Solution Explanation: Output Contest Matches

This problem asks to simulate a contest match between teams, pairing the strongest with the weakest iteratively until a single winner remains. The solution uses a bottom-up approach, constructing the match results from the individual team pairings.

Approach:

The core idea is to represent the teams as strings and recursively pair them. We start with an array of strings representing team numbers. In each iteration:

1. We pair the strongest (lowest index) with the weakest (highest index) teams, creating a string representation like "(1,n),(2,n-1),...".
2. We replace the original team strings with the newly formed pairs, effectively reducing the number of teams by half.
3. We repeat steps 1 and 2 until only one string (the final match result) remains.

Time Complexity Analysis:

The algorithm iterates through the teams, reducing their number by half in each iteration. This corresponds to the logarithm of the number of teams (base 2). The inner loop iterates through roughly half the remaining teams in each iteration. Therefore, the total number of operations is approximately:

n/2 + n/4 + n/8 + ... + 1 ≈ n

This is a geometric series which sums to approximately n. Therefore, the overall time complexity is O(n), where n is the number of teams. The log n iterations are dominated by the n operations within each iteration. This differs from a strict O(n log n) because the operations within each iteration are linear with respect to the remaining number of teams, not the original n.

Space Complexity Analysis:

The algorithm uses an array s to store the team pairings. The size of this array is initially n and is halved in each iteration. The space used is dominated by the largest size of the array, which is n. Therefore, the space complexity is O(n).

Code Explanation (Python):

class Solution:
    def findContestMatch(self, n: int) -> str:
        s = [str(i + 1) for i in range(n)]  # Initialize team strings
        while n > 1:
            for i in range(n >> 1):  # Iterate through half the teams
                s[i] = f"({s[i]},{s[n - i - 1]})"  # Pair strongest and weakest
            n >>= 1  # Halve the number of teams
        return s[0]  # Return the final match result

The code directly implements the steps described in the approach. The >> 1 operator is a bitwise right shift, equivalent to integer division by 2. This efficient way to halve the number of teams. The f-string formatting creates the parenthesized pairs concisely.

Other Languages (Java, C++, Go, TypeScript):

The solutions in other languages follow the same algorithm and have similar structures. The main differences lie in syntax and data type handling. The time and space complexity remain the same. The key is the iterative pairing of teams and halving the array size until a single winner remains.