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2660. Determine the Winner of a Bowling Game

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 {
    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);
    }
};

  • 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)


  • 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
        }
    }
}

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