**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. Constraints 0 ≤ n ≤ 100,000 where n is the length of intervals Example 1 Input intervals = [ [1, 9], [2, 4], [4, 6], [5, 9] ] Output [0, 1, 1, 2] Example 2 Input intervals = [ [1, 9], [2, 6], [6, 7], [3, 10] ] Output [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];
else
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]) {
available_rooms.insert(booked_rooms.begin()->second);
booked_rooms.erase(booked_rooms.begin());
}
// assign rooms
ans[v[2]] = *available_rooms.begin();
booked_rooms.insert({v[1], *available_rooms.begin()});
available_rooms.erase(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;
this.st = 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 (!this.st)
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);
}
Arrays.sort(events);
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 (e.st)
++curr;
else
--curr;
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 (e.st) {
int x = available.first();
available.remove(x);
ans[e.idx] = x;
} else
available.add(ans[e.idx]);
}
return ans;
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, intervals):
pq = list(range(len(intervals))) # Priority queue storing available rooms
heapify(pq)
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
else:
heappush(pq, res[idx]) # Add the current room back to available rooms
return res
```

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