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2660. Determine the Winner of a Bowling Game
Description
You are given two 0indexed 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 x_{i}
pins in the i^{th}
turn. The value of the i^{th}
turn for the player is:
2x_{i}
if the player hit10
pins in any of the previous two turns. Otherwise, It is
x_{i}
.
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, and0
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[i1] == 10)  (i > 1 && arr[i2] == 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 } } }