Find if k bookings possible with given arrival and departure times

Difficulty Level : Medium
Last Updated : 14 Jun, 2021

A hotel manager has to process N advance bookings of rooms for the next season. His hotel has K rooms. Bookings contain an arrival date and a departure date. He wants to find out whether there are enough rooms in the hotel to satisfy the demand.

The idea is to sort the arrays and keep track of overlaps.

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Examples:

Input :  Arrivals :   [1 3 5]
         Departures : [2 6 8]
         K: 1
Output: False
Hotel manager needs at least two
rooms as the second and third 
intervals overlap.

Approach 1
The idea is store arrival and departure times in an auxiliary array with an additional marker to indicate whether the time is arrival or departure. Now sort the array. Process the sorted array, for every arrival increment active bookings. And for every departure, decrement. Keep track of maximum active bookings. If the count of active bookings at any moment is more than k, then return false. Else return true.

Below is the implementation of the above approach:

C++

// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
 
bool areBookingsPossible(int arrival[],
                         int departure[], int n, int k)
{
    vector<pair<int, int> > ans;
 
    // create a common vector both arrivals
    // and departures.
    for (int i = 0; i < n; i++) {
        ans.push_back(make_pair(arrival[i], 1));
        ans.push_back(make_pair(departure[i], 0));
    }
 
    // sort the vector
    sort(ans.begin(), ans.end());
 
    int curr_active = 0, max_active = 0;
 
    for (int i = 0; i < ans.size(); i++) {
 
        // if new arrival, increment current
        // guests count and update max active
        // guests so far
        if (ans[i].second == 1) {
            curr_active++;
            max_active = max(max_active,
                             curr_active);
        }
 
        // if a guest departs, decrement
        // current guests count.
        else
            curr_active--;
    }
 
    // if max active guests at any instant
    // were more than the available rooms,
    // return false. Else return true.
    return (k >= max_active);
}
 
int main()
{
    int arrival[] = { 1, 3, 5 };
    int departure[] = { 2, 6, 8 };
    int n = sizeof(arrival) / sizeof(arrival[0]);
    cout << (areBookingsPossible(arrival,
                                 departure, n, 1)
                 ? "Yes\n"
                 : "No\n");
    return 0;
}

Java

// Java implementation of the above approach
import java.io.*;
import java.util.*;
 
// User defined Pair class
class Pair {
  int x;
  int y;
 
  // Constructor
  public Pair(int x, int y)
  {
    this.x = x;
    this.y = y;
  }
}
 
// class to define user defined conparator
class Compare {
 
  static void compare(Pair arr[], int n)
  {
 
    // Comparator to sort the pair according to second element
    Arrays.sort(arr, new Comparator<Pair>() {
      @Override public int compare(Pair p1, Pair p2)
      {
        return p1.x - p2.x;
      }
    });
  }
}
 
class GFG
{
  static boolean areBookingsPossible(int arrival[],
                                     int departure[],
                                     int n, int k)
  {
    Pair ans[] = new Pair[2*n];
 
    // create a common vector both arrivals
    // and departures.
    int j = 0;
    for (int i = 0; i < n; i++)
    {
      ans[i + j] = new Pair(arrival[i], 1);
      ans[i + j + 1] = new Pair(departure[i], 0);
      j++;
    }
 
    // sort the vector
    Compare obj = new Compare();
    obj.compare(ans, 2 * n);     
    int curr_active = 0, max_active = 0;    
    for (int i = 0; i < 2 * n; i++)
    {
 
      // if new arrival, increment current
      // guests count and update max active
      // guests so far
      if (ans[i].y == 1)
      {
        curr_active++;
        max_active = Math.max(max_active,
                              curr_active);
      }
 
      // if a guest departs, decrement
      // current guests count.
      else
        curr_active--;
    }
 
    // if max active guests at any instant
    // were more than the available rooms,
    // return false. Else return true.
    return (k >= max_active);
  }
 
  // Driver code
  public static void main(String[] args)
  {
    int arrival[] = { 1, 3, 5 };
    int departure[] = { 2, 6, 8 };
    int n = arrival.length;
    System.out.println(areBookingsPossible(arrival, departure, n, 1) ? "Yes" : "No");
  }
}
 
// This code is contributed by divyeshrabadiya07

Python3

# Python3 code for the above approach.
def areBookingsPossible(arrival, departure, n, k):
 
    ans = []
 
    # Create a common vector both arrivals
    # and departures.
    for i in range(0, n):
        ans.append((arrival[i], 1))
        ans.append((departure[i], 0))
 
    # Sort the vector
    ans.sort()
    curr_active, max_active = 0, 0
 
    for i in range(0, len(ans)):
 
        # If new arrival, increment current
        # guests count and update max active
        # guests so far
        if ans[i][1] == 1:
            curr_active += 1
            max_active = max(max_active,
                             curr_active)
 
        # if a guest departs, decrement
        # current guests count.
        else:
            curr_active -= 1
 
    # If max active guests at any instant
    # were more than the available rooms,
    # return false. Else return true.
    return k >= max_active
 
# Driver Code
if __name__ == "__main__":
 
    arrival = [1, 3, 5]
    departure = [2, 6, 8]
    n = len(arrival)
     
    if areBookingsPossible(arrival,
                           departure, n, 1):
        print("Yes")
    else:
        print("No")
     
# This code is contributed by Rituraj Jain

Javascript

<script>
 
// JavaScript implementation of the above approach
 
function areBookingsPossible(arrival, departure, n, k)
{
    var ans = [];
 
    // create a common vector both arrivals
    // and departures.
    for (var i = 0; i < n; i++) {
        ans.push([arrival[i], 1]);
        ans.push([departure[i], 0]);
    }
 
    // sort the vector
    ans.sort();
 
    var curr_active = 0, max_active = 0;
 
    for (var i = 0; i < ans.length; i++) {
 
        // if new arrival, increment current
        // guests count and update max active
        // guests so far
        if (ans[i][1] == 1) {
            curr_active++;
            max_active = Math.max(max_active,
                             curr_active);
        }
 
        // if a guest departs, decrement
        // current guests count.
        else
            curr_active--;
    }
 
    // if max active guests at any instant
    // were more than the available rooms,
    // return false. Else return true.
    return (k >= max_active);
}
 
var arrival = [1, 3, 5];
var departure = [2, 6, 8];
var n = arrival.length;
document.write(areBookingsPossible(arrival,
                             departure, n, 1)
             ? "Yes"
             : "No");
 
</script>

Output

No

Time Complexity: O(n Log n)
Auxiliary Space: O(n)

Approach 2
The idea is to simply sort the 2 arrays (Array for arrival dates and Array for departure dates) first.
Now, the next step would be to check how many overlaps are present in one particular range. If the number of overlaps are greater than the number of rooms, we can say that we have less rooms to accommodate guests.

So, for a particular range where arrival date(ith of Arrival array) being the start date and departure date(ith of departure array) being the end date, overlap can be only possible if the next arrival dates(from i+1th) are less than end date of the range and greater than or equal to start date of the range (Since this is a sorted array, we don’t need to take care about the latter condition).

Considering the fact, that we have sorted array, we directly need to check if the next Kth (i+Kth) arrival date falls in the range, if it does, all the dates before that arrival date will also fall in the taken range, resulting in K+1 overlaps with the range in question, hence exceeding the number of rooms.

Following is the Implementation of above Approach –

C++

// C++ code implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
 
string areBookingsPossible(int A[], int B[],
                           int K, int N)
{
    sort(A, A + N);
    sort(B, B + N);
     
    for(int i = 0; i < N; i++)
    {
        if (i + K < N && A[i + K] < B[i])
        {
            return "No";
        }
    }
    return "Yes";
}
 
// Driver Code
int main()
{
    int arrival[] = { 1, 2, 3 };
    int departure[] = { 2, 3, 4 };
    int N = sizeof(arrival) / sizeof(arrival[0]);
    int K = 1;
   
    cout << (areBookingsPossible(
        arrival, departure, K, N));
 
    return 0;
}
 
// This code is contributed by rag2127

Java

// Java code implementation of the above approach
import java.util.*;
class GFG
{
      
static String areBookingsPossible(int []A, int []B,
                                  int K)
{
    Arrays.sort(A);
    Arrays.sort(B);  
    for(int i = 0; i < A.length; i++)
    {
        if (i + K < A.length && A[i + K] < B[i])
        {
            return "No";
        }
    }
    return "Yes";
}
  
// Driver Code
public static void main(String []s)
{
    int []arrival = { 1, 2, 3 };
    int []departure = { 2, 3, 4 };
    int K = 1;    
    System.out.print(areBookingsPossible(
        arrival, departure, K));
}
}
 
// This code is contributed by Pratham76

Python

# Python Code Implementation of the above approach
def areBookingsPossible(A, B, K):
        A.sort()
        B.sort()
        for i in range(len(A)):
            if i+K < len(A) and A[i+K] < B[i] :
                return "No"
        return "Yes"
 
if __name__ == "__main__":
    arrival = [1, 2, 3] 
    departure = [2, 3, 4]  
    K = 1
    print areBookingsPossible(arrival,departure,K)
 
# This code was contributed by Vidhi Modi

C#

// C# code implementation of the above approach
using System;
 
class GFG{
     
static string areBookingsPossible(int []A, int []B,
                                  int K)
{
    Array.Sort(A);
    Array.Sort(B);
     
    for(int i = 0; i < A.Length; i++)
    {
        if (i + K < A.Length && A[i + K] < B[i])
        {
            return "No";
        }
    }
    return "Yes";
}
 
// Driver Code
static void Main()
{
    int []arrival = { 1, 2, 3 };
    int []departure = { 2, 3, 4 };
    int K = 1;
     
    Console.Write(areBookingsPossible(
        arrival, departure, K));
}
}
 
// This code is contributed by rutvik_56

Javascript

<script>
     
// Javascript code implementation of
// the above approach
function areBookingsPossible(A, B, K, N)
{
    A.sort();
    B.sort();
 
    for(let i = 0; i < N; i++)
    {
        if (i + K < N && A[i + K] < B[i])
        {
            return "No";
        }
    }
    return "Yes";
}
 
// Driver code
let arrival = [ 1, 2, 3 ];
let departure = [ 2, 3, 4 ];
let N = arrival.length;
let K = 1;
     
document.write(areBookingsPossible(
    arrival, departure, K, N));
   
// This code is contributed by suresh07
   
</script>

Output

Yes

Time Complexity: O(n Log n)
Auxiliary Space: O(n) used by Python sort

My Personal Notes

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