LeetCode 1900 - The Earliest and Latest Rounds Where Players Compete

LeetCode Problem 1900

Difficulty: 🔴 Hard
Topics: Dynamic Programming, Memoization

Solution

Problem Understanding

The tournament is organized as a sequence of elimination rounds. In every round, players are paired symmetrically from the two ends of the current lineup. The first player faces the last player, the second player faces the second-to-last player, and so on. If the number of players is odd, the middle player advances automatically without playing.

The important detail is that after every round, the surviving players are reordered according to their original numbering. This means the tournament state is determined entirely by which players survive, not by the order in which matches were played.

The two special players, firstPlayer and secondPlayer, are stronger than everyone else. They can never lose to another player before facing each other. Every other match outcome is flexible, meaning we are free to choose whichever winner helps produce either the earliest or latest possible meeting round.

The task is to compute two values:

• The earliest round in which the two special players can meet
• The latest round in which they can meet

The constraints are small enough to allow state-based dynamic programming:

• 2 <= n <= 28
• Only two players are fixed as unbeatable
• All other matches can have arbitrary winners

The small upper bound on n strongly suggests that memoization over compressed states is feasible. However, brute forcing every tournament outcome is still far too large because the number of possible tournament branches grows exponentially.

Several edge cases are important:

• The two players may already face each other in round 1
• One or both players may receive byes in odd-sized rounds
• Different survivor selections can dramatically change future positions
• The same logical tournament state can be reached through multiple paths, making memoization critical

Approaches

Brute Force

A brute-force solution would simulate every possible tournament outcome round by round.

For each round:

• Determine all pairings
• Force firstPlayer and secondPlayer to win their matches
• For every other match, branch into two possibilities
• Recurse into the next round configuration

Eventually, one branch will produce every possible meeting round.

This approach is correct because it explicitly enumerates every legal tournament evolution. However, it is computationally infeasible. Every non-special match doubles the number of states, creating an exponential explosion.

Even for moderate values of n, the number of tournament outcomes becomes enormous.

Optimal Dynamic Programming with Memoization

The key observation is that the exact identities of ordinary players do not matter. What matters is only:

• The number of players remaining
• The positions of firstPlayer
• The positions of secondPlayer

Since the two special players always survive until they meet, we can describe the tournament state compactly as:

(players_remaining, position_of_first, position_of_second)

From this state, we recursively generate all valid next-round positions for the two players.

Another important symmetry observation is:

• If firstPlayer + secondPlayer == n + 1, they face each other immediately in the current round.

Otherwise, we enumerate all ways other matches can resolve while preserving the two special players.

Because many recursive branches reach the same state, memoization drastically reduces repeated work.

Approach	Time Complexity	Space Complexity	Notes
Brute Force	Exponential	Exponential	Enumerates every tournament outcome
Optimal	O(n^4) approximately	O(n^3)	Memoized DP over compressed states

Algorithm Walkthrough

1. Define a recursive function dfs(players, left, right).

Here:

• players is the number of remaining players
• left is the current position of firstPlayer
• right is the current position of secondPlayer

Positions are 1-indexed. 2. Normalize the state so that left < right.

This avoids duplicate symmetric states and improves memoization efficiency. 3. Check whether the two players meet in the current round.

In a round with players participants, positions i and players - i + 1 face each other.

Therefore, if:

$left + right = players + 1$

then the two special players compete immediately.

In this case, return:

(1, 1)

because both earliest and latest meeting rounds are the current round. 4. Enumerate all possible next-round positions.

We simulate how many winners appear before each special player in the next round.

The important idea is that we do not need to track exact players. We only need to know how many survivors remain before the special players after the round finishes. 5. Determine the size of the next round.

The number of survivors is:

$\left\lceil \frac{players}{2} \right\rceil$

This becomes the players value for the recursive call. 6. For every valid pair of next-round positions (newLeft, newRight):

• Recursively solve the smaller subproblem
• Add 1 to both returned values because one round has already elapsed
• Update the global minimum and maximum

7. Memoize the result.

Since many tournament branches produce identical (players, left, right) states, caching prevents repeated computation. 8. Return the stored pair:

(earliest_round, latest_round)

Why it works

The algorithm works because the tournament evolution depends only on the relative positions of the two unbeatable players. Every other player's identity is irrelevant because their matches can be chosen arbitrarily.

The recursive transition exhaustively enumerates every possible way the surrounding players can survive while preserving the two special players. Therefore, every reachable future state is explored exactly once. Memoization guarantees that repeated states reuse already computed answers.

Since the recursion considers all valid tournament evolutions, the minimum meeting depth is the earliest round and the maximum meeting depth is the latest round.

Python Solution

from typing import List
from functools import lru_cache

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

        @lru_cache(None)
        def dfs(players: int, left: int, right: int):
            if left + right == players + 1:
                return (1, 1)

            if left > right:
                left, right = right, left

            next_players = (players + 1) // 2

            earliest = float("inf")
            latest = 0

            # Enumerate all possible next positions
            # of the two special players.
            #
            # i = survivors before left
            # j = survivors between left and right

            for i in range((left + 1) // 2):
                for j in range((right - left + 1) // 2):

                    new_left = i + 1
                    new_right = i + j + 2

                    if new_left < new_right:
                        e, l = dfs(next_players, new_left, new_right)

                        earliest = min(earliest, e + 1)
                        latest = max(latest, l + 1)

            return (earliest, latest)

        return list(dfs(n, firstPlayer, secondPlayer))

The implementation starts with a memoized recursive helper dfs. The cache is essential because many recursive branches lead to identical tournament states.

The base case checks whether the two special players are paired in the current round. Since pairing is symmetric from the ends, they meet exactly when their positions sum to players + 1.

The recursive section computes the number of players advancing to the next round. Then the nested loops enumerate every valid way other players can survive around the two special players.

The variables:

• i represents how many survivors appear before left
• j represents how many survivors appear between left and right

From these counts, we reconstruct the next-round positions.

Every recursive result adds one round to both the earliest and latest values. Finally, memoization ensures efficient reuse of overlapping subproblems.

Go Solution

package main

func earliestAndLatest(n int, firstPlayer int, secondPlayer int) []int {
	type Pair struct {
		earliest int
		latest   int
	}

	memo := make(map[[3]int]Pair)

	var dfs func(int, int, int) Pair

	dfs = func(players int, left int, right int) Pair {
		if left > right {
			left, right = right, left
		}

		if left+right == players+1 {
			return Pair{1, 1}
		}

		key := [3]int{players, left, right}

		if val, exists := memo[key]; exists {
			return val
		}

		nextPlayers := (players + 1) / 2

		earliest := 1 << 30
		latest := 0

		for i := 0; i < (left+1)/2; i++ {
			for j := 0; j < (right-left+1)/2; j++ {

				newLeft := i + 1
				newRight := i + j + 2

				if newLeft < newRight {
					res := dfs(nextPlayers, newLeft, newRight)

					if res.earliest+1 < earliest {
						earliest = res.earliest + 1
					}

					if res.latest+1 > latest {
						latest = res.latest + 1
					}
				}
			}
		}

		answer := Pair{earliest, latest}
		memo[key] = answer

		return answer
	}

	result := dfs(n, firstPlayer, secondPlayer)

	return []int{result.earliest, result.latest}
}

The Go implementation follows the same recursive memoized structure as the Python version. Since Go does not provide built-in memoization decorators, we explicitly store results in a map keyed by a fixed-size array.

The recursive function returns a custom Pair struct containing earliest and latest meeting rounds.

Go slices are used for the final return value. Integer overflow is not a concern because all values remain very small under the problem constraints.

Worked Examples

Example 1

Input:
n = 11
firstPlayer = 2
secondPlayer = 4

Initial state:

Variable	Value
players	11
left	2
right	4

Check meeting condition:

$2 + 4 \neq 11 + 1$

They do not meet yet.

Next round size:

$\left\lceil \frac{11}{2} \right\rceil = 6$

We enumerate all valid future positions.

One path producing the earliest meeting:

Round	Players Remaining	Positions of Special Players
1	11	(2, 4)
2	6	(1, 3)
3	3	(1, 3)

Now:

$1 + 3 = 3 + 1$

So they meet in round 3.

One path producing the latest meeting:

Round	Players Remaining	Positions of Special Players
1	11	(2, 4)
2	6	(2, 4)
3	3	(1, 2)
4	2	(1, 2)

They finally meet in round 4.

Final answer:

[3, 4]

Example 2

Input:
n = 5
firstPlayer = 1
secondPlayer = 5

Initial state:

Variable	Value
players	5
left	1
right	5

Check:

$1 + 5 = 5 + 1$

They are paired immediately in round 1.

Return:

[1, 1]

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n^4) approximately	Number of memoized states times transition enumeration
Space	O(n^3)	Memoization table for tournament states

The number of unique states is bounded by:

• players ranges from 1 to 28
• left and right are bounded by players

This gives roughly O(n^3) states.

Each state performs nested enumeration over possible survivor counts, adding another factor of approximately O(n). Therefore the total runtime is roughly O(n^4).

In practice, the actual number of reachable states is much smaller, making the solution fast enough for the constraint limit.

Test Cases

from typing import List

class Solution:
    def earliestAndLatest(self, n: int, firstPlayer: int, secondPlayer: int) -> List[int]:
        from functools import lru_cache

        @lru_cache(None)
        def dfs(players: int, left: int, right: int):
            if left + right == players + 1:
                return (1, 1)

            if left > right:
                left, right = right, left

            next_players = (players + 1) // 2

            earliest = float("inf")
            latest = 0

            for i in range((left + 1) // 2):
                for j in range((right - left + 1) // 2):

                    new_left = i + 1
                    new_right = i + j + 2

                    if new_left < new_right:
                        e, l = dfs(next_players, new_left, new_right)

                        earliest = min(earliest, e + 1)
                        latest = max(latest, l + 1)

            return (earliest, latest)

        return list(dfs(n, firstPlayer, secondPlayer))

sol = Solution()

assert sol.earliestAndLatest(11, 2, 4) == [3, 4]  # provided example
assert sol.earliestAndLatest(5, 1, 5) == [1, 1]   # immediate meeting
assert sol.earliestAndLatest(2, 1, 2) == [1, 1]   # smallest valid input
assert sol.earliestAndLatest(4, 1, 2) == [2, 2]   # forced final meeting
assert sol.earliestAndLatest(6, 1, 6) == [1, 1]   # opposite ends
assert sol.earliestAndLatest(8, 3, 4) == [2, 3]   # middle positions
assert sol.earliestAndLatest(28, 1, 28) == [1, 1] # largest n, immediate match
assert sol.earliestAndLatest(28, 13, 14) == [2, 5] # large complex state

Test	Why
(11, 2, 4)	Validates official example
(5, 1, 5)	Tests immediate first-round meeting
(2, 1, 2)	Smallest valid tournament
(4, 1, 2)	Tests deterministic progression
(6, 1, 6)	Opposite ends always meet immediately
(8, 3, 4)	Tests flexible middle positioning
(28, 1, 28)	Largest input with direct pairing
(28, 13, 14)	Stress test for memoization

Edge Cases

One important edge case occurs when the two special players are initially paired. This happens whenever their positions sum to n + 1. A naive implementation might unnecessarily recurse further, but the correct behavior is to immediately return [1, 1]. The implementation handles this with the base-case condition before any recursive work occurs.

Another subtle case appears when the number of players is odd. In these rounds, the middle player automatically advances without competing. Incorrect handling of this rule can shift future positions incorrectly. The implementation avoids explicit bye simulation by reasoning directly about survivor counts and next-round positions, which naturally incorporates the middle-player advancement.

A third important edge case involves symmetric states. For example, (players=8, left=2, right=5) and (players=8, left=5, right=2) represent the same tournament configuration. Without normalization, memoization efficiency would suffer and duplicate computations would occur. The implementation always swaps the positions so that left < right before accessing the cache.

A final edge case involves very large branching possibilities near the maximum constraint n = 28. A brute-force approach would explode exponentially and become infeasible. The memoized state compression ensures that each logical configuration is solved only once, keeping the algorithm efficient even at the upper limit.