During the NBA playoffs, we always set the rather strong team to play with the rather weak team, like making the rank 1 team play with the rank nth team, which is a good strategy to make the contest more interesting.
Given n teams, return their final contest matches in the form of a string.
The n teams are labeled from 1 to n, which represents their initial rank (i.e., Rank 1 is the strongest team and Rank n is the weakest team).
We will use parentheses '(', and ')' and commas ',' to represent the contest team pairing. We use the parentheses for pairing and the commas for partition. During the pairing process in each round, you always need to follow the strategy of making the rather strong one pair with the rather weak one.
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].Problem Overview: You are given n teams labeled from 1 to n, where n is a power of two. In each round of the tournament, the strongest team plays the weakest (1 vs n), the second strongest plays the second weakest (2 vs n-1), and so on. The task is to generate the final contest match representation as a single string with nested parentheses showing how matches are formed until only one bracket remains.
Approach 1: Recursive Simulation (O(n log n) time, O(n) space)
This method models the tournament structure directly using recursion. Start with a list of team labels represented as strings. Pair the first and last teams, wrap them in parentheses like (a,b), and build a new list representing the next round. Recursively repeat this process until only one string remains. The key insight is that each round halves the number of teams while embedding previous match results inside new parentheses, naturally forming the bracket structure.
The recursion depth equals the number of tournament rounds, which is log n. Each round processes all current teams once to create pairings. Since string concatenation grows with each round, the total runtime is approximately O(n log n), while space remains O(n) to store the intermediate bracket strings.
Approach 2: Iterative Simulation (O(n log n) time, O(n) space)
The iterative version avoids recursion and performs the same bracket construction using a loop. Start with an array of strings ["1", "2", ..., "n"]. During each round, iterate from both ends using two pointers: one at the beginning and one at the end. Combine each pair into a match string using (left,right) and store it in a new list representing the next round.
This process continues until the list contains only one element. Using a two‑pointer pattern keeps the pairing logic simple and clearly reflects the tournament rule of strongest versus weakest. The algorithm is essentially a controlled simulation of tournament rounds, with heavy use of string construction to build the final bracket.
Recommended for interviews: The iterative simulation approach is typically what interviewers expect. It clearly demonstrates understanding of the pairing rule and keeps the implementation concise. Showing the recursive version first can highlight structural insight into tournament trees, but the iterative solution communicates stronger control over state updates and space usage.
We can use an array s of length n to store the ID of each team, and then simulate the process of the matches.
In each round of matches, we pair up the first n elements in array s two by two, and then store the ID of the winners in the first n/2 positions of array s. After that, we halve n and continue to the next round of matches, until n is reduced to 1. At this point, the first element in array s is the final match-up scheme.
The time complexity is O(n log n), and the space complexity is O(n). Here, n is the number of teams.
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| Approach | Time | Space | When to Use |
|---|---|---|---|
| Recursive Simulation | O(n log n) | O(n) | When modeling the tournament as a recursive bracket tree |
| Iterative Simulation (Two Pointers) | O(n log n) | O(n) | Best for interviews and production code due to simpler control flow |
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