The Earliest and Latest Rounds Where Players Compete

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: [3,4]

💡 Note: Round 1: [1,2,3,4,5] → pairs (1,5), (2,4), 3 advances. Players 2 and 4 can't meet this round since they're not paired. Best case: they meet in round 3. Worst case: round 4.

Example 2 — Adjacent Players

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

› Output: [2,3]

💡 Note: Round 1: [1,2,3,4,5,6] → pairs (1,6), (2,5), (3,4). Players 2 and 5 are paired immediately, so they meet in round 1. Wait, that's round 1, but answer is [2,3], so they meet later.

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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Understanding the Visualization

Initial Setup

n=5 players [1,2,3,4,5], target players 2 and 4

Round Simulation

Each round pairs front/back players, winners advance

Find Meeting Point

Track earliest/latest rounds where players 2 and 4 compete

Key Takeaway

🎯 Key Insight: We can manipulate all non-target matches to control when our two special players finally compete

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

Approach	Time	Space	Notes
✓ Memoization with State Caching	O(2ᵏ × S)	O(S × n)	Cache results for player configurations to avoid redundant calculations
Brute Force Recursion	O(2ⁿ)	O(n)	Try all possible tournament outcomes recursively without memoization

Memoization with State Caching — Algorithm Steps

Convert player list to hashable state
Check cache before computing
Recursively simulate with memoization
Cache and return results

Visualization

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Step-by-Step Walkthrough

State Key

Create unique key from player set + round

Check Cache

Look up result if state seen before

Compute & Cache

Calculate result and store for future use

Code -

solution.c — C

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

#define MAX_CACHE 10000

typedef struct {
    char key[200];
    int early;
    int late;
    int valid;
} CacheEntry;

CacheEntry cache[MAX_CACHE];
int cacheSize = 0;

int findInCache(const char* key, int* early, int* late) {
    for (int i = 0; i < cacheSize; i++) {
        if (cache[i].valid && strcmp(cache[i].key, key) == 0) {
            *early = cache[i].early;
            *late = cache[i].late;
            return 1;
        }
    }
    return 0;
}

void addToCache(const char* key, int early, int late) {
    if (cacheSize < MAX_CACHE) {
        strcpy(cache[cacheSize].key, key);
        cache[cacheSize].early = early;
        cache[cacheSize].late = late;
        cache[cacheSize].valid = 1;
        cacheSize++;
    }
}

char* playersToKey(int* players, int size, int roundNum) {
    static char key[200];
    sprintf(key, "%d", roundNum);
    strcat(key, "|");
    for (int i = 0; i < size; i++) {
        char temp[20];
        sprintf(temp, "%d,", players[i]);
        strcat(key, temp);
    }
    return key;
}

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

Result simulate(int* players, int size, int roundNum, int first, int second) {
    Result result;
    char* key = playersToKey(players, size, roundNum);
    
    if (findInCache(key, &result.early, &result.late)) {
        return result;
    }
    
    if (size == 1) {
        result.early = INT_MAX;
        result.late = -1;
        addToCache(key, result.early, result.late);
        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;
        addToCache(key, result.early, result.late);
        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 {
            nextPlayers[nextSize++] = left; // Default choice
        }
    }
    
    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);
    addToCache(key, result.early, result.late);
    
    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(2ᵏ × S)

k is max rounds, S is number of unique states - much better than pure brute force

✓ Linear Growth

Space Complexity

O(S × n)

Cache stores S states, each requiring O(n) space for player list

⚡ Linearithmic Space

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Medium Frequency

~35 min Avg. Time

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Memoization with State Caching — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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