What is the time complexity of Find The First Player to win K Games in a Row?

The optimal solution for Find The First Player to win K Games in a Row runs in O(n²) time and uses O(1) space. The time complexity is O(n²) because in the worst case, we may have to simulate n games for each player, leading to n * n comparisons.

What algorithm or data structure is used to solve Find The First Player to win K Games in a Row?

Find The First Player to win K Games in a Row uses the following concepts: Array, Simulation. The recommended patterns to study are: Queue, Simulation.

What are the interview tips for Find The First Player to win K Games in a Row?

Clearly explain your thought process while coding. Consider edge cases like k >= n early on. Use comments to clarify complex logic.

What are common mistakes when solving Find The First Player to win K Games in a Row?

Failing to handle the case where k >= n. Not using a queue for efficient player management.

Find The First Player to win K Games in a Row — LeetCode #3175 (Medium)

Tags: Array, Simulation

Related patterns: Queue, Simulation

Brute Force approach

Time complexity: O(n²). Space complexity: O(1).

We can simulate the competition by repeatedly comparing the first two players in the queue. The winner stays at the front, and the loser goes to the back. We keep track of how many games the winner has won consecutively until one player wins k games in a row.

The time complexity is O(n²) because in the worst case, we may have to simulate n games for each player, leading to n * n comparisons.

Step 1: Initialize a queue with player indices and a variable to track consecutive wins.
Step 2: While no player has won k games, compare the first two players in the queue.
Step 3: Update the queue based on the winner and loser, and increment the win count for the winner.

1. Initial queue: [0, 1, 2, 3, 4], current winner: -1, consecutive wins: 0 2. Players 0 (skill 4) and 1 (skill 2) play: Winner is 0. Queue: [0, 2, 3, 4, 1], current winner: 0, consecutive wins: 1 3. Players 0 (skill 4) and 2 (skill 6) play: Winner is 2. Queue: [2, 3, 4, 1, 0], current winner: 2, consecutive wins: 1 4. Players 2 (skill 6) and 3 (skill 3) play: Winner is 2. Queue: [2, 4, 1, 0, 3], current winner: 2, consecutive wins: 2 5. Player 2 has won 2 games in a row, return index 2.

Common Mistakes

Failing to handle the case where k >= n.
Not using a queue for efficient player management.

Interview Tips

Clearly explain your thought process while coding.
Consider edge cases like k >= n early on.
Use comments to clarify complex logic.

#3175

Find The First Player to win K Games in a Row

Medium

Array↗Simulation↗Queue↗Simulation↗

LeetCode ↗

Approaches

💡

Intuition

Time O(n²)Space O(1)

⚙️

Algorithm

3 steps

1Step 1: Initialize a queue with player indices and a variable to track consecutive wins.
2Step 2: While no player has won k games, compare the first two players in the queue.
3Step 3: Update the queue based on the winner and loser, and increment the win count for the winner.

solution.py30 lines

1from collections import deque
2
3def first_player_to_win_k_games(skills, k):
4    n = len(skills)
5    queue = deque(range(n))
6    consecutive_wins = 0
7    current_winner = None
8
9    while True:
10        player1 = queue.popleft()
11        player2 = queue.popleft()
12
13        if skills[player1] > skills[player2]:
14            winner = player1
15            loser = player2
16        else:
17            winner = player2
18            loser = player1
19
20        if current_winner == winner:
21            consecutive_wins += 1
22        else:
23            current_winner = winner
24            consecutive_wins = 1
25
26        if consecutive_wins == k:
27            return winner
28
29        queue.append(winner)
30        queue.append(loser)

ℹ

Complexity note: The time complexity is O(n²) because in the worst case, we may have to simulate n games for each player, leading to n * n comparisons.

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