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 ith 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 ticketsint 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 codeint main(){ int seats[] = { 1, 2, 4 }; int M = sizeof(seats) / sizeof(int); int N = 3; cout << maxAmount(N, M, seats); return 0;} |
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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 |
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Python 3
# Python 3 implementation of the approach# Function to return the maximum amount# that can be collected by selling ticketsdef 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 codeif __name__ == '__main__': seats = [1, 2, 4] M = len(seats) N = 3 print(maxAmount(N, M, seats))# This code is contributed by Surendra_Gangwar |
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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. |
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Output:
9
