Hotel Room Assignments - Google Top Interview Questions

Problem Statement :

You are given a two-dimensional list of integers intervals where intervals[i] contains two numbers start and end representing a half-open interval [start, end). 

Each element represents requested times for a hotel room. Hotel rooms are assigned at the time that the booking starts.

 If requests intervals[i] and intervals[j] have the same starting time, then earlier index min(i, j) is assigned a room first.

We assign rooms starting from 0 and in general use the smallest non-negative integer that's currently unoccupied.

For each element in intervals return the assigned room for that request.


0 ≤ n ≤ 100,000 where n is the length of intervals

Example 1


intervals = [

    [1, 9],

    [2, 4],

    [4, 6],

    [5, 9]



[0, 1, 1, 2]

Example 2


intervals = [

    [1, 9],

    [2, 6],

    [6, 7],

    [3, 10]



[0, 1, 1, 2]

Solution :


                        Solution in C++ :

vector<int> solve(vector<vector<int>>& A) {
    set<int> available_rooms;          // set of available rooms
    set<pair<int, int>> booked_rooms;  // {end time, room assigned}

    for (int i = 0; i < A.size(); i++) A[i].push_back(i), available_rooms.insert(i);

    // If requests intervals[i] and intervals[j] have the same starting time, then earlier index
    // min(i, j) is assigned a room first.
    sort(begin(A), end(A), [&](auto& a, auto& b) {
        if (a[0] == b[0])
            return a[2] < b[2];
            return a[0] < b[0];

    vector<int> ans(A.size());  // store rooms assigned to each index

    for (auto& v : A) {
        // make free rooms available
        while (!booked_rooms.empty() && booked_rooms.begin()->first <= v[0]) {
        // assign rooms
        ans[v[2]] = *available_rooms.begin();
        booked_rooms.insert({v[1], *available_rooms.begin()});
    return ans;

                        Solution in Java :

import java.util.*;

class Solution {
    static class Event implements Comparable<Event> {
        int time; // Time of event
        boolean st; // Is starting event or ending event ?
        int idx; // Index in original array

        public Event(int time, boolean st, int idx) {
            this.time = time;
   = st;
            this.idx = idx;

        // Sort events by time, if to events have same time then give more priority to ending event.

        public int compareTo(Event e) {
            if (e.time == this.time) {
                if (!
                    return -1;
                return 1;
            return this.time - e.time;

    public int[] solve(int[][] intervals) {
        int n = intervals.length;
        Event[] events = new Event[n * 2];

        for (int i = 0, j = 0; i < n; ++i, j += 2) {
            events[j] = new Event(intervals[i][0], true, i);
            events[j + 1] = new Event(intervals[i][1], false, i);


        int[] ans = new int[n];
        int curr = 0;
        int cnt = 0;

        // Line sweep algorithm
        // Count total rooms to be alloted

        for (int i = 0; i < 2 * n; ++i) {
            Event e = events[i];
            if (
            cnt = Math.max(cnt, curr);

        // Store available rooms at any instant
        // Initially all rooms are available

        TreeSet<Integer> available = new TreeSet<>();
        for (int i = 0; i < cnt; ++i) available.add(i);

        // For every starting event assign first available room
        // For every ending event free up its alloted room

        for (int i = 0; i < 2 * n; ++i) {
            Event e = events[i];
            if ( {
                int x = available.first();
                ans[e.idx] = x;
            } else
        return ans;

                        Solution in Python : 
class Solution:
    def solve(self, intervals):
        pq = list(range(len(intervals)))  # Priority queue storing available rooms
        events = []  # This will store the start and end times for each interval
        for index, interval in enumerate(intervals):
            events.append((interval[0], 1, index))
            events.append((interval[1], 0, index))
        res = [0] * len(intervals)  # Stores the output
        for time, event, idx in sorted(events):
            if event:  # We are processing the start of an inteval
                res[idx] = heappop(pq)  # Get smallest available room
                heappush(pq, res[idx])  # Add the current room back to available rooms
        return res

View More Similar Problems

Print in Reverse

Given a pointer to the head of a singly-linked list, print each data value from the reversed list. If the given list is empty, do not print anything. Example head* refers to the linked list with data values 1->2->3->Null Print the following: 3 2 1 Function Description: Complete the reversePrint function in the editor below. reversePrint has the following parameters: Sing

View Solution →

Reverse a linked list

Given the pointer to the head node of a linked list, change the next pointers of the nodes so that their order is reversed. The head pointer given may be null meaning that the initial list is empty. Example: head references the list 1->2->3->Null. Manipulate the next pointers of each node in place and return head, now referencing the head of the list 3->2->1->Null. Function Descriptio

View Solution →

Compare two linked lists

You’re given the pointer to the head nodes of two linked lists. Compare the data in the nodes of the linked lists to check if they are equal. If all data attributes are equal and the lists are the same length, return 1. Otherwise, return 0. Example: list1=1->2->3->Null list2=1->2->3->4->Null The two lists have equal data attributes for the first 3 nodes. list2 is longer, though, so the lis

View Solution →

Merge two sorted linked lists

This challenge is part of a tutorial track by MyCodeSchool Given pointers to the heads of two sorted linked lists, merge them into a single, sorted linked list. Either head pointer may be null meaning that the corresponding list is empty. Example headA refers to 1 -> 3 -> 7 -> NULL headB refers to 1 -> 2 -> NULL The new list is 1 -> 1 -> 2 -> 3 -> 7 -> NULL. Function Description C

View Solution →

Get Node Value

This challenge is part of a tutorial track by MyCodeSchool Given a pointer to the head of a linked list and a specific position, determine the data value at that position. Count backwards from the tail node. The tail is at postion 0, its parent is at 1 and so on. Example head refers to 3 -> 2 -> 1 -> 0 -> NULL positionFromTail = 2 Each of the data values matches its distance from the t

View Solution →

Delete duplicate-value nodes from a sorted linked list

This challenge is part of a tutorial track by MyCodeSchool You are given the pointer to the head node of a sorted linked list, where the data in the nodes is in ascending order. Delete nodes and return a sorted list with each distinct value in the original list. The given head pointer may be null indicating that the list is empty. Example head refers to the first node in the list 1 -> 2 -

View Solution →