Maximize the profit after selling the tickets

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: 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
Java
Python3
C#
Javascript
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std; // Function to return the maximized profitint 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 codeint 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 approachimport java.util.*; class GFG { // Function to return the maximized profitstatic 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 Codepublic 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 approachimport 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 Codeseats = [2, 3, 4, 5, 1]k = len(seats)n = 6print(maxProfit(seats, k, n)) '''Code is written by Rajat Kumar (GLAU)'''
// C# implementation of the approachusing System;using System.Collections.Generic; class GFG{     // Function to return the maximized profitstatic 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 Codestatic 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>
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++
Java
Python3
C#
Javascript
#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 approachdef 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 = 6print(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>
22
Time Complexity: O(k logk), where k is the size of the given array of seatsAuxiliary Space: O(1)
ankthon
rutvik_56
divyeshrabadiya07
decode2207
parthagarwal1962000
rajsanghavi9
shivanisinghss2110
rrrtnx
sumitgumber28
rajatkumargla19
suryap
shinjanpatra
singhh3010
Microsoft
priority-queue
Arrays
Heap
Microsoft
Arrays
Heap
priority-queue
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