Problem

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.

Examples

Example 1:

1
2
3
4
5
6
7

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:

1
2
3
4
5
6
7

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].

Solution

Method 1 – Simulation with Pairing

Intuition

The problem is about simulating the tournament pairing process. In each round, the strongest and weakest teams are paired, and the winners move to the next round. We keep pairing until one match remains, building the string representation recursively or iteratively.

Approach

  1. Start with a list of team labels as strings from 1 to n.
  2. In each round, pair the first and last, second and second-last, etc., and wrap each pair in parentheses.
  3. Replace the list with the new list of pairs and repeat until only one string remains.
  4. Return the final string.

Code

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15

class Solution {
public:
    string findContestMatch(int n) {
        vector<string> ans;
        for (int i = 1; i <= n; ++i) ans.push_back(to_string(i));
        while (ans.size() > 1) {
            vector<string> nxt;
            int sz = ans.size();
            for (int i = 0; i < sz/2; ++i)
                nxt.push_back("(" + ans[i] + "," + ans[sz-1-i] + ")");
            ans = nxt;
        }
        return ans[0];
    }
};
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12

func findContestMatch(n int) string {
    ans := make([]string, n)
    for i := 0; i < n; i++ { ans[i] = strconv.Itoa(i+1) }
    for len(ans) > 1 {
        nxt := make([]string, len(ans)/2)
        for i := 0; i < len(ans)/2; i++ {
            nxt[i] = "(" + ans[i] + "," + ans[len(ans)-1-i] + ")"
        }
        ans = nxt
    }
    return ans[0]
}
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14

class Solution {
    public String findContestMatch(int n) {
        List<String> ans = new ArrayList<>();
        for (int i = 1; i <= n; i++) ans.add(String.valueOf(i));
        while (ans.size() > 1) {
            List<String> nxt = new ArrayList<>();
            int sz = ans.size();
            for (int i = 0; i < sz/2; i++)
                nxt.add("(" + ans.get(i) + "," + ans.get(sz-1-i) + ")");
            ans = nxt;
        }
        return ans.get(0);
    }
}
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12

class Solution {
    fun findContestMatch(n: Int): String {
        var ans = MutableList(n) { (it+1).toString() }
        while (ans.size > 1) {
            val nxt = MutableList(ans.size/2) { "" }
            for (i in 0 until ans.size/2)
                nxt[i] = "(" + ans[i] + "," + ans[ans.size-1-i] + ")"
            ans = nxt
        }
        return ans[0]
    }
}
1
2
3
4
5
6

class Solution:
    def findContestMatch(self, n: int) -> str:
        ans = [str(i) for i in range(1, n+1)]
        while len(ans) > 1:
            ans = [f"({ans[i]},{ans[-1-i]})" for i in range(len(ans)//2)]
        return ans[0]
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14

impl Solution {
    pub fn find_contest_match(n: i32) -> String {
        let mut ans: Vec<String> = (1..=n).map(|x| x.to_string()).collect();
        while ans.len() > 1 {
            let sz = ans.len();
            let mut nxt = Vec::with_capacity(sz/2);
            for i in 0..sz/2 {
                nxt.push(format!("({},{} )", ans[i], ans[sz-1-i]));
            }
            ans = nxt;
        }
        ans[0].replace(" ", "")
    }
}
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13

class Solution {
    findContestMatch(n: number): string {
        let ans: string[] = Array.from({length: n}, (_, i) => (i+1).toString());
        while (ans.length > 1) {
            let nxt: string[] = [];
            for (let i = 0; i < ans.length/2; i++) {
                nxt.push(`(${ans[i]},${ans[ans.length-1-i]})`);
            }
            ans = nxt;
        }
        return ans[0];
    }
}

Complexity

  • ⏰ Time complexity: O(n log n), because in each round the number of teams halves and we do O(n) work per round.
  • 🧺 Space complexity: O(n log n), for storing all intermediate match strings.