2660

2660. Determine the Winner of a Bowling Game (Easy)

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

Companies: Amazon, DE Shaw

Related Topics:
Array, Simulation

Similar Questions:

• High Five (Easy)

Solution 1.

// OJ: https://leetcode.com/problems/determine-the-winner-of-a-bowling-game
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
    int score(vector<int> &A) {
        int a = -1, b = 1, ans = 0;
        for (int n : A) {
            if (a == 10 || b == 10) ans += 2 * n;
            else ans += n;
            a = b;
            b = n;
        }
        return ans;
    }
public:
    int isWinner(vector<int>& A, vector<int>& B) {
        int a = score(A), b = score(B);
        return a > b ? 1 : (a < b ? 2 : 0);
    }
};