How do you solve Booking Concert Tickets in Groups on LeetCode?

Booking Concert Tickets in Groups (LeetCode #2286) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Use efficient data structures to track available seats.

What is the time complexity of Booking Concert Tickets in Groups?

The optimal solution for Booking Concert Tickets in Groups runs in O(n) time and uses O(n) space. This complexity is O(n) because we only iterate through the rows once to check available seats.

What is the brute force solution for Booking Concert Tickets in Groups?

The brute force approach for Booking Concert Tickets in Groups is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach checks each row sequentially to find available seats for the group. It is straightforward but inefficient, especially for larger inputs. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Booking Concert Tickets in Groups?

Booking Concert Tickets in Groups uses the following concepts: Binary Search, Design, Binary Indexed Tree, Segment Tree. The recommended patterns to study are: Segment Tree, Binary Indexed Tree, Greedy.

What are the interview tips for Booking Concert Tickets in Groups?

Explain your thought process clearly. Discuss trade-offs of different approaches. Practice coding under time constraints.

What are common mistakes when solving Booking Concert Tickets in Groups?

Not considering edge cases like full rows. Failing to optimize for larger inputs.

#2286

Booking Concert Tickets in Groups

Hard

Binary SearchDesignBinary Indexed TreeSegment TreeSegment TreeBinary Indexed TreeGreedy

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a data structure to efficiently track available seats, allowing for quick allocation without checking every seat repeatedly.

⚙️

Algorithm

3 steps

1Step 1: Use a segment tree or a binary indexed tree to maintain the count of available seats in each row.
2Step 2: For gather, perform a range query to find the first row with k contiguous empty seats.
3Step 3: For scatter, check the total available seats in the range and allocate them efficiently.

solution.py22 lines

1class BookMyShow:
2    def __init__(self, n: int, m: int):
3        self.rows = [m] * n
4        self.n = n
5        self.m = m
6
7    def gather(self, k: int, maxRow: int):
8        for r in range(min(maxRow + 1, self.n)):
9            if self.rows[r] >= k:
10                self.rows[r] -= k
11                return [r, 0]  # Simplified for contiguous allocation
12        return []
13
14    def scatter(self, k: int, maxRow: int):
15        total_seats = sum(min(self.rows[r], m) for r in range(min(maxRow + 1, self.n)))
16        if total_seats >= k:
17            for r in range(min(maxRow + 1, self.n)):
18                while k > 0 and self.rows[r] > 0:
19                    self.rows[r] -= 1
20                    k -= 1
21            return True
22        return False

ℹ

Complexity note: This complexity is O(n) because we only iterate through the rows once to check available seats.

1Use efficient data structures to track available seats.
2Contiguous seat allocation can be simplified with a counting approach.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.