Output Contest Matches

Problem Statement

In the context of a tournament-style competition, such as the NBA playoffs, an interesting strategy involves pairing teams so that the strongest plays against the weakest, the second strongest against the second weakest, and so on. This approach aims to make the matches more engaging and balanced. We're tasked with a challenge where given n teams, each labeled from 1 (strongest) to n (weakest), we must arrange them in such pairing patterns and depict this as a single string representation.

Each match or pairing is represented using parentheses '(' and ')' and separated by commas ','. The process of generating this string involves multiple rounds of pairing, where each consecutive round takes the winners of the previous round until a final matchup is formed. The ultimate goal is to return a string that represents the entire sequence of matches from the start to the final.

Examples

Example 1

Input:

n = 4

Output:

"((1,4),(2,3))"

Explanation:

In the first round, we pair the team 1 and 4, the teams 2 and 3 together, as we need to make the strong team and weak team together.
And we got (1, 4),(2, 3).
In the second round, the winners of (1, 4) and (2, 3) need to play again to generate the final winner, so you need to add the paratheses outside them.
And we got the final answer ((1,4),(2,3)).

Example 2

Input:

n = 8

Output:

"(((1,8),(4,5)),((2,7),(3,6)))"

Explanation:

First round: (1, 8),(2, 7),(3, 6),(4, 5)
Second round: ((1, 8),(4, 5)),((2, 7),(3, 6))
Third round: (((1, 8),(4, 5)),((2, 7),(3, 6)))
Since the third round will generate the final winner, you need to output the answer (((1,8),(4,5)),((2,7),(3,6))).

Constraints

• n == 2x where x in in the range [1, 12].

Approach and Intuition

Given the problem's constraints and requirements, the solution involves several key steps, each building upon the previous to generate the final output.

1. Initialization: Start by representing each team as a string number of its rank. This provides a base for the first round of pairing.

2. Pairing Process:

   • In each round, pair the first team in the current list with the last, the second with the second last, and so forth until you reach the midpoint of the list.
   • Combine these teams into pairs using the designated string format - for instance, "1" and "n" become "(1,n)".
   • After each round of pairing, the results become the input list for the next round.

3. Building up the Rounds:

   • After forming the initial pairs, each subsequent round takes the results of the previous round and pairs them similarly until only one pair remains.
   • During this process, enclose each new pair from the previous results within additional parentheses - representing the progression in tournament rounds.

4. Completion:

   • The process repeats until the list of results contains only a single string, which represents the final layout of matches from the initial to the last round.
   • This final string is returned as the solution.

The idea is to use a structured approach to iteratively reduce the number of teams/participants by half each round, simulating the tournament's progression until the champion emerges. The process visually and structurally organizes the participants in a manner illustrating each match in each round, providing a clear view of the entire tournament's progression. Each operation's time complexity effectively involves string manipulations and list management, optimized by the constraints given (the maximum n is considerably manageable).

Solutions

class MatchFinder {
    int[] players;
    int index;
    StringBuilder result;
    public String generateMatch(int size) {
        players = new int[size];
        index = 0;
        result = new StringBuilder();
        generate(size, Integer.numberOfTrailingZeros(size));
        return result.toString();
    }
    
    public void generate(int size, int level) {
        if (level == 0) {
            int x = Integer.lowestOneBit(index);
            players[index] = x > 0 ? size / x + 1 - players[index - x] : 1;
            result.append("" + players[index++]);
        } else {
            result.append("(");
            generate(size, level - 1);
            result.append(",");
            generate(size, level - 1);
            result.append(")");
        }
    }
}