Booking Concert Tickets in Groups

Design a ticketing system to allocate concert seats in specific groupings while efficiently handling seat reservations.

binary search design binary indexed tree segment tree

Problem Statement

A concert hall has n rows numbered from 0 to n - 1, with each row containing m seats numbered from 0 to m - 1. Your task is to implement a ticketing system that handles seat allocation in two different ways. First, you need to support gathering a group of k consecutive seats in a specific row. Second, you need to support scattering groups across multiple rows, where each member of the group is seated in a separate row but the group size remains intact. The system should efficiently manage seat reservations for large groups while optimizing for the smallest seat number in each row.

The BookMyShow class must be implemented with the following methods: 'gather(k, maxRow)' to reserve k consecutive seats in a row starting from row 0 to row maxRow, and 'scatter(k, maxRow)' to reserve k seats in separate rows, one in each row from row 0 to maxRow. The system must work efficiently even with up to 50,000 calls to gather and scatter, where the number of seats m and rows n can be large.

Examples

Example 1

Input: See original problem statement.

Output: See original problem statement.

Input ["BookMyShow", "gather", "gather", "scatter", "scatter"] [[2, 5], [4, 0], [2, 0], [5, 1], [5, 1]] Output [null, [0, 0], [], true, false]

Explanation BookMyShow bms = new BookMyShow(2, 5); // There are 2 rows with 5 seats each bms.gather(4, 0); // return [0, 0] // The group books seats [0, 3] of row 0. bms.gather(2, 0); // return [] // There is only 1 seat left in row 0, // so it is not possible to book 2 consecutive seats. bms.scatter(5, 1); // return True // The group books seat 4 of row 0 and seats [0, 3] of row 1. bms.scatter(5, 1); // return False // There is only one seat left in the hall.

Constraints

1 <= n <= 5 * 104
1 <= m, k <= 109
0 <= maxRow <= n - 1
At most 5 * 104 calls in total will be made to gather and scatter.

Solution Approach

Binary Search for Seat Allocation

The problem can be approached using binary search over the valid range of seat allocations. For 'gather', binary search helps quickly determine if there is a contiguous block of k seats in any of the rows. For 'scatter', binary search finds the smallest row with available seats to place each group member.

Using Binary Indexed Tree (BIT)

A Binary Indexed Tree can be employed to track the availability of seats across rows efficiently. This data structure helps update and query available seat counts in logarithmic time, allowing us to quickly determine where to allocate seats without needing to scan through every row.

Segment Tree for Range Queries

A Segment Tree can be used to manage seat availability across multiple rows and quickly query for available seat blocks. By storing the number of available seats in each segment, we can optimize the scatter method to find seats across rows, minimizing the number of checks needed.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the method chosen for seat allocation. With binary search and a BIT, both 'gather' and 'scatter' operations can be executed in O(log n) time. Segment Tree-based approaches can achieve similar results, with time complexities of O(log n) for each query and update operation. Space complexity is O(n) for both the BIT and Segment Tree approaches, where n is the number of rows in the concert hall.

What Interviewers Usually Probe

Evaluate how well the candidate implements binary search to find seat allocations within rows.
Assess whether the candidate effectively utilizes Binary Indexed Tree or Segment Tree for optimization.
Check how the candidate handles edge cases, such as when there are very few available seats or when no valid allocation is possible.

Common Pitfalls or Variants

Common pitfalls

Forgetting to handle edge cases where no valid seat group can be found within the given rows.
Inefficient seat allocation methods that don’t scale well when n and m are large.
Not optimizing seat tracking with data structures like Binary Indexed Tree or Segment Tree, leading to excessive computation times.

Follow-up variants

Allow partial group reservations where a group can be scattered but still need to sit together in the same row if possible.
Consider adding a feature where groups can reserve seats for future dates, requiring additional optimizations for date management.
Support for dynamic updates to the number of available seats, allowing rows to be added or removed during runtime.

FAQ

What is the core pattern in the 'Booking Concert Tickets in Groups' problem?

The core pattern is binary search over the valid answer space for seat allocation, along with optimization techniques like Binary Indexed Tree and Segment Tree.

How can Binary Indexed Tree help in this problem?

A Binary Indexed Tree allows efficient tracking of seat availability across rows and provides fast updates and queries, making it ideal for seat allocation.

What happens if we don't use binary search or data structures like BIT?

Without binary search or efficient data structures, the solution will become too slow, especially when handling up to 50,000 calls for seat allocation in large halls.

Can we solve the problem without using a Segment Tree?

While a Segment Tree is an optimal choice for range queries, it is possible to solve the problem using other methods, like Binary Indexed Tree or direct binary search over rows.

What are the time and space complexities of this problem?

The time complexity depends on the approach, but it is generally O(log n) for both 'gather' and 'scatter' operations using Binary Search and BIT/Segment Tree. The space complexity is O(n), where n is the number of rows.

terminal

Solution

Solution 1: Segment Tree

From the problem description, we can deduce the following:

class Node:
    __slots__ = "l", "r", "s", "mx"

    def __init__(self):
        self.l = self.r = 0
        self.s = self.mx = 0


class SegmentTree:
    def __init__(self, n, m):
        self.m = m
        self.tr = [Node() for _ in range(n << 2)]
        self.build(1, 1, n)

    def build(self, u, l, r):
        self.tr[u].l, self.tr[u].r = l, r
        if l == r:
            self.tr[u].s = self.tr[u].mx = self.m
            return
        mid = (l + r) >> 1
        self.build(u << 1, l, mid)
        self.build(u << 1 | 1, mid + 1, r)
        self.pushup(u)

    def modify(self, u, x, v):
        if self.tr[u].l == x and self.tr[u].r == x:
            self.tr[u].s = self.tr[u].mx = v
            return
        mid = (self.tr[u].l + self.tr[u].r) >> 1
        if x <= mid:
            self.modify(u << 1, x, v)
        else:
            self.modify(u << 1 | 1, x, v)
        self.pushup(u)

    def query_sum(self, u, l, r):
        if self.tr[u].l >= l and self.tr[u].r <= r:
            return self.tr[u].s
        mid = (self.tr[u].l + self.tr[u].r) >> 1
        v = 0
        if l <= mid:
            v += self.query_sum(u << 1, l, r)
        if r > mid:
            v += self.query_sum(u << 1 | 1, l, r)
        return v

    def query_idx(self, u, l, r, k):
        if self.tr[u].mx < k:
            return 0
        if self.tr[u].l == self.tr[u].r:
            return self.tr[u].l
        mid = (self.tr[u].l + self.tr[u].r) >> 1
        if self.tr[u << 1].mx >= k:
            return self.query_idx(u << 1, l, r, k)
        if r > mid:
            return self.query_idx(u << 1 | 1, l, r, k)
        return 0

    def pushup(self, u):
        self.tr[u].s = self.tr[u << 1].s + self.tr[u << 1 | 1].s
        self.tr[u].mx = max(self.tr[u << 1].mx, self.tr[u << 1 | 1].mx)


class BookMyShow:
    def __init__(self, n: int, m: int):
        self.n = n
        self.tree = SegmentTree(n, m)

    def gather(self, k: int, maxRow: int) -> List[int]:
        maxRow += 1
        i = self.tree.query_idx(1, 1, maxRow, k)
        if i == 0:
            return []
        s = self.tree.query_sum(1, i, i)
        self.tree.modify(1, i, s - k)
        return [i - 1, self.tree.m - s]

    def scatter(self, k: int, maxRow: int) -> bool:
        maxRow += 1
        if self.tree.query_sum(1, 1, maxRow) < k:
            return False
        i = self.tree.query_idx(1, 1, maxRow, 1)
        for j in range(i, self.n + 1):
            s = self.tree.query_sum(1, j, j)
            if s >= k:
                self.tree.modify(1, j, s - k)
                return True
            k -= s
            self.tree.modify(1, j, 0)
        return True


# Your BookMyShow object will be instantiated and called as such:
# obj = BookMyShow(n, m)
# param_1 = obj.gather(k,maxRow)
# param_2 = obj.scatter(k,maxRow)

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