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

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 

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

# Python3 implementation of the approach
import heapq
# 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 = seats
    # for maintaining the property of max heap
    heapq._heapify_max(q)
    # 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[0] > 0 then update the collected
    # amount with the top of the priority
    # queue
    while ticketSold < N and q[0] > 0:
        # updating ans
        # with maximum number of tickets
        ans += q[0]
        q[0] -= 1
        if q[0] == 0:
            break
        # for maintaining the property of max heap
        heapq._heapify_max(q)
        ticketSold += 1
    return ans


# Driver Code
seats = [1, 2, 4]
M = len(seats)
N = 3
print(maxAmount(M, N, seats))

'''Code is written by Rajat Kumar (GLAU)'''

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:

9

Time Complexity: O(N*logM) where N is tickets to be sold.

Explanation: In the condition of the while loop:  ticketSold < N and q.top() >0, Value of N is determining the time required and here we need to maintain heap every time that needs O(Log(M))….so overall complexity will be O(N* Log(m)) where M is the size of given array.

Auxiliary Space: O(M) to store heap q.