Tournament Winners - Practice Coding Problems

Tournament Winners - Problem

Database Hard

Imagine you're organizing a multi-group tournament where players compete within their groups. Each player belongs to exactly one group, and matches are played only between players from the same group.

Your task: Determine the winner of each group based on total points scored across all matches.

You have two tables:

Players: Contains player_id and their group_id
Matches: Contains match details with first_player, second_player, and their respective scores

Winner Rules:

The player with the highest total points in their group wins
In case of a tie, the player with the lowest player_id wins

Return the winner of each group. The result can be in any order.

Input & Output

example_1.sql — Basic Tournament

$ Input: Players: [[1,2],[2,1],[3,1],[4,1],[5,2]]<br>Matches: [[1,3,4,5,2],[2,4,3,8,4],[3,1,5,6,3]]

› Output: [[1,3],[2,1]]

💡 Note: Group 1: Player 3 scored 8+4=12 points, Player 4 scored 2+8=10 points → Player 3 wins
Group 2: Player 1 scored 6 points, Player 5 scored 3 points → Player 1 wins

example_2.sql — Tie Breaker

$ Input: Players: [[1,1],[2,1],[3,2]]<br>Matches: [[1,1,2,10,10],[2,3,3,5,0]]

› Output: [[1,1],[2,3]]

💡 Note: Group 1: Both players scored 10 points, but Player 1 has lower ID → Player 1 wins
Group 2: Player 3 scored 5 points → Player 3 wins

example_3.sql — Multiple Matches

$ Input: Players: [[1,1],[2,1]]<br>Matches: [[1,1,2,8,12],[2,2,1,15,7],[3,1,2,6,9]]

› Output: [[1,2]]

💡 Note: Player 1: 8+7+6=21 points
Player 2: 12+15+9=36 points → Player 2 wins Group 1

Constraints

1 ≤ players.length ≤ 5 × 10⁴
1 ≤ matches.length ≤ 10⁵
1 ≤ player_id ≤ 10⁵
1 ≤ group_id ≤ 10³
0 ≤ first_score, second_score ≤ 100
All player_id values are unique
Players in each match belong to the same group

Visualization

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Understanding the Visualization

Setup Phase

Create scorecards for all players and organize them by house/group

Match Processing

As each match completes, immediately update player scores and check for new house leaders

Champion Selection

Final champions emerge based on highest scores, with lowest ID breaking ties

Results

Each house has exactly one champion, determined efficiently in a single pass

Key Takeaway

🎯 Key Insight: Process matches once while maintaining group leaders in hash maps, avoiding the need for multiple passes through the data

Asked in

G Google 35 a Amazon 28 f Facebook 22 ⊞ Microsoft 18

The optimal solution uses a single-pass hash table approach to process matches efficiently. Key insights: UNION ALL in SQL to combine player-score pairs, window functions for ranking, and hash maps in programming languages for O(1) lookups. The algorithm runs in O(m + n log n) time where m is matches and n is players.

Common Approaches

Approach	Time	Space	Notes
✓ One-Pass Hash Table (Optimal SQL)	O(m + n log n)	O(n)	Calculate scores and find winners in a single SQL query with UNION and window functions
Brute Force SQL (Multiple Subqueries)	O(n² * m)	O(1)	Calculate total scores for each player using multiple subqueries

One-Pass Hash Table (Optimal SQL) — Algorithm Steps

Use UNION ALL to create all player-score pairs from matches
Group by player to calculate total scores
Use window functions to rank players within groups
Filter to get only the top-ranked player from each group

Visualization

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Step-by-Step Walkthrough

Initialize Tracking

Set up hash maps for scores and group mappings

Process Matches

Single pass through matches, updating scores and group winners

Return Winners

Extract winners from group tracking map

Code -

solution.c — C

// Optimal C solution
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>

// Simple hash table implementation for demonstration
struct HashNode {
    int key;
    int value;
    struct HashNode* next;
};

struct Result* tournamentWinners(int players[][2], int playersSize, int matches[][5], int matchesSize, int* returnSize) {
    // Implementation would require creating hash table from scratch
    // Following the same logic as other optimal solutions
    
    struct Result* winners = malloc(sizeof(struct Result) * playersSize);
    *returnSize = 0;
    
    // Hash table operations and winner finding logic here
    // (Simplified for brevity)
    
    return winners;
}

Time & Space Complexity

Time Complexity

⏱️

O(m + n log n)

O(m) to scan matches, O(n log n) for sorting within groups

⚡ Linearithmic

Space Complexity

O(n)

Space for intermediate results and grouping

⚡ Linearithmic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

One-Pass Hash Table (Optimal SQL) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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