Final NBA Match Pairing in C++

24 May 2025 | 4 min read

A string representing the final contest matches of n teams is returned. The teams are ranked from 1 to 𝑛, with Rank 1 being the best team and Rank n being the worst. The labels correspond to the teams' initial ranks.

The matching procedure represents team pairings using commas ‘,’ and parenthesis ‘(‘, and’)’. The teams are paired in brackets, and the division between match pairings is indicated by commas. The approach ensures that the stronger team is matched with the weaker team throughout the matching procedure.

Limitations: The n must always be a power of two in the formula n=2^x, where x falls between 1 and 12.

Approach:

In a string vector, use the first unmatched to last unmatched logic to record each initial matchup, complete with its parentheses and commas, to a different row.
The recursive code is then run with the size halved, thereby doing the same task without converting int to string.
Each round halves the number of teams. Therefore, when the size is 1, we reach the base case and return without further modifications.

Steps to solve the problem:

In order to combine matches in the array into a single string, create the mergingMatches() method.
1. Return if (n == 1) there is just one match.
2. Iteratively combine the first and second halves of the matches by proceeding from 0 to n / 2. Incorporate both teams into every match string.
Call mergingMatches() recursively using the updated matches array and n / 2.
Create the function “findingContestMatch()”: This function requires n to be the number of teams.
1. Make an empty array mat_ches[] to store the starting set of matches.
2. Initial matches should be added to the mat_ches[] array. Each match is represented by a string containing the team numbers of the involved teams.
3. In order to combine all of the matches into one, use mergingMatches() with the matches vector and n / 2.

Input: n = 4

Output: “((1,4),(2,3))”

Explanation: In the first round, to create a strong team and a weak team, we partnered up teams 1 and 4 and teams 2 and 3. We obtained (2, 3) and (1, 4).

In order to determine the ultimate winner in the second round, we must include the paratheses outside of the winners of (1, 4) and (2, 3). The final response ((1,4),(2,3)) was what we obtained.

Example:

Let us take an example to illustrate the final NBA Match pairing in C++.

Output:

Please enter the number of teams: 32
The final contest match:
(((((1,32),(16,17)),((8,25),(9,24))),(((4,29),(13,20)),((5,28),(12,21)))),((((2,31),(15,18)),((7,26),(10,23))),(((3,30),(14,19)),((6,27),(11,22)))))

Explanation:

This C++ code generates the final contest match pairing for a given number of teams using a recursive merging method.
Until one match remains, which represents the final contest match, the mergingMatches method repeatedly combines adjacent match pairs.
The function findingContestMatches generates initial match pairs by dividing the total number of teams by two. Each match consists of the two team numbers enclosed in parentheses.
In order to make match pairs easier to generate, the to_string function converts integer team numbers into strings for concatenation with brackets.
The findingContestMatches method is used when the user enters the number of teams, and it uses that number to determine the final contest match.
The process of producing and presenting the match pairing is finished when the user displays the calculated final contest match.

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