LeetCode in Kotlin

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:

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

Return

Example 1:

Input: player1 = [4,10,7,9], player2 = [6,5,2,3]

Output: 1

Explanation: The score of player1 is 4 + 10 + 27 + 29 = 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 + 210 + 22 = 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:

Solution

class Solution {
    fun isWinner(player1: IntArray, player2: IntArray): Int {
        var p1Score = 0
        var p2Score = 0
        var isTen = 0
        for (score in player1) {
            p1Score += if (isTen > 0) 2 * score else score
            if (isTen > 0) isTen--
            if (score == 10) isTen = 2
        }
        isTen = 0
        for (score in player2) {
            p2Score += if (isTen > 0) 2 * score else score
            if (isTen > 0) isTen--
            if (score == 10) isTen = 2
        }
        return when {
            p1Score == p2Score -> 0
            p1Score > p2Score -> 1
            else -> 2
        }
    }
}