The Earliest and Latest Rounds Where Players Compete

The Earliest and Latest Rounds Where Players Compete - Problem

Tournament Bracket Simulation

Imagine a thrilling elimination tournament with n players standing in a row, numbered 1 to n based on their initial positions. This isn't your typical bracket tournament - it has a unique twist!

How the Tournament Works:
• In each round, the i-th player from the front competes against the i-th player from the back
• If there's an odd number of players, the middle player automatically advances
• Winners are re-arranged in their original numerical order for the next round

Special Players: Two players (firstPlayer and secondPlayer) are tournament champions who can beat anyone except each other. When other players compete, either could win (your choice!).

Your Mission: Determine the earliest and latest possible rounds where these two champions will face each other.

Example: With players [1,2,4,6,7], player 1 fights player 7, player 2 fights player 6, and player 4 advances automatically.

Input & Output

example_1.py — Basic Tournament

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

› Output: [3, 4]

💡 Note: With 11 players, players 2 and 4 can meet as early as round 3 (if other eliminations position them optimally) or as late as round 4 (worst case positioning). The tournament structure and elimination patterns determine these bounds.

example_2.py — Immediate Competition

$ Input: n = 5, firstPlayer = 1, secondPlayer = 5

› Output: [1, 1]

💡 Note: Players 1 and 5 are at opposite ends of the 5-player lineup, so they compete directly in round 1. The middle player (3) advances automatically. Result: they meet in round 1 in all scenarios.

example_3.py — Maximum Delay

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

› Output: [2, 3]

💡 Note: With 6 players [1,2,3,4,5,6], round 1 matches are: 1vs6, 2vs5, 3vs4. Players 2 and 5 could meet immediately (round 1), but since they're champions, one wins. Actually, they meet in round 2 at earliest, round 3 at latest depending on other eliminations.

Constraints

2 ≤ n ≤ 1000
1 ≤ firstPlayer < secondPlayer ≤ n
The tournament structure guarantees champions will eventually meet
firstPlayer and secondPlayer are guaranteed to be different

Visualization

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Understanding the Visualization

Initial Setup

All players lined up, champions identified at positions 2 and 4

Round 1 Matches

Players pair up: 1st vs last, 2nd vs 2nd-last, etc.

Position Tracking

After eliminations, track new positions of champions

Recursive Solution

Continue until champions meet, exploring all possibilities

Key Takeaway

🎯 Key Insight: By tracking only champion positions and using memoization, we transform an exponential problem into a polynomial one, efficiently finding when these tournament legends will clash!

Asked in

G Google 25 f Meta 18 a Amazon 15 ⊞ Microsoft 12

This tournament simulation problem is optimally solved using dynamic programming with memoization. The key insight is to track only the positions of the two champion players rather than simulating the entire tournament. By memoizing results for each state (pos1, pos2, total_players), we reduce the time complexity from exponential O(2^n) to polynomial O(n³). The algorithm recursively explores all possible elimination scenarios while caching results to avoid redundant calculations.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force Recursion	O(2^n)	O(n log n)	Recursively simulate all possible tournament outcomes
Dynamic Programming with Memoization (Optimal)	O(n^3)	O(n^3)	Use memoized recursion tracking only champion positions to avoid redundant calculations

Brute Force Recursion — Algorithm Steps

Start with initial player positions and round 1
For each round, identify who fights whom
If our champions fight each other, record the round number
Otherwise, try all possible outcomes for other fights
Recursively continue to next round with winners
Track minimum and maximum rounds where champions meet

Visualization

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Step-by-Step Walkthrough

Initial Setup

Start with all players [1,2,3,4,5] and our champions at positions 2 and 4

Round 1 Matches

Player 1 vs 5, Player 2 vs 4 (champions meet!), Player 3 advances

Explore All Paths

For non-champion matches, try both possible winners

Track Results

Record when champions meet in each path

Code -

solution.c — C

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

int compare(const void *a, const void *b) {
    return (*(int*)a - *(int*)b);
}

void generateOutcomes(int *matches, int matchCount, int idx, int *winners, int winnerCount,
                     int *players, int playerCount, int round, int first, int second,
                     int *minRound, int *maxRound);

void simulate(int *players, int playerCount, int round, int first, int second,
             int *minRound, int *maxRound) {
    // Check if champions compete this round
    for (int i = 0; i < playerCount / 2; i++) {
        int left = players[i];
        int right = players[playerCount - 1 - i];
        if ((left == first && right == second) || (left == second && right == first)) {
            if (round < *minRound) *minRound = round;
            if (round > *maxRound) *maxRound = round;
            return;
        }
    }
    
    // Generate matches (stored as pairs in flat array)
    int *matches = malloc(playerCount * sizeof(int));
    int matchCount = 0;
    for (int i = 0; i < playerCount / 2; i++) {
        matches[matchCount++] = players[i];
        matches[matchCount++] = players[playerCount - 1 - i];
    }
    
    int *winners = malloc(playerCount * sizeof(int));
    generateOutcomes(matches, matchCount / 2, 0, winners, 0, players, playerCount,
                    round, first, second, minRound, maxRound);
    
    free(matches);
    free(winners);
}

void generateOutcomes(int *matches, int matchCount, int idx, int *winners, int winnerCount,
                     int *players, int playerCount, int round, int first, int second,
                     int *minRound, int *maxRound) {
    if (idx == matchCount) {
        if (playerCount % 2 == 1) {
            winners[winnerCount++] = players[playerCount / 2];
        }
        qsort(winners, winnerCount, sizeof(int), compare);
        simulate(winners, winnerCount, round + 1, first, second, minRound, maxRound);
        return;
    }
    
    int left = matches[idx * 2];
    int right = matches[idx * 2 + 1];
    
    int candidates[2];
    int candidateCount = 0;
    
    if (left == first || left == second) {
        candidates[candidateCount++] = left;
    } else if (right == first || right == second) {
        candidates[candidateCount++] = right;
    } else {
        candidates[candidateCount++] = left;
        candidates[candidateCount++] = right;
    }
    
    for (int i = 0; i < candidateCount; i++) {
        winners[winnerCount] = candidates[i];
        generateOutcomes(matches, matchCount, idx + 1, winners, winnerCount + 1,
                        players, playerCount, round, first, second, minRound, maxRound);
    }
}

int* earliestAndLatest(int n, int firstPlayer, int secondPlayer, int* returnSize) {
    int *players = malloc(n * sizeof(int));
    for (int i = 0; i < n; i++) {
        players[i] = i + 1;
    }
    
    int minRound = INT_MAX, maxRound = 0;
    simulate(players, n, 1, firstPlayer, secondPlayer, &minRound, &maxRound);
    
    int *result = malloc(2 * sizeof(int));
    result[0] = minRound;
    result[1] = maxRound;
    *returnSize = 2;
    
    free(players);
    return result;
}

Time & Space Complexity

Time Complexity

⏱️

O(2^n)

In worst case, we explore 2^k possibilities where k is the number of fights between non-champions across all rounds

⚠ Quadratic Growth

Space Complexity

O(n log n)

Recursion stack depth is proportional to number of rounds, which is O(log n), and each call stores O(n) player information

⚡ Linearithmic Space

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force Recursion — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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