The Earliest and Latest Rounds Where Players Compete - Problem

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.

Input & Output

Example 1 — Basic Tournament

$ Input: n = 5, firstPlayer = 2, secondPlayer = 4

› Output: [1,1]

💡 Note: Round 1: [1,2,3,4,5] → pairs (1,5), (2,4), 3 advances. Players 2 and 4 are paired directly in round 1, so they meet immediately.

Example 2 — Adjacent Players

$ Input: n = 6, firstPlayer = 2, secondPlayer = 5

› Output: [1,1]

💡 Note: Round 1: [1,2,3,4,5,6] → pairs (1,6), (2,5), (3,4). Players 2 and 5 are paired directly in round 1, so they meet immediately.

Example 3 — Early Meeting

$ Input: n = 4, firstPlayer = 1, secondPlayer = 4

› Output: [1,1]

💡 Note: Round 1: [1,2,3,4] → pairs (1,4), (2,3). Players 1 and 4 are directly paired in round 1, so earliest and latest both equal 1.

Constraints

3 ≤ n ≤ 50
1 ≤ firstPlayer, secondPlayer ≤ n
firstPlayer ≠ secondPlayer

Visualization

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Asked in

G Google 15 f Meta 12 a Amazon 8

The key insight is that we can control all non-target player matches to either accelerate or delay when our two target players meet. Best approach uses memoization to cache tournament states and avoid redundant calculations. Time: O(2^k × S), Space: O(S × n) where k is rounds and S is unique states.

Common Approaches

✓ Brute Force Recursion

⏱️ Time: O(2ⁿ) Space: O(n)

Simulate every possible tournament progression by trying all combinations of match winners. For each round, determine who advances and recursively check when our target players meet.

Memoization with State Caching

⏱️ Time: O(2ᵏ × S) Space: O(S × n)

Same recursive approach but cache results based on current player set and round number. This eliminates redundant computations when the same player configuration is reached through different paths.

Brute Force Recursion — Algorithm Steps

Simulate tournament round by round
Try all possible winners for non-target player matches
Track earliest and latest rounds where target players compete

Visualization

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Step-by-Step Walkthrough

Current Round

Check if target players compete this round

Try Outcomes

For other matches, try both possible winners

Recurse

Continue simulation with winners from this round

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <limits.h>

typedef struct {
    int early;
    int late;
} Result;

Result simulate(int* players, int size, int roundNum, int first, int second) {
    Result result;
    if (size == 1) {
        result.early = INT_MAX;
        result.late = -1;
        return result;
    }
    
    int pos1 = -1, pos2 = -1;
    for (int i = 0; i < size; i++) {
        if (players[i] == first) pos1 = i;
        else if (players[i] == second) pos2 = i;
    }
    
    if (pos1 != -1 && pos2 != -1 && pos1 + pos2 == size - 1) {
        result.early = roundNum;
        result.late = roundNum;
        return result;
    }
    
    int* nextPlayers = malloc(size * sizeof(int));
    int nextSize = 0;
    
    for (int i = 0; i < size / 2; i++) {
        int left = players[i];
        int right = players[size - 1 - i];
        
        if (left == first || left == second) {
            nextPlayers[nextSize++] = left;
        } else if (right == first || right == second) {
            nextPlayers[nextSize++] = right;
        } else {
            // For simplicity, just choose left player
            nextPlayers[nextSize++] = left;
        }
    }
    
    if (size % 2 == 1) {
        nextPlayers[nextSize++] = players[size / 2];
    }
    
    // Sort nextPlayers
    for (int i = 0; i < nextSize - 1; i++) {
        for (int j = i + 1; j < nextSize; j++) {
            if (nextPlayers[i] > nextPlayers[j]) {
                int temp = nextPlayers[i];
                nextPlayers[i] = nextPlayers[j];
                nextPlayers[j] = temp;
            }
        }
    }
    
    result = simulate(nextPlayers, nextSize, roundNum + 1, first, second);
    free(nextPlayers);
    return result;
}

int* solution(int n, int firstPlayer, int secondPlayer) {
    int* players = malloc(n * sizeof(int));
    for (int i = 0; i < n; i++) {
        players[i] = i + 1;
    }
    
    Result result = simulate(players, n, 1, firstPlayer, secondPlayer);
    free(players);
    
    int* answer = malloc(2 * sizeof(int));
    answer[0] = result.early;
    answer[1] = result.late;
    return answer;
}

int main() {
    int n, firstPlayer, secondPlayer;
    scanf("%d %d %d", &n, &firstPlayer, &secondPlayer);
    
    int* result = solution(n, firstPlayer, secondPlayer);
    printf("[%d,%d]\n", result[0], result[1]);
    
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(2ⁿ)

Each non-target match has 2 possible outcomes, leading to exponential combinations

✓ Linear Growth

Space Complexity

O(n)

Recursion stack depth proportional to number of rounds

⚡ Linearithmic Space

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The Earliest and Latest Rounds Where Players Compete - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force Recursion — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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