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‎dsa-solutions/lc-solutions/2200-2300/2286 - Booking Concert Tickets in Groups.md‎

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---
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id: Booking-concert-tickets-in-groups
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title: Booking Concert Tickets in Groups
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sidebar_label: 2286 Booking Concert Tickets in Groups
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tags:
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- Java
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- Binary Search
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- Design
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- Segmnent Tree
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- Binary Indexed Tree
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description: "This document provides a solution where we need to design a ticketing system that can allocate seats."
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---
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## Problem
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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:
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- If a group of $k$ spectators can sit **together** in a row.
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- If **every** member of a group of $k$ spectators can get a seat. They may or **may not** sit together.
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Note that the spectators are very picky. Hence:
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- 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.
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- 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.
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Implement the $BookMyShow$ class:
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- $BookMyShow(int n, int m)$ Initializes the object with $n$ as number of rows and $m$ as number of seats per row.
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- $int[] gather(int k, int maxRow)$ 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.
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- $boolean scatter(int k, int maxRow)$ 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 $false$.
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### Examples
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**Example 1:**
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```
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Input: ["BookMyShow", "gather", "gather", "scatter", "scatter"]
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[[2, 5], [4, 0], [2, 0], [5, 1], [5, 1]]
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Output: [null, [0, 0], [], true, false]
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Explanation:
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BookMyShow bms = new BookMyShow(2, 5); // There are 2 rows with 5 seats each
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bms.gather(4, 0); // return [0, 0]
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// The group books seats [0, 3] of row 0.
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bms.gather(2, 0); // return []
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// There is only 1 seat left in row 0,
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// so it is not possible to book 2 consecutive seats.
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bms.scatter(5, 1); // return True
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// The group books seat 4 of row 0 and seats [0, 3] of row 1.
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bms.scatter(5, 1); // return False
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// There is only one seat left in the hall.
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```
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### Constraints
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- $1 <= n <= 5 * 10^4$
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- $1 <= m, k <= 10^9$
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- $0 <= maxRow <= n - 1$
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- At most $5 * 10^4$ calls in total will be made to $gather$ and $scatter$.
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---
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## Approach
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To solve the problem, we need to understand the nature of the allowed moves:
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1. **Segment Tree**:
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- **Build**: Construct the segment tree with initial values, where each node stores the sum and maximum number of seats in the range it represents.
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- **Update**: Update the segment tree when seats are booked, modifying the relevant nodes to reflect the new seat counts.
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- **Queries**:
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- **Gather Query**: Find a row with at least **'k'** seats available within the first **'maxRow'** rows.
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- **Sum Query**: Compute the total number of seats available within the first **'maxRow'** rows.
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2. **BookMyShow Class**:
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- **Constructor**: Initialize the segment tree and an array to keep track of seats available in each row.
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- **Gather**: Check if there is a row with at least **'k'** seats and update the tree and row seat count accordingly.
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- **Scatter**: Allocate **'k'**
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## Solution for Booking Concert Tickets in Groups
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- The problem revolves around efficiently managing and querying seat allocations in a theater. The segment tree is used to handle range queries and updates swiftly, ensuring that operations like finding available seats or updating seat counts are performed in logarithmic time.
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#### Code in Java
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```java
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class BookMyShow {
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static class SegTree{
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long sum[]; // store sum of seats in a range
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long segTree[]; // store maximum seats in a range
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int m, n;
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public SegTree(int n, int m) {
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this.m = m;
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this.n = n;
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segTree = new long[4*n];
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sum = new long[4*n];
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build(0, 0, n-1, m);
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}
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private void build(int index, int lo, int hi, long val){
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if(lo == hi){
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segTree[index] = val; // initialize segement tree with initial seat capacity
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sum[index] = val; // initialize "sum" with initial seat capacity of a row
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return;
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}
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int mid = (lo + hi)/2;
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build(2*index +1, lo, mid, val); // build left sub tree
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build(2*index +2, mid+1, hi, val); // build right sub tree
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segTree[index] = Math.max(segTree[2*index + 1], segTree[2*index + 2]); // maximum seats in a row for subtrees
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sum[index] = sum[2*index + 1] + sum[2*index + 2]; // sum of seats in a range
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}
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private void update(int index, int lo, int hi, int pos, int val){
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/**
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Method to update segment tree based on the available seats in a row
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**/
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if(lo == hi){
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segTree[index] = val;
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sum[index] = val;
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return;
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}
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int mid = (lo + hi) / 2;
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if (pos <= mid) { // position to update is in left
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update(2 * index + 1, lo, mid, pos, val);
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} else { // position to update is in right
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update(2 * index + 2, mid+1, hi, pos, val);
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}
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// update segment tree and "sum" based on the update in "pos" index
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segTree[index] = Math.max(segTree[2*index + 1] , segTree[2*index + 2]);
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sum[index] = sum[2*index + 1] + sum[2*index + 2];
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}
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public void update(int pos, int val){
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update(0, 0, n - 1 , pos, val);
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}
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public int gatherQuery(int k, int maxRow){
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return gatherQuery(0, 0, n - 1 , k, maxRow);
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}
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private int gatherQuery(int index, int lo, int hi, int k, int maxRow){
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/**
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Method to check if seats are available in a single row
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**/
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if(segTree[index] < k || lo > maxRow)
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return -1;
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if(lo == hi) return lo;
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int mid = (lo + hi) / 2;
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int c = gatherQuery(2*index + 1, lo, mid, k, maxRow);
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if(c == -1){
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c = gatherQuery(2*index + 2, mid +1, hi, k, maxRow);
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}
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return c;
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}
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public long sumQuery(int k, int maxRow){
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return sumQuery(0, 0, n-1, k, maxRow);
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}
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private long sumQuery(int index, int lo, int hi, int l, int r){
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if(lo > r || hi < l ) return 0; // not in range
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if(lo >= l && hi <= r) return sum[index]; // in range
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int mid = (lo + hi)/2;
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return sumQuery(2*index+1, lo, mid, l, r) + sumQuery(2*index+2, mid+1, hi, l, r);
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}
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}
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SegTree segTree;
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int[] rowSeats; // stores avaiable seats in a row, helps to find the vacant seat in a row
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public BookMyShow(int n, int m) {
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segTree = new SegTree(n, m);
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rowSeats = new int[n];
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Arrays.fill(rowSeats, m); // initialize vacant seats count to "m" for all the rows
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}
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public int[] gather(int k, int maxRow) {
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int row = segTree.gatherQuery(k, maxRow); // find row which has k seats
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if(row == -1) return new int[]{}; // can't find a row with k seats
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int col = segTree.m - rowSeats[row]; // find column in the row which has k seats
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rowSeats[row] -= k; // reduce the seats
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segTree.update(row, rowSeats[row]); // update the segment tree
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return new int[]{row, col};
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}
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public boolean scatter(int k, int maxRow) {
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long sum = segTree.sumQuery(0, maxRow); // find the sum for the given range [0, maxRow]
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if(sum < k) return false; // can't find k seats in [0, maxRow]
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for(int i=0; i<=maxRow && k !=0 ; i++){
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if(rowSeats[i] > 0){ // if current row has seats then allocate those seats
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long t = Math.min(rowSeats[i], k);
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rowSeats[i] -= t;
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k -= t;
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segTree.update(i,rowSeats[i]); // update the segment tree
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}
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}
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return true;
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}
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}
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```
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### Complexity Analysis
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#### Time Complexity: $O(logn)$
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> **Reason**: Segment tree operations are build: $O(n)$, update: $O(log n)$, gather query: $O(log n)$, sum query: $O(log n)$; BookMyShow operations are constructor: $O(n)$, gather: $O(log n)$, scatter: $O(n log n)$.
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#### Space Complexity: $O(n)$
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> **Reason**: The space complexity is $O(n)$, The **'rowSeats'** array requires O(n) space to store the seat counts for each row.
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# References
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- **LeetCode Problem:** [Booking Concert Tickets in Groups](https://leetcode.com/problems/booking-concert-tickets-in-groups/description/)
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- **Solution Link:** [Booking Concert Tickets in Groups Solution on LeetCode](https://leetcode.com/problems/booking-concert-tickets-in-groups/solutions/)
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- **Authors LeetCode Profile:** [Vivek Vardhan](https://leetcode.com/u/vivekvardhan43862/)

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