#1900

Hard

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

The Earliest and Latest Rounds Where Players Compete

This problem requires finding the earliest and latest rounds where two players compete using dynamic programming with state transitions.

dynamic programming memoization

Problem Statement

In a tournament with n players, players are initially arranged in a single row and numbered from 1 to n. In each round, the ith player from the front competes against the ith player from the back. If the number of players is odd, the middle player automatically advances. After each round, the winners are re-seated in their initial order, and the next round continues with the winners.

Given the positions of two players, firstPlayer and secondPlayer, the task is to determine both the earliest and the latest rounds in which they will compete. The solution should model each round using dynamic programming and memoization to track the progression of players through the rounds.

Examples

Example 1

Input: n = 11, firstPlayer = 2, secondPlayer = 4

Output: [3,4]

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]

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

Solution Approach

State Transition Dynamic Programming

This problem can be solved efficiently using dynamic programming, where you simulate the tournament rounds. For each round, you compute the matchups using bitmasks to represent the state of the players remaining in the tournament. Transition between states to determine the round in which the two players will meet.

Memoization to Avoid Redundant Calculations

Memoization is used to store results of subproblems (specific matchups) and prevent redundant calculations. This significantly speeds up the solution, especially when simulating multiple rounds with a large number of players.

Brute Force Simulation with Bitmasks

A brute force approach using bitmasks can be applied to simulate each round. By representing each player's presence with a bitmask, you can efficiently track the players that remain after each round. The bitmask approach makes it possible to simulate multiple rounds and track the players' progress.

Complexity Analysis

Metric	Value
Time	O(n^4 \log n)
Space	O(n^2 \log n)

The time complexity is O(n^4 log n) due to the need to simulate each round and perform bitmask-based operations. Space complexity is O(n^2 log n) to store the state transitions and intermediate results during simulation.

What Interviewers Usually Probe

Strong understanding of dynamic programming and state transitions is required.
Ability to apply memoization to optimize the solution and avoid recalculating results.
Comfortable with bitmasking and simulating multiple rounds efficiently.

Common Pitfalls or Variants

Common pitfalls

Misunderstanding how to simulate the rounds efficiently with dynamic programming.
Failing to memoize intermediate results, leading to unnecessary recalculations.
Overcomplicating the solution when a brute force bitmasking approach can be applied.

Follow-up variants

Implementing the solution without dynamic programming for simplicity, which may lead to high time complexity.
Solving the problem using a simulation-based approach without bitmasking.
Optimizing the solution by considering fewer rounds or limiting unnecessary simulation steps.

FAQ

How does dynamic programming help in solving the Earliest and Latest Rounds Where Players Compete problem?

Dynamic programming allows for efficient simulation of the rounds by breaking down the problem into smaller subproblems and solving each one, saving time by avoiding redundant computations.

What is bitmasking and how does it apply to this problem?

Bitmasking is a technique used to represent sets of players. Each bit in a bitmask indicates whether a player is still in the game. This technique is crucial for efficiently tracking the progression of players through each round.

How can I efficiently simulate the rounds in this problem?

By using bitmasks to represent the set of players in each round, you can simulate the rounds and track which players advance, allowing you to determine when the two players meet.

What is memoization and why is it important for this problem?

Memoization stores the results of subproblems to avoid redundant calculations. In this problem, it helps optimize the simulation of rounds by storing previously computed matchups, leading to faster execution.

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

The time complexity is O(n^4 log n), which comes from simulating multiple rounds with bitmasking and state transitions. The memoization technique helps to manage the large number of possible states.

terminal

Solution

Solution 1: Memoization + Binary Enumeration

We define a function $\text{dfs}(l, r, n)$, which represents the earliest and latest rounds where players numbered $l$ and $r$ compete among $n$ players in the current round.

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

31

32

33

34

35

36

@cache
def dfs(l: int, r: int, n: int):
    if l + r == n - 1:
        return [1, 1]
    res = [inf, -inf]
    m = n >> 1
    for i in range(1 << m):
        win = [False] * n
        for j in range(m):
            if i >> j & 1:
                win[j] = True
            else:
                win[n - 1 - j] = True
        if n & 1:
            win[m] = True
        win[n - 1 - l] = win[n - 1 - r] = False
        win[l] = win[r] = True
        a = b = c = 0
        for j in range(n):
            if j == l:
                a = c
            if j == r:
                b = c
            if win[j]:
                c += 1
        x, y = dfs(a, b, c)
        res[0] = min(res[0], x + 1)
        res[1] = max(res[1], y + 1)
    return res


class Solution:
    def earliestAndLatest(
        self, n: int, firstPlayer: int, secondPlayer: int
    ) -> List[int]:
        return dfs(firstPlayer - 1, secondPlayer - 1, n)

Continue Practicing

#1857 largest color value in a directed graph #1815 maximum number of groups getting fresh donuts #2019 the score of students solving math expression