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Count maximum number of cars parked at the same time

  • Difficulty Level : Medium
  • Last Updated : 23 Aug, 2021

Given a 2d array arr[][] with each row representing a pair representing entry and exit time of a car in a parking lot, the task is to calculate the maximum number of cars that can be parked at the same time.

Examples:

Input: arr[][] = {{1012, 1136}, {1317, 1417}, {1015, 1020}}
Output: 2
Explanation:
1st car entered at 10:12 and exits at 11:36 and 3rd car entered at 10:15 and exits at 10:20.
Therefore, 1st and 3rd car are present at the same time.

Input: arr[][] = {{1120, 1159}, {1508, 1529}, {1508, 1527}, {1503, 1600}, {1458, 1629}, {1224, 1313}}
Output: 4
Explanation: 2nd, 3rd, 4th and 5th cars are present at the same time.

Approach: The idea is to use Kadane’s algorithm to solve this problem. Follow the steps below to solve the problem:



  • Initialize a vector of pairs to store the entry or exit time as the first element of a pair and true as the second element of a pair, if corresponding time stored is entry time. Otherwise, store as false.
  • Sort the vector in non-decreasing order of time.
  • Initialize two variables, say curMax, to look for all true contiguous segments of the array, and maxFinal, to keep track of longest true contiguous segment among all true segments.
  • Iterate over the range [0, 2*N – 1]:
    • If a car entered, increment curMax by 1.
    • Otherwise:
      • If curMax > maxFinal, update maxFinal = curMax.
      • Whenever a car exits, subtract curMax by 1.
  • Print maxFinal as the answer.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to count maximum number
// of cars parked at the same
int maxCars(int arr[][2], int N)
{
    // Stores info about
    // entry and exit times
    pair<int, bool> a[2 * N];
 
    // Convert given array to new array
    for (int i = 0; i < N; i++) {
        a[2 * i] = { arr[i][0], true };
        a[2 * i + 1] = { arr[i][1], false };
    }
 
    // Sort array in ascending
    // order of time
    sort(a, a + 2 * N);
 
    // Stores current maximum
    // at every iteration
    int curMax = 0;
 
    // Stores final maximum number
    // of cars parked at any time
    int maxFinal = 0;
 
    // Traverse the array
    for (int i = 0; i < 2 * N; ++i) {
 
        // When car entered
        if (a[i].second) {
            curMax++;
        }
 
        // When car exits
        else {
            if (curMax > maxFinal) {
                maxFinal = curMax;
            }
            curMax--;
        }
    }
 
    // Print the answer
    cout << maxFinal;
}
 
// Driver Code
int main()
{
    // Given array
    int arr[][2] = { { 1012, 1136 },
                     { 1317, 1412 },
                     { 1015, 1020 } };
 
    // Size of the array
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    maxCars(arr, N);
 
    return 0;
}

Java




// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG{
 
// Pair class
static class pair
{
    int first;
    boolean second;
 
    pair(int first, boolean second)
    {
        this.first = first;
        this.second = second;
    }
}
 
// Function to count maximum number
// of cars parked at the same
static void maxCars(int arr[][], int N)
{
     
    // Stores info about
    // entry and exit times
    pair a[] = new pair[2 * N];
 
    // Convert given array to new array
    for(int i = 0; i < N; i++)
    {
        a[2 * i] = new pair(arr[i][0], true);
        a[2 * i + 1] = new pair(arr[i][1], false);
    }
 
    // Sort array in ascending
    // order of time
    Arrays.sort(a, (p1, p2) -> p1.first - p2.first);
 
    // Stores current maximum
    // at every iteration
    int curMax = 0;
 
    // Stores final maximum number
    // of cars parked at any time
    int maxFinal = 0;
 
    // Traverse the array
    for(int i = 0; i < 2 * N; ++i)
    {
         
        // When car entered
        if (a[i].second)
        {
            curMax++;
        }
 
        // When car exits
        else
        {
            if (curMax > maxFinal)
            {
                maxFinal = curMax;
            }
            curMax--;
        }
    }
 
    // Print the answer
    System.out.println(maxFinal);
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given array
    int arr[][] = { { 1012, 1136 },
                    { 1317, 1412 },
                    { 1015, 1020 } };
 
    // Size of the array
    int N = arr.length;
 
    // Function Call
    maxCars(arr, N);
}
}
 
// This code is contributed by Kingash

Python3




# Python3 program for the above approach
 
# Function to count maximum number
# of cars parked at the same
def maxCars(arr, N):
   
    # Stores info about
    # entry and exit times
    a = [[0,True] for i in range(2 * N)]
 
    # Convert given array to new array
    for i in range(N):
        a[2 * i] = [arr[i][0], True]
        a[2 * i + 1] = [arr[i][1], False]
 
    # Sort array in ascending
    # order of time
    a = sorted(a)
 
    # Stores current maximum
    # at every iteration
    curMax = 0
 
    # Stores final maximum number
    # of cars parked at any time
    maxFinal = 0
 
    # Traverse the array
    for i in range(2*N):
 
        # When car entered
        if (a[i][1]):
            curMax += 1
        # When car exits
        else:
            if (curMax > maxFinal):
                maxFinal = curMax
            curMax -= 1
 
    # Print answer
    print (maxFinal)
 
# Driver Code
if __name__ == '__main__':
    # Given array
    arr= [ [ 1012, 1136 ],
         [ 1317, 1412 ],
         [ 1015, 1020 ]]
 
    # Size of the array
    N = len(arr)
 
    # Function Call
    maxCars(arr, N)
 
# This code is contributed by mohit kumar 29.

C#




// C# program for the above approach
using System;
class GFG
{
    // Function to count maximum number
    // of cars parked at the same
    static void maxCars(int[,] arr, int N)
    {
       
        // Stores info about
        // entry and exit times
        Tuple<int, bool>[] a = new Tuple<int, bool>[2 * N];
      
        // Convert given array to new array
        for (int i = 0; i < N; i++) {
            a[2 * i] = new Tuple<int, bool>(arr[i,0], true);
            a[2 * i + 1] = new Tuple<int, bool>(arr[i,1], false);
        }
      
        // Stores current maximum
        // at every iteration
        int curMax = 1;
      
        // Stores final maximum number
        // of cars parked at any time
        int maxFinal = 0;
      
        // Traverse the array
        for (int i = 0; i < 2 * N; ++i) {
      
            // When car entered
            if (a[i].Item2) {
                curMax++;
            }
      
            // When car exits
            else {
                if (curMax > maxFinal) {
                    maxFinal = curMax;
                }
                curMax--;
            }
        }
      
        // Print the answer
        Console.WriteLine(maxFinal);
    }
 
  static void Main ()
  {
    // Given array
    int[,] arr = { { 1012, 1136 },
                    { 1317, 1412 },
                    { 1015, 1020 } };
  
    // Size of the array
    int N = 2;
  
    // Function Call
    maxCars(arr, N);
  }
}
 
// This code is contributed by suresh07.

Javascript




<script>
 
// JavaScript program for the above approach
 
// Pair class
class pair
{
    constructor(first,second)
    {
        this.first = first;
        this.second = second;
    }
}
 
// Function to count maximum number
// of cars parked at the same
function maxCars(arr,N)
{
    // Stores info about
    // entry and exit times
    let a = new Array(2 * N);
  
    // Convert given array to new array
    for(let i = 0; i < N; i++)
    {
        a[2 * i] = new pair(arr[i][0], true);
        a[2 * i + 1] = new pair(arr[i][1], false);
    }
  
    // Sort array in ascending
    // order of time
    a.sort(function(p1, p2){return p1.first - p2.first});
  
    // Stores current maximum
    // at every iteration
    let curMax = 0;
  
    // Stores final maximum number
    // of cars parked at any time
    let maxFinal = 0;
  
    // Traverse the array
    for(let i = 0; i < 2 * N; ++i)
    {
          
        // When car entered
        if (a[i].second)
        {
            curMax++;
        }
  
        // When car exits
        else
        {
            if (curMax > maxFinal)
            {
                maxFinal = curMax;
            }
            curMax--;
        }
    }
  
    // Print the answer
    document.write(maxFinal+"<br>");
}
 
// Driver Code
let arr=[[ 1012, 1136 ],
                    [ 1317, 1412 ],
                    [ 1015, 1020 ]];
// Size of the array
let N = arr.length;
 
// Function Call
maxCars(arr, N);
 
 
// This code is contributed by unknown2108
 
</script>
Output: 
2

 

Time Complexity: O(N)
Auxiliary Space: O(N)

 

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