How do you solve Determine the Winner of a Bowling Game on LeetCode?

Determine the Winner of a Bowling Game (LeetCode #2660) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Understanding how previous scores affect current scores is crucial.

What is the time complexity of Determine the Winner of a Bowling Game?

The optimal solution for Determine the Winner of a Bowling Game runs in O(n) time and uses O(1) space. The complexity is O(n) because we are only iterating through the players' scores once, checking conditions in constant time.

What is the brute force solution for Determine the Winner of a Bowling Game?

The brute force approach for Determine the Winner of a Bowling Game is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we will iterate through each player's scores and calculate their total score based on the rules provided. We will check the previous two turns to see if they hit 10 pins, which will affect the current score. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Determine the Winner of a Bowling Game?

Determine the Winner of a Bowling Game uses the following concepts: Array, Simulation. The recommended patterns to study are: Simulation, Array.

What are the interview tips for Determine the Winner of a Bowling Game?

Clarify the rules before starting to code. Think about edge cases, like when all scores are zeros. Explain your thought process as you code.

What are common mistakes when solving Determine the Winner of a Bowling Game?

Not checking the previous two scores correctly. Overcomplicating the score calculation.

#2660

Determine the Winner of a Bowling Game

Easy

ArraySimulationSimulationArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

This approach uses a single loop to calculate the scores for both players in one pass, checking the last two turns for each player to determine if the current score should be doubled. This is more efficient than the brute force method.

⚙️

Algorithm

5 steps

1Step 1: Initialize scores for both players to 0.
2Step 2: Loop through each turn for both players.
3Step 3: For each turn, check the previous two turns to see if either hit 10 pins. If so, double the current turn's score; otherwise, use the normal score.
4Step 4: Sum the scores for both players.
5Step 5: Compare the scores and return 1 if player 1 wins, 2 if player 2 wins, or 0 if it's a draw.

solution.py26 lines

1# Full working Python code
2
3def determine_winner(player1, player2):
4    score1, score2 = 0, 0
5    n = len(player1)
6    for i in range(n):
7        if i > 1 and (player1[i-1] == 10 or player1[i-2] == 10):
8            score1 += 2 * player1[i]
9        elif i > 0 and player1[i-1] == 10:
10            score1 += 2 * player1[i]
11        else:
12            score1 += player1[i]
13
14        if i > 1 and (player2[i-1] == 10 or player2[i-2] == 10):
15            score2 += 2 * player2[i]
16        elif i > 0 and player2[i-1] == 10:
17            score2 += 2 * player2[i]
18        else:
19            score2 += player2[i]
20
21    if score1 > score2:
22        return 1
23    elif score2 > score1:
24        return 2
25    else:
26        return 0

ℹ

Complexity note: The complexity is O(n) because we are only iterating through the players' scores once, checking conditions in constant time.

1Understanding how previous scores affect current scores is crucial.
2Simulating the game turn by turn helps clarify the scoring rules.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.