Leetcode #2286: Booking Concert Tickets in Groups
In this guide, we solve Leetcode #2286 Booking Concert Tickets in Groups in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Design, Binary Indexed Tree, Segment Tree, Binary Search
Intuition
The problem structure suggests a monotonic decision, which makes binary search a natural fit.
By halving the search space each step, we reach the answer efficiently.
Approach
Search either directly on a sorted array or on the answer space using a check function.
Each check is fast, and the logarithmic search keeps the overall runtime low.
Steps:
- Define the search bounds.
- Check the mid point condition.
- Narrow the bounds until convergence.
Example
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.
Python Solution
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)
Complexity
The time complexity is O(log n) or O(n log n). The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.