What is the time complexity of The Earliest and Latest Rounds Where Players Compete?

The optimal solution for The Earliest and Latest Rounds Where Players Compete runs in O(log n) time and uses O(1) space. The time complexity is O(log n) because we are effectively halving the number of players in each round, similar to how binary search works.

What is the brute force solution for The Earliest and Latest Rounds Where Players Compete?

The brute force approach for The Earliest and Latest Rounds Where Players Compete is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach simulates each round of the tournament until the two players compete against each other. This method is straightforward but inefficient as it involves checking all players and their outcomes in each round. The optimal Optimal Solution approach improves this to O(log n).

What algorithm or data structure is used to solve The Earliest and Latest Rounds Where Players Compete?

The Earliest and Latest Rounds Where Players Compete uses the following concepts: Dynamic Programming, Memoization. The recommended patterns to study are: Binary Search, Dynamic Programming.

What are the interview tips for The Earliest and Latest Rounds Where Players Compete?

Always clarify the rules of the tournament before diving into coding. Think about edge cases, such as the smallest and largest possible values for n. Practice explaining your thought process as you code, as this can help clarify your understanding.

What are common mistakes when solving The Earliest and Latest Rounds Where Players Compete?

Failing to recognize that players are paired from both ends of the list can lead to incorrect simulations. Not considering the case when the number of players is odd, which allows a middle player to advance automatically.

#1900

The Earliest and Latest Rounds Where Players Compete

Hard

Dynamic ProgrammingMemoizationBinary SearchDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(log n)Space O(1)

The optimal solution leverages the properties of binary representation of player indices to determine the rounds directly. By calculating the rounds based on the positions of the players, we can efficiently find when they will compete without simulating every round.

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Algorithm

3 steps

1Step 1: Calculate the positions of firstPlayer and secondPlayer in binary form.
2Step 2: Determine the earliest round by finding the first round where their positions differ.
3Step 3: Determine the latest round by finding the last round where their positions can still be aligned before they meet.

solution.py18 lines

1def earliestAndLatest(n, firstPlayer, secondPlayer):
2    def getRound(player):
3        round = 0
4        while player > 1:
5            player = (player + 1) // 2
6            round += 1
7        return round
8
9    earliest = 0
10    latest = 0
11
12    while firstPlayer != secondPlayer:
13        earliest += 1
14        firstPlayer = (firstPlayer + 1) // 2
15        secondPlayer = (secondPlayer + 1) // 2
16        latest += 1
17
18    return [earliest, latest]

ℹ

Complexity note: The time complexity is O(log n) because we are effectively halving the number of players in each round, similar to how binary search works.

1Understanding how players are paired in each round is crucial for determining when two specific players will compete.
2Using binary representation helps in efficiently calculating rounds without simulating every match.

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