Maximize the profit after selling the tickets

Given an array seats[] where seat[i] is the number of vacant seats in the ith row in a stadium for a cricket match. There are N people in a queue waiting to buy the tickets. Each seat costs equal to the number of vacant seats in the row it belongs to. The task is to maximize the profit by selling the tickets to N people.

Examples:

Input: seats[] = {2, 1, 1}, N = 3
Output: 4
Person 1: Sell the seat in the row with
2 vacant seats, seats = {1, 1, 1}
Person 2: All the rows have 1 vacant
seat each, seats[] = {0, 1, 1}
Person 3: seats[] = {0, 0, 1}

Input: seats[] = {2, 3, 4, 5, 1}, N = 6
Output: 22

Approach: In order to maximize the profit, the ticket must be for the seat in a row which has the maximum number of vacant seats and the number of vacant seats in that row will be decrement by 1 as one of the seats has just been sold. All the persons can be sold a seat ticket until there are vacant seats. This can be computed efficiently with the help of a priority_queue.



Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the maximized profit
int maxProfit(int seats[], int k, int n)
{
  
    // Push all the vacant seats
    // in a priority queue
    priority_queue<int> pq;
    for (int i = 0; i < k; i++)
        pq.push(seats[i]);
  
    // To store the maximized profit
    int profit = 0;
  
    // To count the people that
    // have been sold a ticket
    int c = 0;
    while (c < n) {
  
        // Get the maximimum number of
        // vacant seats for any row
        int top = pq.top();
  
        // Remove it from the queue
        pq.pop();
  
        // If there are no vacant seats
        if (top == 0)
            break;
  
        // Update the profit
        profit = profit + top;
  
        // Push the updated status of the
        // vacant seats in the current row
        pq.push(top - 1);
  
        // Update the count of persons
        c++;
    }
    return profit;
}
  
// Driver code
int main()
{
    int seats[] = { 2, 3, 4, 5, 1 };
    int k = sizeof(seats) / sizeof(int);
    int n = 6;
  
    cout << maxProfit(seats, k, n);
  
    return 0;
}

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Java

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// Java implementation of the approach 
import java.util.*;
  
class GFG {
  
// Function to return the maximized profit 
static int maxProfit(int seats[], int k, int n) 
      
    // Push all the vacant seats 
    // in a priority queue 
    PriorityQueue<Integer> pq;
    pq = new PriorityQueue<>(Collections.reverseOrder());
      
    for(int i = 0; i < k; i++) 
       pq.add(seats[i]); 
  
    // To store the maximized profit 
    int profit = 0
  
    // To count the people that 
    // have been sold a ticket 
    int c = 0
    while (c < n)
    
  
        // Get the maximimum number of 
        // vacant seats for any row 
        int top = pq.remove(); 
  
        // If there are no vacant seats 
        if (top == 0
            break
  
        // Update the profit 
        profit = profit + top; 
  
        // Push the updated status of the 
        // vacant seats in the current row 
        pq.add(top - 1); 
  
        // Update the count of persons 
        c++; 
    
      
    return profit; 
      
// Driver Code
public static void main(String args[])
{
    int seats[] = { 2, 3, 4, 5, 1 }; 
    int k = seats.length;
    int n = 6;
  
    System.out.println(maxProfit(seats, k ,n));
}
}
  
// This code is contributed by rutvik_56

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Python3

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# Python3 implementation of the approach 
  
# Function to return the maximized profit 
def maxProfit(seats, k, n) : 
  
    # Push all the vacant seats 
    # in a priority queue 
    pq = []; 
    for i in range(k) :
        pq.append(seats[i]);
      
    # for maintaining the property of max heap
    pq.sort(reverse = True);
      
    # To store the maximized profit 
    profit = 0
  
    # To count the people that 
    # have been sold a ticket 
    c = 0
    while (c < n) :
          
        # for maintaining the property of max heap
        pq.sort(reverse = True);
          
        # Get the maximimum number of 
        # vacant seats for any row 
        top = pq[0];
          
        # Remove it from the queue
        pq.pop(0);
          
        # If there are no vacant seats
        if (top == 0) :
            break;
              
        # Update the profit
        profit = profit + top;
          
        # Push the updated status of the
        # vacant seats in the current row
        pq.append(top - 1);
          
        # Update the count of persons
        c += 1
      
    return profit; 
  
# Driver code 
if __name__ == "__main__"
  
    seats = [ 2, 3, 4, 5, 1 ]; 
    k = len(seats); 
    n = 6
  
    print(maxProfit(seats, k, n)); 
  
# This code is contributed by AnkitRai01

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Output:

22

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Improved By : AnkitRai01, rutvik_56