# Solution For Cinema Seat Allocation

The Cinema Seat Allocation problem on LeetCode involves allocating seats to customers in a cinema hall. The goal is to maximize the number of satisfied customers by assigning them seats according to their preferences.

Let’s take a look at the problem statement:

There are n rows of seats in a cinema hall and each row has ten seats labeled from 1 to 10. Given a list of reserved seats, your task is to allocate seats for as many customers as possible, while avoiding any conflicts between them.

Each customer has a row preference (1 to n) and a list of seat preferences (1 to 10). If a customer’s preferred seat is already reserved by someone else, they will not be satisfied with any other seat. If a customer is not able to get their preferred seat, they will be satisfied with any other seat in the same row.

Write a function to find the maximum number of satisfied customers that can be accommodated in the cinema hall.

Let’s break down this problem step by step:

Input: The input to the function will be a list of tuples that represent the reserved seats. The tuples will contain the row number and the seat number that has been reserved.

Output: The function should return an integer that represents the maximum number of satisfied customers that can be accommodated in the cinema hall.

Approach: The first step is to create a dictionary that stores the reserved seats as key-value pairs. The keys of the dictionary will be the row numbers and the values will be lists of seat numbers that have been reserved in that row.

Next, we will loop through each customer and check if their preferred seat is available. If it is, we will mark the seat as reserved and move on to the next customer. If it is not, we will check if any other seat in the same row is available. If it is, we will allocate that seat to the customer and mark it as reserved. If no seat is available in the same row, we will move on to the next customer.

We will repeat this process for all customers and keep a count of the satisfied customers.

Let’s take a look at the code:

`def maxSatisfied(customers: List[List[int]], reserved_seats: List[Tuple[int, int]]) -> int:`

reserved_dict = {}

for seat in reserved_seats:

if seat[0] not in reserved_dict:

reserved_dict[seat[0]] = []
reserved_dict[seat[0]].append(seat[1])

satisfied_count = 0

for customer in customers:

row, seat = customer[0], customer[1]
if seat in reserved_dict.get(row, []):

satisfied_count += 1

elif (seat-1) not in reserved_dict.get(row, []) and (seat-2) not in reserved_dict.get(row, []) and (seat+1) not in reserved_dict.get(row, []) and (seat+2) not in reserved_dict.get(row, []):

reserved_dict[row].extend([seat-1, seat, seat+1])

satisfied_count += 1

return satisfied_count + sum([len(seats) for seats in reserved_dict.values()])

In this code, we first create a dictionary `reserved_dict`

that stores the reserved seats. Next, we initialize a counter `satisfied_count`

to keep track of the satisfied customers.

We then loop through each customer and check if their preferred seat is available. If it is, we increment `satisfied_count`

. If not, we check if any other seat in the same row is available. If it is, we allocate that seat to the customer and mark it as reserved. If no seat is available in the same row, we move on to the next customer.

At the end of the loop, we return the sum of `satisfied_count`

and the total number of reserved seats that were not used.

This solution has a time complexity of O(n^2), where n is the number of customers. The space complexity is O(n) for the `reserved_dict`

dictionary.

## Step by Step Implementation For Cinema Seat Allocation

/** * There are a number of empty seats, and each seat is identified by a number. * * Write a function that, given the array of numbers representing the seat numbers, * will return the total number of ways that the cinema seating can be arranged, * so that no two people are sitting next to each other. * * The array of numbers will contain integers in the range 1 to 10^9. * * Example * * For seatNumbers = [1, 2, 3, 4, 5], the output should be * cinemaSeatAllocation(seatNumbers) = 10. * * Here are all the possible ways: * * [1, 2, 3, 4, 5] * [1, 3, 2, 4, 5] * [1, 3, 4, 2, 5] * [1, 3, 4, 5, 2] * [1, 4, 3, 2, 5] * [1, 4, 3, 5, 2] * [1, 4, 5, 3, 2] * [1, 4, 5, 2, 3] * [1, 5, 4, 2, 3] * [1, 5, 4, 3, 2] */ int cinemaSeatAllocation(int[] seatNumbers) { // calculate the factorial of the length of the array // this will give us the total number of ways the seats can be arranged int length = seatNumbers.length; int factorial = 1; for (int i = 2; i <= length; i++) { factorial *= i; } // now we need to remove the arrangements where two people are sitting next to each other // to do this, we will create a Set of integers, which will not allow duplicates // we will iterate through the seatNumbers array, and for each number, we will check if the number before it and the number after it are in the Set // if they are not, we will add the current number to the Set // if they are, we will remove the current number from the Set // at the end, the size of the Set will give us the number of ways that no two people are sitting next to each other Setset = new HashSet<>(); for (int i = 0; i < length; i++) { int current = seatNumbers[i]; int before = i - 1 >= 0 ? seatNumbers[i - 1] : -1; int after = i + 1 < length ? seatNumbers[i + 1] : -1; if (!set.contains(before) && !set.contains(after)) { set.add(current); } else { set.remove(current); } } return factorial / set.size(); }

This is a Python solution for the leetcode problem "cinema seat allocation". The output should only consist of exact code with comments and nothing else. def maxNumberOfFamilies(n, reservedSeats): # create a set of all the reserved seats reserved = set() for r, c in reservedSeats: reserved.add((r, c)) # create a set of all the possible seat arrangements possible = set() for i in range(1, n+1): # seats in the same row can't be together, so we need to check for each row for j in range(1, 11): # check if the current seat is available if (i, j) not in reserved: # check if the seats to the right and left are available if (i, j-1) not in reserved and (i, j+1) not in reserved: # if so, add this seat arrangement to the set possible.add((i, j, i, j+1)) # check if the seats to the right and left are available elif (i, j-1) not in reserved and (i, j+2) not in reserved: # if so, add this seat arrangement to the set possible.add((i, j, i, j+2)) # return the number of possible seat arrangements return len(possible)

var maxNumberOfFamilies = function(n, reservedSeats) { // create a hashmap to store the seats that are reserved // key will be the row number and value will be an array of the column numbers that are reserved let hashmap = {}; for (let i = 0; i < reservedSeats.length; i++) { let row = reservedSeats[i][0]; let col = reservedSeats[i][1]; if (!hashmap[row]) { hashmap[row] = [col]; } else { hashmap[row].push(col); } } // iterate through the hashmap and check how many families can be seated // a family can be seated if there are 4 consecutive empty seats together let numFamilies = 0; for (let key in hashmap) { let reserved = hashmap[key]; let canSit1 = canSit(reserved, 2, 5); let canSit2 = canSit(reserved, 4, 7); let canSit3 = canSit(reserved, 6, 9); if (canSit1 || canSit2 || canSit3) { numFamilies++; } } // for each row that doesn't have any reserved seats, we can seat 2 families numFamilies += (n - Object.keys(hashmap).length) * 2; return numFamilies; }; // helper function to check if 4 consecutive seats are available var canSit = function(reserved, start, end) { for (let i = start; i <= end; i++) { if (reserved.includes(i)) { return false; } } return true; }

There are a number of ways to approach this problem. One approach would be to use a greedy algorithm, where we iterate through the rows of seats and try to allocate seats to as many people as possible. Another approach would be to use a dynamic programming algorithm, where we try to find the optimal way to allocate seats. In this solution, we will use a greedy algorithm. We will iterate through the rows of seats and try to allocate seats to as many people as possible. We will keep track of the number of people who have been allocated a seat, and the number of people who have not been allocated a seat. If there are more people who have not been allocated a seat than there are seats available, we will not be able to allocate a seat to everyone and we will return false. Otherwise, we will return true. /* This is a C++ solution to the leetcode problem "cinema seat allocation". The problem is as follows: There are a number of rows of seats in a cinema. Each row has a certain number of seats. People queue up to buy tickets. The tickets are sold one by one, starting from the front of the queue. Each person buys a ticket and is given a seat. If there are no more seats available in the row, the person leaves the queue. The goal is to allocate seats to as many people as possible. */ #include#include using namespace std; bool cinemaSeatAllocation(vector & row, int n) { // keep track of the number of people who have been allocated a seat int numAllocated = 0; // keep track of the number of people who have not been allocated a seat int numNotAllocated = 0; // iterate through the rows of seats for (int i = 0; i < row.size(); i++) { // try to allocate seats to as many people as possible int seatsAvailable = row[i]; int numToAllocate = min(seatsAvailable, n - numAllocated); numAllocated += numToAllocate; numNotAllocated += n - numAllocated; // if there are more people who have not been allocated a seat than there are seats available, we cannot allocate a seat to everyone if (numNotAllocated > seatsAvailable) { return false; } } return true; } int main() { // test case 1 vector row1 = {3, 2, 2, 1}; int n1 = 4; bool result1 = cinemaSeatAllocation(row1, n1); cout << "Result 1: " << result1 << endl; // test case 2 vector row2 = {3, 2, 2, 1, 0}; int n2 = 4; bool result2 = cinemaSeatAllocation(row2, n2); cout << "Result 2: " << result2 << endl; return 0; }

using System; class GFG { // Function to calculate the number // of ways to allocate seats static int ways(int n, int m) { // Base cases if (n == 1 || m == 1) return 1; // Recursive calls return ways(n - 1, m) + ways(n, m - 1); } // Driver code public static void Main() { int n = 3, m = 3; Console.Write(ways(n, m)); } } // This code is contributed by Smitha Dinesh Semwal