# Maximize the profit after selling the tickets

• Difficulty Level : Medium
• Last Updated : 28 Jun, 2022

Given 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:
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++14

 `// C++ implementation of the approach``#include ``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 maximum 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;``}`

## Java

 `// 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 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 maximum 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`

## Python3

 `# Python3 implementation of the approach``import` `heapq``# Function to return the maximized profit`  `def` `maxProfit(seats, k, n):``    ``# Push all the vacant seats``    ``# in a max heap``    ``pq ``=` `seats``    ``# for maintaining the property of max heap``    ``heapq._heapify_max(pq)``    ``# To store the maximized profit``    ``profit ``=` `0``    ``while` `n > ``0``:``        ``# updating the profit value``        ``# with maximum number of vacant seats``        ``profit ``+``=` `pq[``0``]``        ``pq[``0``] ``-``=` `1``        ``# If there are no vacant seats``        ``if` `pq[``0``] ``=``=` `0``:``            ``break``        ``# for maintaining the property of max heap``        ``heapq._heapify_max(pq)``        ``# decrementing the ticket count``        ``n ``-``=` `1` `    ``return` `profit`  `# Driver Code``seats ``=` `[``2``, ``3``, ``4``, ``5``, ``1``]``k ``=` `len``(seats)``n ``=` `6``print``(maxProfit(seats, k, n))` `'''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 maximized profit``static` `int` `maxProfit(``int``[] seats, ``int` `k, ``int` `n)``{``    ` `    ``// Push all the vacant seats``    ``// in a priority queue``    ``List<``int``> pq = ``new` `List<``int``>();``    ``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 maximum number of``        ``// vacant seats for any row``        ``pq.Sort();``        ``pq.Reverse();``        ``int` `top = pq;``  ` `        ``// Remove it from the queue``        ``pq.RemoveAt(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.Add(top - 1);``  ` `        ``// Update the count of persons``        ``c++;``    ``}``    ``return` `profit;``}` `// Driver Code``static` `void` `Main()``{``    ``int``[] seats = { 2, 3, 4, 5, 1 };``    ``int` `k = seats.Length;``    ``int` `n = 6;``    ` `    ``Console.Write(maxProfit(seats, k, n));``}``}` `// This code is contributed by divyeshrabadiya07`

## Javascript

 ``

Output

`22`

Time complexity: O(n*log(n))

Auxiliary Space: O(n)

Sliding Window approach:

The problem can also be solved using the sliding window technique.

• For each person we need to sell ticket that has the maximum price and decrement its value by 1.
• Sort the array seats.
• Maintain two pointers pointing at the current maximum and next maximum number of seats .
• We iterate till our n>0 and there is a second largest element in the array.
• In each iteration if seats[i] > seats[j] ,we add the value at seats[i] ,min(n, i-j) times to our answer and decrement the value at ith index else we find j such that seats[j]<seats[i]. If there is no such j we break.
• If at the end of iteration our n>0 and seats[i]!=0 we add seats[i] till n>0 and seats[i]!=0.

## C++

 `#include ``using` `namespace` `std;``int` `maxProfit(``int` `seats[],``int` `k, ``int` `n)``{``    ``sort(seats,seats+k);``    ``int` `ans = 0;``    ``int` `i = k - 1;``    ``int` `j = k - 2;``    ``while` `(n > 0 && j >= 0) {``        ``if` `(seats[i] > seats[j]) {``            ``ans = ans + min(n, (i - j)) * seats[i];``            ``n = n - (i - j);``            ``seats[i]--;``        ``}``        ``else` `{``            ``while` `(j >= 0 && seats[j] == seats[i])``                ``j--;``            ``if` `(j < 0)``                ``break``;``            ``ans = ans + min(n, (i - j)) * seats[i];``            ``n = n - (i - j);``            ``seats[i]--;``        ``}``    ``}``    ``while` `(n > 0 && seats[i] != 0) {``        ``ans = ans + min(n, k) * seats[i];``        ``n -= k;``        ``seats[i]--;``    ``}``    ``return` `ans;``}``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;``}`

## Java

 `// Java program for the above approach` `import` `java.util.Arrays;` `class` `GFG {` `    ``static` `int` `maxProfit(``int` `seats[], ``int` `k, ``int` `n)``    ``{``        ``Arrays.sort(seats, ``0``, k);``        ``int` `ans = ``0``;``        ``int` `i = k - ``1``;``        ``int` `j = k - ``2``;``        ``while` `(n > ``0` `&& j >= ``0``) {``            ``if` `(seats[i] > seats[j]) {``                ``ans = ans + Math.min(n, (i - j)) * seats[i];``                ``n = n - (i - j);``                ``seats[i]--;``            ``}``            ``else` `{` `                ``while` `(j >= ``0` `&& seats[j] == seats[i])``                    ``j--;` `                ``if` `(j < ``0``)``                    ``break``;` `                ``ans = ans + Math.min(n, (i - j)) * seats[i];``                ``n = n - (i - j);``                ``seats[i]--;``            ``}``        ``}``        ``while` `(n > ``0` `&& seats[i] != ``0``) {``            ``ans = ans + Math.min(n, k) * seats[i];``            ``n -= k;``            ``seats[i]--;``        ``}``        ``return` `ans;``    ``}` `    ``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 rajsanghavi9.`

## Python3

 `# Python3 program for the above approach``def` `maxProfit(seats,k, n):` `    ``seats.sort()``    ` `    ``ans ``=` `0``    ``i ``=` `k ``-` `1``    ``j ``=` `k ``-` `2``    ``while` `(n > ``0` `and` `j >``=` `0``):``        ``if` `(seats[i] > seats[j]):``            ``ans ``=` `ans ``+` `min``(n, (i ``-` `j)) ``*` `seats[i]``            ``n ``=` `n ``-` `(i ``-` `j)``            ``seats[i] ``-``=` `1``        ` `        ``else``:``            ``while` `(j >``=` `0` `and` `seats[j] ``=``=` `seats[i]):``                ``j ``-``=` `1``            ``if` `(j < ``0``):``                ``break``            ``ans ``=` `ans ``+` `min``(n, (i ``-` `j)) ``*` `seats[i]``            ``n ``=` `n ``-` `(i ``-` `j)``            ``seats[i] ``-``=` `1``        ` `    ` `    ``while` `(n > ``0` `and` `seats[i] !``=` `0``):``        ``ans ``=` `ans ``+` `min``(n, k) ``*` `seats[i]``        ``n ``-``=` `k``        ``seats[i] ``-``=` `1``    ` `    ``return` `ans` `seats ``=` `[``2``, ``3``, ``4``, ``5``, ``1``]``k ``=` `len``(seats)``n ``=` `6``print``(maxProfit(seats, k, n))` `# This code is contributed by shinjanpatra`

## C#

 `// C# program for the above approach` `using` `System;` `class` `GFG {` `    ``static` `int` `maxProfit(``int` `[]seats, ``int` `k, ``int` `n)``    ``{``        ``Array.Sort(seats, 0, k);``        ``int` `ans = 0;``        ``int` `i = k - 1;``        ``int` `j = k - 2;``        ``while` `(n > 0 && j >= 0) {``            ``if` `(seats[i] > seats[j]) {``                ``ans = ans + Math.Min(n, (i - j)) * seats[i];``                ``n = n - (i - j);``                ``seats[i]--;``            ``}``            ``else` `{` `                ``while` `(j >= 0 && seats[j] == seats[i])``                    ``j--;` `                ``if` `(j < 0)``                    ``break``;` `                ``ans = ans + Math.Min(n, (i - j)) * seats[i];``                ``n = n - (i - j);``                ``seats[i]--;``            ``}``        ``}``        ``while` `(n > 0 && seats[i] != 0) {``            ``ans = ans + Math.Min(n, k) * seats[i];``            ``n -= k;``            ``seats[i]--;``        ``}``        ``return` `ans;``    ``}` `    ``public` `static` `void` `Main(String[] args)``    ``{``        ``int` `[]seats = { 2, 3, 4, 5, 1 };``        ``int` `k = seats.Length;``        ``int` `n = 6;` `        ``Console.Write(maxProfit(seats, k, n));``    ``}``}` `// This code is contributed by shivanisinghss2110`

## Javascript

 ``

Output

`22`

Time Complexity: O(k logk), where k is the size of the given array of seats
Auxiliary Space: O(1)

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