    }

    int encode(int x, int y) {
        return (x << 8) | y;
    }

    vector<int> decode(int val) {
        return {val >> 8, val & 255};
    }
};

var f [30][30][31]int

func earliestAndLatest(n int, firstPlayer int, secondPlayer int) []int {
	return dfs(firstPlayer-1, secondPlayer-1, n)
}

func dfs(l, r, n int) []int {
	if f[l][r][n] != 0 {
		return decode(f[l][r][n])
	}
	if l+r == n-1 {
		f[l][r][n] = encode(1, 1)
		return []int{1, 1}
	}

	min, max := 1<<30, -1<<31
	m := n >> 1

	for i := 0; i < (1 << m); i++ {
		win := make([]bool, n)
		for j := 0; j < m; j++ {
			if (i>>j)&1 == 1 {
				win[j] = true
			} else {
				win[n-1-j] = true
			}
		}
		if n&1 == 1 {
			win[m] = true
		}
		win[n-1-l] = false
		win[n-1-r] = false
		win[l] = true
		win[r] = true

		a, b, c := 0, 0, 0
		for j := 0; j < n; j++ {
			if j == l {
				a = c
			}
			if j == r {
				b = c
			}
			if win[j] {
				c++
			}
		}

		t := dfs(a, b, c)
		if t[0]+1 < min {
			min = t[0] + 1
		}
		if t[1]+1 > max {
			max = t[1] + 1
		}
	}

	f[l][r][n] = encode(min, max)
	return []int{min, max}
}

func encode(x, y int) int {
	return (x << 8) | y
}

func decode(val int) []int {
	return []int{val >> 8, val & 255}
}

function earliestAndLatest(n: number, firstPlayer: number, secondPlayer: number): number[] {
    return dfs(firstPlayer - 1, secondPlayer - 1, n);
}

const f: number[][][] = Array.from({ length: 30 }, () =>
    Array.from({ length: 30 }, () => Array(31).fill(0)),
);

function dfs(l: number, r: number, n: number): number[] {
    if (f[l][r][n] !== 0) {
        return decode(f[l][r][n]);
    }
    if (l + r === n - 1) {
        f[l][r][n] = encode(1, 1);
        return [1, 1];
    }

    let min = Number.MAX_SAFE_INTEGER;
    let max = Number.MIN_SAFE_INTEGER;
    const m = n >> 1;

    for (let i = 0; i < 1 << m; i++) {
        const win: boolean[] = Array(n).fill(false);
        for (let j = 0; j < m; j++) {
            if ((i >> j) & 1) {
                win[j] = true;
            } else {
                win[n - 1 - j] = true;
            }
        }

        if (n & 1) {
            win[m] = true;
        }

        win[n - 1 - l] = false;
        win[n - 1 - r] = false;
        win[l] = true;
        win[r] = true;

        let a = 0,
            b = 0,
            c = 0;
        for (let j = 0; j < n; j++) {
            if (j === l) a = c;
            if (j === r) b = c;
            if (win[j]) c++;
        }

        const t = dfs(a, b, c);
        min = Math.min(min, t[0] + 1);
        max = Math.max(max, t[1] + 1);
    }

    f[l][r][n] = encode(min, max);
    return [min, max];
}

function encode(x: number, y: number): number {
    return (x << 8) | y;
}

function decode(val: number): number[] {
    return [val >> 8, val & 255];
}

1900 - The Earliest and Latest Rounds Where Players Compete

1900. The Earliest and Latest Rounds Where Players Compete

Description

Solutions

All Problems

All Solutions

Welcome to Subscribe On Youtube

1900. The Earliest and Latest Rounds Where Players Compete

Description

Solutions

All Problems

All Solutions