Find if k bookings possible with given arrival and departure times - GeeksforGeeks

19 Mar, 2022
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.
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++
Java
Python3
Javascript
// 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 implementation of the above approachimport java.io.*;import java.util.*; // User defined Pair classclass Pair {  int x;  int y;   // Constructor  public Pair(int x, int y)  {    this.x = x;    this.y = y;  }} // class to define user defined conparatorclass 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 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 Codeif __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
<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>
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++
Java
Python
C#
Javascript
// 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 Codeint 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 code implementation of the above approachimport 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 Codepublic 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 Code Implementation of the above approachdef 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# code implementation of the above approachusing 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 Codestatic 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
<script>     // Javascript code implementation of// the above approachfunction 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 codelet 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>
Yes
Time Complexity: O(n Log n) Auxiliary Space: O(n) used by Python sort
Approach 3
The idea is to add the arrival and departure times of every booking to a list and then add it to another list of lists. After that, we will sort the list on the basis of increasing order of arrival times. In case, two bookings have the same arrival times, then we will sort it on the basis of early departure times.
Create a priority queue (min-heap), and traverse through each booking of the sorted array. For each traversal push the departure time in the priority queue and decrement the value of k (number of rooms). If the value of k becomes zero, it means we have no available rooms, then we will extract the minimum departure time of the room with the help of our priority queue. 
We can accept the booking only and only if that minimum departure time is less than equal to the arrival time of the booking or else it’s impossible to accept the booking.  
Following is the Implementation of above Approach –
Java
/*package whatever //do not write package name here */ import java.util.*; class GFG {    public static boolean areBookingsPossible(int arrival[], int departure[], int n, int k) {        List<List<Integer>> list = new ArrayList<>();        for(int i = 0; i < arrival.length; i++) {            List<Integer> li = new ArrayList<>();            li.add(arrival[i]);            li.add(departure[i]);            list.add(li);        }         //sorting the List        Collections.sort(list, new Comp());        PriorityQueue<Integer> pq = new PriorityQueue<>();         for(int i = 0; i < list.size(); i++) {            if(list.get(i).get(0) != list.get(i).get(1)) {                if(k > 0) {                    k--;                    pq.add(list.get(i).get(1));                } else {                    if(pq.peek() <= list.get(i).get(0)) {                        pq.remove();                        pq.add(list.get(i).get(1));                    } else {                        return false;                    }                }            }        }         return true;    }         // 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");    }} class Comp implements Comparator<List<Integer>> {    public int compare (List<Integer> l1, List<Integer> l2) {        if(l1.get(0) < l2.get(0)) {            return -1;        } else if(l1.get(0) == l2.get(0)) {            if(l1.get(1) < l2.get(1)) {                return -1;            } else {                return 1;            }        } else {            return 1;        }    }} //This code is contributed by darshanagrawalla06
Time Complexity: O(n Log n) 
Auxiliary Space: O(k) 
rituraj_jain
MRINALIMANGAL
Vidhi Modi
rutvik_56
pratham76
divyesh072019
divyeshrabadiya07
suresh07
rrrtnx
darshanagrawalla06
Arrays
Greedy
Sorting
Arrays
Greedy
Sorting
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