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2286. Booking Concert Tickets in Groups

Difficulty Topics

Description

A concert hall has n rows numbered from 0 to n - 1, each with m seats, numbered from 0 to m - 1. You need to design a ticketing system that can allocate seats in the following cases:

  • If a group of k spectators can sit together in a row.
  • If every member of a group of k spectators can get a seat. They may or may not sit together.

Note that the spectators are very picky. Hence:

  • They will book seats only if each member of their group can get a seat with row number less than or equal to maxRow. maxRow can vary from group to group.
  • In case there are multiple rows to choose from, the row with the smallest number is chosen. If there are multiple seats to choose in the same row, the seat with the smallest number is chosen.

Implement the BookMyShow class:

  • BookMyShow(int n, int m) Initializes the object with n as number of rows and m as number of seats per row.
  • int[] gather(int k, int maxRow) Returns an array of length 2 denoting the row and seat number (respectively) of the first seat being allocated to the k members of the group, who must sit together. In other words, it returns the smallest possible r and c such that all [c, c + k - 1] seats are valid and empty in row r, and r <= maxRow. Returns [] in case it is not possible to allocate seats to the group.
  • boolean scatter(int k, int maxRow) Returns true if all k members of the group can be allocated seats in rows 0 to maxRow, who may or may not sit together. If the seats can be allocated, it allocates k seats to the group with the smallest row numbers, and the smallest possible seat numbers in each row. Otherwise, returns false.

 

Example 1:

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

booking-concert-tickets-in-groups.py
class Node:
    def __init__(self, start, end):
        self.left = None
        self.right = None
        self.start = start
        self.end = end
        self.total = 0
        self.mx = 0

class SegTree:

    def __init__(self, n, val):
        self.root = self.buildTree(0, n, val)

    def buildTree(self, start, end, val):
        if start > end: return None

        if start == end:
            node = Node(start, end)
            node.total = val
            node.mx = val

            return node

        node = Node(start, end)
        mid = start + (end - start) // 2
        node.left = self.buildTree(start, mid, val)
        node.right = self.buildTree(mid + 1, end, val)
        node.total = node.left.total + node.right.total
        node.mx = max(node.left.mx, node.right.mx)

        return node

    def update(self, index: int, val: int) -> None:
        self.updateR(self.root, index, val)

    def updateR(self, node, index, val):

        if node.start == node.end:
            node.total -= val
            node.mx -= val
        else:
            mid = node.start + (node.end - node.start) // 2
            if index <= mid:
                self.updateR(node.left, index, val)
            else:
                self.updateR(node.right, index, val)

            node.total = node.left.total + node.right.total
            node.mx = max(node.left.mx, node.right.mx)

    def sumRange(self, left: int, right: int) -> int:
        return self.sumRangeR(self.root, left, right)

    def sumRangeR(self, node, left, right):   
        if not node or node.start > right or node.end < left: return 0

        if node.start >= left and node.end <= right: return node.total

        l = self.sumRangeR(node.left, left, right)
        r = self.sumRangeR(node.right, left, right)

        return l + r

    def maxQuery(self, k, maxRow, seats):

        def maxQueryHelper(node):
            if node.start == node.end:
                if node.end > maxRow or node.total < k:
                    return []

                if node.end <= maxRow and node.total >= k:
                    return [node.end, seats - node.total]

            return maxQueryHelper(node.left) if node.left.mx >= k else maxQueryHelper(node.right)

        return maxQueryHelper(self.root)

class BookMyShow:

    def __init__(self, n: int, m: int):
        self.seg = SegTree(n - 1, m)  
        self.seats = [m] * n
        self.m = m
        self.n = n
        self.startRow = 0

    def gather(self, k: int, maxRow: int) -> List[int]:
        res = self.seg.maxQuery(k, maxRow, self.m)

        if res:
            row = res[0]
            self.seg.update(row, k)
            self.seats[row] -= k

        return res


    def scatter(self, k: int, maxRow: int) -> bool:
        if self.seg.sumRange(0, maxRow) < k:
            return False

        index = self.startRow
        curr = 0

        while curr < k:
            rest = k - curr

            if rest >= self.seats[index]:
                self.seg.update(index, self.seats[index])
                curr += self.seats[index]
                self.seats[index] = 0
                index += 1
                self.startRow = index
            else:
                self.seg.update(index, rest)
                self.seats[index] -= rest
                curr += rest

        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)