Maximize the profit after selling the tickets
Last Updated :
12 Jul, 2025
Given array seats[] where seats[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++14
// 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 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 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 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 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# 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[0];
// 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
<script>
// Javascript implementation of the approach
// Function to return the maximized profit
function maxProfit(seats, k, n)
{
// Push all the vacant seats
// in a priority queue
let priorityQueue = counter.map((item) => item);
// To store the maximized profit
let profit = 0;
while (n != 0)
{
// Get the maximum number of
// vacant seats for any row
priorityQueue.sort((a,b) => b - a);
let top = priorityQueue[0];
// Remove it from the queue
priorityQueue.shift();
// 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
priorityQueue.push(top - 1);
// Update the count of persons
n--;
}
return profit;
}
let seats = [ 2, 3, 4, 5, 1 ];
let k = seats.length;
let n = 6;
document.write(maxProfit(seats, k, n));
</script>
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 <bits/stdc++.h>
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 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 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# 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
<script>
function maxProfit(seats,k, n)
{
seats.sort();
var ans = 0;
var i = k - 1;
var 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;
}
var seats = [2, 3, 4, 5, 1];
var k = seats.length;
var n = 6;
document.write(maxProfit(seats, k, n));
// This code is contributed by rrrtnx.
</script>
Time Complexity: O(k logk), where k is the size of the given array of seats
Auxiliary Space: O(1)
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