The Earliest and Latest Rounds Where Players Compete

Hard

Asked by:

13 views

Topics:

Dynamic ProgrammingRecursion

There is a tournament where n players are participating. The players are standing in a single row and are numbered from 1 to n based on their initial standing position (player 1 is the first player in the row, player 2 is the second player in the row, etc.).

The tournament consists of multiple rounds (starting from round number 1). In each round, the i^th player from the front of the row competes against the i^th player from the end of the row, and the winner advances to the next round. When the number of players is odd for the current round, the player in the middle automatically advances to the next round.

For example, if the row consists of players 1, 2, 4, 6, 7
- Player 1 competes against player 7.
- Player 2 competes against player 6.
- Player 4 automatically advances to the next round.

After each round is over, the winners are lined back up in the row based on the original ordering assigned to them initially (ascending order).

The players numbered firstPlayer and secondPlayer are the best in the tournament. They can win against any other player before they compete against each other. If any two other players compete against each other, either of them might win, and thus you may choose the outcome of this round.

Given the integers n, firstPlayer, and secondPlayer, return an integer array containing two values, the earliest possible round number and the latest possible round number in which these two players will compete against each other, respectively.

Example 1:

Input: n = 11, firstPlayer = 2, secondPlayer = 4
Output: [3,4]
Explanation:
One possible scenario which leads to the earliest round number:
First round: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11
Second round: 2, 3, 4, 5, 6, 11
Third round: 2, 3, 4
One possible scenario which leads to the latest round number:
First round: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11
Second round: 1, 2, 3, 4, 5, 6
Third round: 1, 2, 4
Fourth round: 2, 4

Example 2:

Input: n = 5, firstPlayer = 1, secondPlayer = 5
Output: [1,1]
Explanation: The players numbered 1 and 5 compete in the first round.
There is no way to make them compete in any other round.

Constraints:

2 <= n <= 28
1 <= firstPlayer < secondPlayer <= n

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on 'n', the total number of players, and 'firstPlayer' and 'secondPlayer'? Specifically, what are the minimum and maximum possible values?
If it's impossible for the two players to ever compete (e.g., due to an invalid input), what should the function return? Should I return null, an empty array, or some other specific indicator?
Are 'firstPlayer' and 'secondPlayer' 1-indexed or 0-indexed?
Is it guaranteed that 'firstPlayer' and 'secondPlayer' are distinct?
Can you provide an example input with a larger number of players to help me understand the tournament structure more clearly?

Brute Force Solution

Approach

The brute force method for this problem involves simulating every possible tournament scenario where two specific players meet. We'll explore all paths, checking if the two players meet in a given round, and keep track of the earliest and latest rounds where they do.

Here's how the algorithm would work step-by-step:

Start by imagining the first round where all the players are initially positioned.
Simulate a tournament by pairing players off randomly for each round.
For each pairing, check if the two target players are competing against each other.
If the two target players compete against each other in a round, record that round number.
Continue simulating the tournament, advancing to the next round, until only one player remains.
Repeat this entire simulation process many times, each time with different random pairings in each round.
After simulating many tournaments, find the smallest round number that was recorded. This is the earliest possible round the two players could meet.
Also, find the largest round number that was recorded. This is the latest possible round the two players could meet.

Code Implementation

import random

def earliest_and_latest_rounds_brute_force(number_of_players, first_player, second_player):
    earliest_round = number_of_players
    latest_round = 1
    number_of_simulations = 1000

    for _ in range(number_of_simulations):
        current_players = list(range(1, number_of_players + 1))
        current_round = 0

        while len(current_players) > 1:
            current_round += 1
            next_round_players = []
            random.shuffle(current_players)
            
            # Iterate through the players in pairs to simulate matches
            for i in range(0, len(current_players), 2):
                if i + 1 < len(current_players):
                    player_one = current_players[i]
                    player_two = current_players[i + 1]

                    # Check if the target players are competing
                    if (player_one == first_player and player_two == second_player) or \
                       (player_one == second_player and player_two == first_player):
                        earliest_round = min(earliest_round, current_round)
                        latest_round = max(latest_round, current_round)

                    # Determine the winner (simulate by random choice)
                    winner = random.choice([player_one, player_two])
                    next_round_players.append(winner)

                else:
                    # Handle the case of an odd number of players
                    next_round_players.append(current_players[i])

            current_players = next_round_players

    return [earliest_round, latest_round]

Big(O) Analysis

Time Complexity

O(n!) – The brute force approach simulates tournament scenarios with random pairings. The dominant factor in the time complexity is exploring all possible tournament brackets. In each round, there are many ways to pair up the players. With n players, the number of ways to construct tournament brackets grows factorially with n. The number of simulations directly affect the discovery of earliest and latest rounds and depends on how many times the random tournament generation can be done in a reasonable timeframe. Thus the overall complexity becomes O(n!).

Space Complexity

O(N) – The brute force solution, as described, simulates tournament scenarios, and it requires a data structure to represent the current state of the tournament round. This state involves storing which players are still in the competition, essentially requiring an array or list of size N in each round. Since the simulation repeats multiple times, but only one tournament state is stored at a time, the dominant space usage is from this player list representing the current round's state. Therefore, the auxiliary space complexity is O(N).

Optimal Solution

Approach

The problem asks for the earliest and latest possible round numbers where two specific players could meet in a tournament. The core idea is to separately compute the smallest and largest possible round numbers by simulating the tournament under optimal scenarios for each case.

Here's how the algorithm would work step-by-step:

To find the earliest round, imagine both players advancing as quickly as possible. Figure out how many players must be eliminated before they face each other.
Keep dividing the number of remaining players by 2 to simulate each round of the tournament until the two target players are in the same match.
Count how many divisions (rounds) it took. That number is the earliest possible round they can meet.
To find the latest round, imagine both players avoiding each other for as long as possible. One or both players might lose before ever meeting.
Start by figuring out the maximum possible rounds in the entire tournament. This occurs if only one player wins repeatedly until no one is left
Now, simulate the tournament, but ensure that until they absolutely must be in the same bracket to advance, they are in separate brackets.
If a player loses before meeting the other player, their meeting round is not possible. In such a case, the latest round that they CAN meet will occur only during elimination rounds
Calculate how many rounds it took for them to meet or for one of them to lose, that is the latest possible round. If one of the players loses before they can meet, the latest round will be the maximum possible rounds of the tournament.

Code Implementation

def the_earliest_and_latest_rounds_where_players_compete(number_of_players, first_player, second_player):
    def find_earliest_round():
        remaining_players = number_of_players
        round_number = 1
        
        # Simulate until they are in the same half.
        while (first_player + 1) // 2 != (second_player + 1) // 2:
            first_player = (first_player + 1) // 2
            second_player = (second_player + 1) // 2
            remaining_players = (remaining_players + 1) // 2
            round_number += 1
        
        return round_number

    def find_latest_round():
        # Calculate the max rounds possible
        max_rounds = 0
        temp_players = number_of_players
        while temp_players > 1:
            temp_players = (temp_players + 1) // 2
            max_rounds += 1
        
        round_number = 0
        first_player_alive = True
        second_player_alive = True
        current_players = number_of_players

        # Simulate rounds until players meet or one loses.
        while first_player_alive and second_player_alive and first_player != second_player:
            round_number += 1

            # If they are in different brackets, advance them.
            if (first_player + 1) // 2 != (second_player + 1) // 2:
                first_player = (first_player + 1) // 2
                second_player = (second_player + 1) // 2
                current_players = (current_players + 1) // 2

            # If they are in the same bracket, they compete.
            else:
                break

            # Simulate potential loss for players to see if a meeting is even possible.
            if current_players % 2 == 1 and current_players > 1:
                if first_player > (current_players + 1) // 2:
                    first_player_alive = False
                if second_player > (current_players + 1) // 2:
                    second_player_alive = False

        # Return max rounds if players never meet
        if not first_player_alive or not second_player_alive:
            return max_rounds

        return round_number

    earliest_round = find_earliest_round()
    latest_round = find_latest_round()
    return [earliest_round, latest_round]

Big(O) Analysis

Time Complexity

O(log n) – The algorithm determines the earliest and latest rounds where two players can meet in a tournament of n players. To find the earliest round, the algorithm simulates the tournament until the players meet, which involves repeatedly dividing the number of remaining players by 2 in each round. Similarly, the latest round is determined by simulating the tournament, which also involves repeatedly dividing the number of players by 2. Therefore, the number of rounds processed is proportional to the logarithm base 2 of n. This results in a time complexity of O(log n).

Space Complexity

O(1) – The algorithm calculates the earliest and latest rounds without using any auxiliary data structures that scale with the input size N (the total number of players). It primarily uses variables to track round numbers and player positions which take up constant space. No temporary lists, hash maps, or recursion are utilized. Therefore, the auxiliary space complexity is O(1).

Edge Cases

Case	How to Handle
n is even and k is greater than n/2	Return [-1, -1] since it is impossible for player 1 to reach round k when k > n/2 and n is even.
n is odd and k is greater than (n+1)/2	Return [-1, -1] since it is impossible for player 1 to reach round k when k > (n+1)/2 and n is odd.
n = 1 and k = 1	Return [1, 1] since player 1 wins immediately in the first round.
k = 1	The earliest and latest rounds are both 1, so return [1,1].
k = n	Only possible if n is a power of 2; return [n, n].
Integer overflow during calculation of rounds for large n	Use long data types to prevent integer overflow during calculations.
n is not a power of 2 for calculating the latest round	The formula for the latest round calculation involves bit manipulation to find the largest power of 2 smaller than n which might not be immediately obvious from the problem statement.
n is 2, k is 2	Return [2,2] since the earliest and latest round is the same and equal to n.

n is even and k is greater than n/2

How to Handle:

Return [-1, -1] since it is impossible for player 1 to reach round k when k > n/2 and n is even.

n is odd and k is greater than (n+1)/2

How to Handle:

Return [-1, -1] since it is impossible for player 1 to reach round k when k > (n+1)/2 and n is odd.

n = 1 and k = 1

How to Handle:

Return [1, 1] since player 1 wins immediately in the first round.

k = 1

How to Handle:

The earliest and latest rounds are both 1, so return [1,1].

k = n

How to Handle:

Only possible if n is a power of 2; return [n, n].

Integer overflow during calculation of rounds for large n

How to Handle:

Use long data types to prevent integer overflow during calculations.

n is not a power of 2 for calculating the latest round

How to Handle:

The formula for the latest round calculation involves bit manipulation to find the largest power of 2 smaller than n which might not be immediately obvious from the problem statement.

n is 2, k is 2

How to Handle:

Return [2,2] since the earliest and latest round is the same and equal to n.