Question

2660. Determine the Winner of a Bowling Game

中文文档

Description

You are given two 0-indexed integer arrays player1 and player2, that represent the number of pins that player 1 and player 2 hit in a bowling game, respectively.

The bowling game consists of n turns, and the number of pins in each turn is exactly 10.

Assume a player hit xi pins in the ith turn. The value of the ith turn for the player is:

• 2xi if the player hit 10 pins in any of the previous two turns.
• Otherwise, It is xi.

The score of the player is the sum of the values of their n turns.

Return

• 1 _if the score of player 1 is more than the score of player 2,_
• 2 _if the score of player 2 is more than the score of player 1, and_
• 0 _in case of a draw._

 

Example 1:

Input: player1 = [4,10,7,9], player2 = [6,5,2,3]
Output: 1
Explanation: The score of player1 is 4 + 10 + 2*7 + 2*9 = 46.
The score of player2 is 6 + 5 + 2 + 3 = 16.
Score of player1 is more than the score of player2, so, player1 is the winner, and the answer is 1.

Example 2:

Input: player1 = [3,5,7,6], player2 = [8,10,10,2]
Output: 2
Explanation: The score of player1 is 3 + 5 + 7 + 6 = 21.
The score of player2 is 8 + 10 + 2*10 + 2*2 = 42.
Score of player2 is more than the score of player1, so, player2 is the winner, and the answer is 2.

Example 3:

Input: player1 = [2,3], player2 = [4,1]
Output: 0
Explanation: The score of player1 is 2 + 3 = 5
The score of player2 is 4 + 1 = 5
The score of player1 equals to the score of player2, so, there is a draw, and the answer is 0.

 

Constraints:

• n == player1.length == player2.length
• 1 <= n <= 1000
• 0 <= player1[i], player2[i] <= 10

Solutions

Solution 1: Simulation

We can define a function $f(arr)$ to calculate the scores of the two players, denoted as $a$ and $b$, respectively, and then return the answer based on the relationship between $a$ and $b$.

The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.

class Solution:
  def isWinner(self, player1: List[int], player2: List[int]) -> int:
    def f(arr: List[int]) -> int:
      s = 0
      for i, x in enumerate(arr):
        k = 2 if (i and arr[i - 1] == 10) or (i > 1 and arr[i - 2] == 10) else 1
        s += k * x
      return s

    a, b = f(player1), f(player2)
    return 1 if a > b else (2 if b > a else 0)

class Solution {
  public int isWinner(int[] player1, int[] player2) {
    int a = f(player1), b = f(player2);
    return a > b ? 1 : b > a ? 2 : 0;
  }

  private int f(int[] arr) {
    int s = 0;
    for (int i = 0; i < arr.length; ++i) {
      int k = (i > 0 && arr[i - 1] == 10) || (i > 1 && arr[i - 2] == 10) ? 2 : 1;
      s += k * arr[i];
    }
    return s;
  }
}

class Solution {
public:
  int isWinner(vector<int>& player1, vector<int>& player2) {
    auto f = [](vector<int>& arr) {
      int s = 0;
      for (int i = 0, n = arr.size(); i < n; ++i) {
        int k = (i && arr[i - 1] == 10) || (i > 1 && arr[i - 2] == 10) ? 2 : 1;
        s += k * arr[i];
      }
      return s;
    };
    int a = f(player1), b = f(player2);
    return a > b ? 1 : (b > a ? 2 : 0);
  }
};

func isWinner(player1 []int, player2 []int) int {
  f := func(arr []int) int {
    s := 0
    for i, x := range arr {
      k := 1
      if (i > 0 && arr[i-1] == 10) || (i > 1 && arr[i-2] == 10) {
        k = 2
      }
      s += k * x
    }
    return s
  }
  a, b := f(player1), f(player2)
  if a > b {
    return 1
  }
  if b > a {
    return 2
  }
  return 0
}

function isWinner(player1: number[], player2: number[]): number {
  const f = (arr: number[]): number => {
    let s = 0;
    for (let i = 0; i < arr.length; ++i) {
      s += arr[i];
      if ((i && arr[i - 1] === 10) || (i > 1 && arr[i - 2] === 10)) {
        s += arr[i];
      }
    }
    return s;
  };
  const a = f(player1);
  const b = f(player2);
  return a > b ? 1 : a < b ? 2 : 0;
}

impl Solution {
  pub fn is_winner(player1: Vec<i32>, player2: Vec<i32>) -> i32 {
    let f = |arr: &Vec<i32>| -> i32 {
      let mut s = 0;
      for i in 0..arr.len() {
        let mut k = 1;
        if (i > 0 && arr[i - 1] == 10) || (i > 1 && arr[i - 2] == 10) {
          k = 2;
        }
        s += k * arr[i];
      }
      s
    };

    let a = f(&player1);
    let b = f(&player2);
    if a > b {
      1
    } else if a < b {
      2
    } else {
      0
    }
  }
}