2660. Determine the Winner of a Bowling Game

EasyArraySimulation
Leetcode Link

Problem Description

In the given problem, we are simulating the scoring process of a bowling game for two players. Each player's progress in the game is recorded in two separate arrays (player1 and player2), which represent the number of pins they knock down in each turn. A complete game consists of n turns, and the maximum number of pins that can be knocked down in a single turn is10.

However, there's a special scoring rule at play. If a player hits all 10 pins (a strike) in a turn, then the score for any of the next two turns is doubled. More specifically, if x_i denotes the number of pins knocked down in turn i, then the score for turn i is:

  • 2 * x_i, if the player scored a strike in either of the two preceding turns (i-1 or i-2).
  • x_i, if the player did not score a strike in the previous two turns.

The total score for a player is the sum of the scores from all turns.

The task is to determine the winner of the game based on their scores; return 1 if player 1 has a higher score, 2 if player 2 has a higher score, and 0 if there is a tie.

Intuition

To solve the problem, we can simulate the process of calculating each player's score turn by turn. We proceed in the following manner:

  1. We write a function f(arr) that calculates the total score for an array representing a player's knocked pins per turn.
  2. During this process, we iterate through the array, and for each turn, we check whether a strike was hit in either of the two previous turns.
  3. If a strike was hit in the previous two turns, we multiply the current turn's pins (x) by 2. Otherwise, we simply add the count of pins.
  4. After we have iterated through all the turns and accumulated the score, we compare the scores of both players.

By utilizing this simple yet effective simulation approach, we can calculate each player's score according to the game's scoring rules. After calculating the scores (a for player 1 and b for player 2) using function f, we then compare scores and return the respective winner: 1 for player 1, 2 for player 2, or 0 in case of a draw.

Solution Approach

The solution implements a function f(arr) which is used to calculate the scores for each player iteratively. The score calculation logic reflects the special rule of the bowling game that states if a player scores a strike (10 pins) in one turn, the score for the next two turns is doubled.

Here is a step-by-step approach to implementing the solution:

  1. Iterate Through Turns:

    • We loop through each turn in the array using enumerate, which gives us both the index (i.e., the turn number) and the score (i.e., the pins knocked down x) in that turn.
  2. Check for Strikes in Previous Turns:

    • For each turn, we check if there was a strike in any of the previous two turns. This is determined using the condition:
      1(i and arr[i - 1] == 10) or (i > 1 and arr[i - 2] == 10)
    • If the current turn index i is non-zero and the score in the immediate previous turn arr[i - 1] equals 10, or if the index is greater than 1 and the score two turns back arr[i - 2] equals 10, we set k to 2; otherwise, k is 1.
  3. Calculate Cumulative Score:

    • We calculate the score for the current turn by multiplying x (the pins knocked down) by k (the factor that may double the score if a strike occurred in one of the two previous turns) and add this to the cumulative score s. This step is repeated for each turn in the array.

    • At the end of iterations, the function returns the total score s.

  4. Compare Players' Scores:

    • We invoke the scoring function f for both player1 and player2, storing the returned values in variables a and b, respectively.
    • Finally, we compare the computed scores a and b to determine the winner. We return:
      • 1 if a > b (player 1 wins),
      • 2 if b > a (player 2 wins), or
      • 0 in case of a draw (equal scores).

This solution takes a straightforward approach to simulating the scoring part of the game, avoiding the complexity of keeping track of strikes and multipliers outside of the immediate iteration. The code structure remains simple, and the logic is encapsulated within a single helper function, emphasizing clarity and maintainability.

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Example Walkthrough

Let's illustrate the solution approach with a small example:

Suppose we have the following pin arrays for two players:

player1 = [4, 10, 5, 1]
player2 = [3, 7, 10, 2]

Now, let's walk through how we would calculate the score for player1 using the function f(arr) as defined in the solution approach:

  1. Turn 1:
    player1 knocks down 4 pins. No previous turn, so score is 4.

  2. Turn 2 (Strike):
    player1 hits a strike with 10 pins. The score for this turn is 10.

  3. Turn 3:
    player1 knocks down 5 pins. There was a strike in the previous turn, so this score is doubled to 10, and the cumulative total now is 24.

  4. Turn 4:
    player1 knocks down 1 pin. There was no strike in the previous turn (turn 3), but there was one in turn 2. Therefore, this score is also doubled to 2, giving us a final total score for player1 of 26.

Following the same process for player2:

  1. Turn 1:
    player2 knocks down 3 pins. The score is 3.

  2. Turn 2:
    player2 knocks down 7 pins. No strike in the previous turn, so the score remains 7, and the cumulative total is 10.

  3. Turn 3 (Strike):
    player2 hits a strike with 10 pins. The score is 10, and the cumulative total is 20.

  4. Turn 4:
    player2 knocks down 2 pins. There was a strike in the previous turn, so this is doubled to 4, giving us a final total score for player2 of 24.

Using the f(arr) function, we calculated the scores for both players: player1 has 26 points and player2 has 24 points. Comparing these scores (a for player1 and b for player2), we determine the winner:

  • player1's score (a) is 26
  • player2's score (b) is 24

Since a > b, player1 is the winner, and therefore the function should return 1.

Solution Implementation

1from typing import List
2
3class Solution:
4    def is_winner(self, player1_scores: List[int], player2_scores: List[int]) -> int:
5        """
6        Determines the winner based on the scores of player1 and player2.
7        Multiplies the score by 2 if the player scored a 10 in either of the two previous attempts.
8      
9        Args:
10        player1_scores (List[int]): List of integers representing player 1's scores.
11        player2_scores (List[int]): List of integers representing player 2's scores.
12      
13        Returns:
14        int: 1 if player 1 wins, 2 if player 2 wins, or 0 for a draw.
15        """
16        # Define a nested function to calculate the total points for a player.
17        def calculate_points(scores: List[int]) -> int:
18            total_points = 0
19            # Iterate through the player's scores to calculate the total points.
20            for i, score in enumerate(scores):
21                # If the current score is preceded by one or two scores of 10, the points for the current score are doubled.
22                multiplier = 2 if (i > 0 and scores[i - 1] == 10) or (i > 1 and scores[i - 2] == 10) else 1
23                total_points += multiplier * score
24            return total_points
25
26        # Calculate total points for both players.
27        player1_total = calculate_points(player1_scores)
28        player2_total = calculate_points(player2_scores)
29      
30        # Determine the winner based on the total points calculated.
31        if player1_total > player2_total:
32            return 1   # Player 1 wins
33        if player2_total > player1_total:
34            return 2   # Player 2 wins
35        return 0       # Draw
36
37# Example usage:
38# solution = Solution()
39# result = solution.is_winner([10, 2, 6], [5, 10, 10])
40# print(result) # The output would be either 1, 2, or 0 based on the scores.
41
1class Solution {
2    // Determines the winning player based on scores
3    public int isWinner(int[] player1Scores, int[] player2Scores) {
4        // Calculate the scores for both players
5        int player1Score = calculateScore(player1Scores);
6        int player2Score = calculateScore(player2Scores);
7      
8        // Return 1 if player 1 wins, 2 if player 2 wins, or 0 for a tie
9        if (player1Score > player2Score) {
10            return 1;
11        } else if (player2Score > player1Score) {
12            return 2;
13        } else {
14            return 0;
15        }
16    }
17
18    // Calculates the score for a player based on the scoring rules
19    private int calculateScore(int[] scores) {
20        int totalScore = 0;
21      
22        // Iterate through the scores array
23        for (int i = 0; i < scores.length; ++i) {
24            // Determine the multiplier based on the previous scores
25            int multiplier = 1;
26            if ((i > 0 && scores[i - 1] == 10) || (i > 1 && scores[i - 2] == 10)) {
27                multiplier = 2;
28            }
29
30            // Update the total score
31            totalScore += multiplier * scores[i];
32        }
33      
34        return totalScore;
35    }
36}
37
1class Solution {
2public:
3    // Method to determine the winner based on the scores of both players
4    int isWinner(vector<int>& player1, vector<int>& player2) {
5        // Calculate the scores of player1 and player2 using the custom scoring fucntion
6        int scorePlayer1 = calculateScore(player1);
7        int scorePlayer2 = calculateScore(player2);
8
9        // Compare scores and return the result indicating the winner or a tie
10        if (scorePlayer1 > scorePlayer2) {
11            return 1; // Player 1 wins
12        } else if (scorePlayer2 > scorePlayer1) {
13            return 2; // Player 2 wins
14        } else {
15            return 0; // It's a tie
16        }
17    }
18
19    // Helper method to calculate the score for a player's array of scores ('arr')
20    int calculateScore(vector<int>& arr) {
21        int scoreSum = 0; // Initialize total score sum to 0
22        // Iterate over all elements in the score array
23        for (int i = 0; i < arr.size(); ++i) {
24            // Determine the multiplier based on previous scores
25            int multiplier = 1; // Default multiplier
26            if ((i > 0 && arr[i - 1] == 10) || (i > 1 && arr[i - 2] == 10)) {
27                // If the previous one or two scores were 10, set multiplier to 2
28                multiplier = 2;
29            }
30            // Calculate score for the current position and add to the total score
31            scoreSum += multiplier * arr[i];
32        }
33        // Return the total score for the player
34        return scoreSum;
35    }
36};
37
1function isWinner(player1: number[], player2: number[]): number {
2    // Function to calculate total score for a player
3    // It doubles the score for the current turn if the previous turn or the one before that was a perfect score.
4    const calculateScore = (scores: number[]): number => {
5        let totalScore = 0;
6
7        // Iterate over the scores to calculate the total
8        for (let turn = 0; turn < scores.length; ++turn) {
9            totalScore += scores[turn];
10
11            // Check the previous or the one before the previous score for a perfect 10
12            // If found, add the current score again as a bonus
13            if ((turn && scores[turn - 1] === 10) || (turn > 1 && scores[turn - 2] === 10)) {
14                totalScore += scores[turn];
15            }
16        }
17        return totalScore;
18    };
19
20    // Calculate total scores for both players
21    const scorePlayer1 = calculateScore(player1);
22    const scorePlayer2 = calculateScore(player2);
23
24    // Determine the winner
25    // If player1's score is higher, return 1; if player2's score is higher, return 2; if it's a tie, return 0.
26    return scorePlayer1 > scorePlayer2 ? 1 : scorePlayer1 < scorePlayer2 ? 2 : 0;
27}
28

Time and Space Complexity

The time complexity of the given code is O(n), where n is the length of the array player1 or player2. This is because the function f(arr: List[int]) -> int iterates through each element of the input array exactly once.

The space complexity of the code is O(1) as it uses a fixed amount of space. The variables s, a, b, along with a few others, do not depend on the size of the input and are only used to store single values or perform basic arithmetic operations no matter how large the input array is.

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