Find the maximum amount that can be collected by selling movie tickets

Difficulty Level : Medium
Last Updated : 15 Apr, 2021

Given an integer N and an array seats[] where N is the number of people standing in a line to buy a movie ticket and seat[i] is the number of empty seats in the i^th row of the movie theater. The task is to find the maximum amount a theater owner can make by selling movie tickets to N people. Price of a ticket is equal to the maximum number of empty seats among all the rows.
Example:

Input: seats[] = {1, 2, 4}, N = 3
Output: 9

Person Empty Seats Ticket Cost
1 1 2 4 4
2 1 2 3 3
3 1 2 2 2
4 + 3 + 2 = 9
Input: seats[] = {2, 3, 5, 3}, N = 4
Output: 15

Person	Empty Seats	Ticket Cost
1	1 2 4	4
2	1 2 3	3
3	1 2 2	2

Approach: This problem can be solved by using a priority queue that will store the count of empty seats for every row and the maximum among them will be available at the top.

Create an empty priority_queue q and traverse the seats[] array and insert all element into the priority_queue.
Initialize two integer variable ticketSold = 0 and ans = 0 that will store the number of tickets sold and the total collection of the amount so far.
Now check while ticketSold < N and q.top() > 0 then remove the top element from the priority_queue and update ans by adding top element of the priority queue. Also store this top value in a variable temp and insert temp – 1 back to the priority_queue.
Repeat these steps until all the people have been sold the tickets and print the final result.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the maximum amount
// that can be collected by selling tickets
int maxAmount(int M, int N, int seats[])
{
 
    // Priority queue that stores
    // the count of empty seats
    priority_queue<int> q;
 
    // Insert each array element
    // into the priority queue
    for (int i = 0; i < M; i++) {
        q.push(seats[i]);
    }
 
    // To store the  total
    // number of tickets sold
    int ticketSold = 0;
 
    // To store the total amount
    // of collection
    int ans = 0;
 
    // While tickets sold are less than N
    // and q.top > 0 then update the collected
    // amount with the top of the priority
    // queue
    while (ticketSold < N && q.top() > 0) {
        ans = ans + q.top();
        int temp = q.top();
        q.pop();
        q.push(temp - 1);
        ticketSold++;
    }
 
    return ans;
}
 
// Driver code
int main()
{
    int seats[] = { 1, 2, 4 };
    int M = sizeof(seats) / sizeof(int);
    int N = 3;
 
    cout << maxAmount(N, M, seats);
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
 
class GFG
{
    static int[] seats = new int[]{ 1, 2, 4 };
 
    // Function to return the maximum amount
    // that can be collected by selling tickets
    public static int maxAmount(int M, int N)
    {
 
        // Priority queue that stores
        // the count of empty seats
        PriorityQueue<Integer> q =
            new PriorityQueue<Integer>(Collections.reverseOrder());
     
        // Insert each array element
        // into the priority queue
        for (int i = 0; i < M; i++)
        {
            q.add(seats[i]);
        }
 
        // To store the total
        // number of tickets sold
        int ticketSold = 0;
     
        // To store the total amount
        // of collection
        int ans = 0;
     
        // While tickets sold are less than N
        // and q.top > 0 then update the collected
        // amount with the top of the priority
        // queue
        while (ticketSold < N && q.peek() > 0)
        {
            ans = ans + q.peek();
            int temp = q.peek();
            q.poll();
            q.add(temp - 1);
            ticketSold++;
        }
        return ans;
    }
     
    // Driver code
    public static void main(String[] args)
    {
        int M = seats.length;
        int N = 3;
     
        System.out.print(maxAmount(M, N));
    }
}
 
// This code is contributed by Sanjit_Prasad

Python 3

# Python 3 implementation of the approach
 
# Function to return the maximum amount
# that can be collected by selling tickets
def maxAmount(M, N, seats):
     
    # Priority queue that stores
    # the count of empty seats
    q = []
 
    # Insert each array element
    # into the priority queue
    for i in range(M):
        q.append(seats[i])
 
    # To store the total
    # number of tickets sold
    ticketSold = 0
 
    # To store the total amount
    # of collection
    ans = 0
 
    # While tickets sold are less than N
    # and q.top > 0 then update the collected
    # amount with the top of the priority
    # queue
    q.sort(reverse = True)
    while (ticketSold < N and q[0] > 0):
        ans = ans + q[0]
        temp = q[0]
        q = q[1:]
        q.append(temp - 1)
        q.sort(reverse = True)
        ticketSold += 1
 
    return ans
 
# Driver code
if __name__ == '__main__':
    seats = [1, 2, 4]
    M = len(seats)
    N = 3
 
    print(maxAmount(N, M, seats))
 
# This code is contributed by Surendra_Gangwar

C#

// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
     
    // Function to return the maximum amount
    // that can be collected by selling tickets
    static int maxAmount(int M, int N, int[] seats)
    {
       
        // Priority queue that stores
        // the count of empty seats
        List<int> q = new List<int>();
       
        // Insert each array element
        // into the priority queue
        for (int i = 0; i < M; i++) {
            q.Add(seats[i]);
        }
         
        q.Sort();
        q.Reverse();
       
        // To store the  total
        // number of tickets sold
        int ticketSold = 0;
       
        // To store the total amount
        // of collection
        int ans = 0;
       
        // While tickets sold are less than N
        // and q.top > 0 then update the collected
        // amount with the top of the priority
        // queue
        while (ticketSold < N && q[0] > 0) {
            ans = ans + q[0];
            int temp = q[0];
            q.RemoveAt(0);
            q.Add(temp - 1);
            q.Sort();
            q.Reverse();
            ticketSold++;
        }  
        return ans;
    }
 
  // Driver code
  static void Main()
  {
    int[] seats = { 1, 2, 4 };
    int M = seats.Length;
    int N = 3;
    Console.WriteLine(maxAmount(N, M, seats));
  }
}
 
// This code is contributed by divyesh072019.

Javascript

<script>
 
    // Javascript implementation of the approach
     
    // Function to return the maximum amount
    // that can be collected by selling tickets
    function maxAmount(M, N, seats)
    {
        
        // Priority queue that stores
        // the count of empty seats
        let q = [];
        
        // Insert each array element
        // into the priority queue
        for (let i = 0; i < M; i++) {
            q.push(seats[i]);
        }
          
        q.sort(function(a, b){return a - b});
        q.reverse();
        
        // To store the  total
        // number of tickets sold
        let ticketSold = 0;
        
        // To store the total amount
        // of collection
        let ans = 0;
        
        // While tickets sold are less than N
        // and q.top > 0 then update the collected
        // amount with the top of the priority
        // queue
        while ((ticketSold < N) && (q[0] > 0)) {
            ans = ans + q[0];
            let temp = q[0];
            q.shift();
            q.push(temp - 1);
            q.sort(function(a, b){return a - b});
            q.reverse();
            ticketSold++;
        } 
        return ans;
    }
     
    let seats = [ 1, 2, 4 ];
    let M = seats.length;
    let N = 3;
    document.write(maxAmount(N, M, seats));
         
</script>

Output:

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