Introduction:
The "Meeting Rooms" problem in C++ is about determining whether a person can attend all scheduled meetings without overlap. Each meeting is represented by a time interval, with a start and end time, and the goal is to check if the meetings conflict in any way.
Suppose you have several meetings throughout the day, each with a start and an end time. If any two meetings overlap, the person won't be able to attend both. The challenge is to examine these intervals and decide whether the person can attend all of them without any conflicts.
Key Concept: Interval Overlap
A meeting interval can be described as [start, end], where:
- The start is the time the meeting begins.
- The end is the time the meeting ends (exclusive, meaning the next meeting can start at the exact end time of the previous one).
Two intervals overlap if and only if one meeting starts before the previous meeting ends:
- Given two intervals [start1, end1] and [start2, end2], overlap occurs if start2 < end1.
To avoid conflicts, we must ensure no two meetings overlap.
Approach 1: Naive Approach
Implementation:
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
struct Interval {
int start;
int end;
};
bool canAttendMeetings(vector<Interval>& meetings) {
// Step 1: Sort meetings based on the start time
sort(meetings.begin(), meetings.end(), [](const Interval& a, const Interval& b) {
return a.start < b.start;
});
// Step 2: Check for overlapping meetings
for (size_t i = 1; i < meetings.size(); ++i) {
// If the start of the current meeting is less than the end of the previous meeting, return false
if (meetings[i].start < meetings[i - 1].end) {
return false;
}
}
return true;
}
int main() {
vector<Interval> meetings = {{1, 3}, {2, 4}, {5, 6}};
if (canAttendMeetings(meetings)) {
cout << "Yes, the person can attend all meetings." << endl;
} else {
cout << "No, the person cannot attend all meetings." << endl;
}
return 0;
}
Output:
No, the person cannot attend all meetings.
Explanation:
Step 1: Structuring the Problem
The first step is to recognize that each meeting has a start time and an end time. These are treated as intervals, with each interval representing a meeting's schedule. The problem becomes checking whether any of these intervals overlap, meaning a person won't be able to attend two overlapping meetings.
Step 2: Sorting the Meetings
To make the overlap detection easier, the meetings are sorted based on their start time. Sorting helps to arrange the meetings in a way that ensures each one is checked against the next one in sequence. This way, once sorted, you only need to compare each meeting with its immediate neighbor, which reduces the complexity.
Step 3: Detecting Overlap
Once the intervals are sorted by their start times, the next step is to check for overlaps. Overlap occurs when one meeting starts before the previous meeting ends.
For example:
- Suppose you have two meetings: one from 1 PM to 3 PM and another from 2 PM to 4 PM. The second meeting starts at 2 PM, which is before the first meeting ends at 3 PM, meaning these two meetings overlap, and the person cannot attend both.
Step 4: Iterating Through the Meetings
After sorting, you move through the list of meetings one by one:
- Start by comparing the second meeting's start time with the first meeting's end time.
- If the start time of the second meeting is less than the end time of the first, an overlap is found, and you can immediately conclude that the person can't attend all meetings.
- If no overlap is detected, continue to compare the next pair of meetings, moving through the list.
Step 5: Returning the Result
The goal is to go through the entire list of meetings, comparing consecutive intervals. If at any point an overlap is found, the function immediately returns "false," indicating that the sessions cannot all be attended. If you make it through the entire list without finding any overlaps, the function returns "true," confirming that it's possible to attend all meetings.
Complexity analysis:
Time Complexity:
Sorting the Intervals:
The first significant operation is sorting the list of meeting intervals by their start time. Sorting generally takes O(nlogn), where n is the number of intervals (or meetings). This is the dominant factor in the time complexity, as sorting is the most time-intensive part of the process.
Checking for Overlaps:
After sorting, the next step is to iterate through the sorted intervals to check for overlaps. This requires a single pass through the list, which takes O(n) time. Each interval is compared with the previous one, so this step is linear in complexity.
Since the sorting operation dominates, the overall time complexity is O(nlogn).
Space Complexity:
In-place Sorting:
If the sorting algorithm used is an in-place sort (such as the C++ std::sort function), no additional space is needed for sorting apart from a constant amount of extra memory. Thus, the sorting step takes O(1) additional space.
Input Storage:
The intervals themselves are stored in a list, which takes O(n) space because there are n intervals.
Auxiliary Space:
The overlap-checking part of the algorithm only uses a few variables to store indices and to compare intervals, which is constant space, i.e., O(1).
Therefore, the overall space complexity is O(n) due to the storage of the input intervals. No extra space beyond the input storage is needed for sorting or checking.
Approach 2: Using a Min-Heap (Priority Queue)
This method ensures that we dynamically manage meeting conflicts by focusing on the meeting that ends the earliest, allowing us to decide whether a new meeting can be attended or not.
Implementation:
#include <iostream>
#include <vector>
#include <queue>
#include <algorithm> // Include this for sort()
using namespace std;
struct Interval {
int start;
int end;
};
// Comparator to sort intervals by their start time
bool compare(const Interval& a, const Interval& b) {
return a.start < b.start;
}
bool canAttendMeetings(vector<Interval>& meetings) {
if (meetings.empty()) {
return true;
}
// Step 1: Sort the meetings by start time
sort(meetings.begin(), meetings.end(), compare);
// Step 2: Use a min-heap (priority queue) to track the end times of meetings
priority_queue<int, vector<int>, greater<int>> minHeap;
// Step 3: Push the end time of the first meeting into the heap
minHeap.push(meetings[0].end);
// Step 4: Iterate through the rest of the meetings
for (size_t i = 1; i < meetings.size(); ++i) {
// Check if the current meeting starts after the earliest ended meeting
if (meetings[i].start >= minHeap.top()) {
// Remove the meeting that ended earliest
minHeap.pop();
}
// Add the current meeting's end time to the heap
minHeap.push(meetings[i].end);
}
// If the heap contains only one meeting at the end, all meetings were attended
return minHeap.size() == 1;
}
int main() {
vector<Interval> meetings = {{1, 3}, {2, 4}, {5, 6}};
if (canAttendMeetings(meetings)) {
cout << "Yes, the person can attend all meetings." << endl;
} else {
cout << "No, the person cannot attend all meetings." << endl;
}
return 0;
}
Output:
No, the person cannot attend all meetings.
Explanation:
Step 1: Problem Understanding
We are given several meetings, each defined by a start and end time. The task is to figure out if a person can attend all of these meetings without any overlap. In this approach, we want to dynamically manage meeting times in a way that lets us quickly check if a new meeting conflicts with previously scheduled ones.
Step 2: Sorting the Meetings by Start Time
The first thing we do is arrange the meetings in order of their start times. Sorting the meetings this way it ensures that each meeting will be compared only with the next one in sequence. This is important because if any conflict exists between two sessions, it will be easiest to spot when they are checked one after another. The sorted list of meetings is the foundation for the next steps.
Step 3: Using a Min-Heap to Track the Earliest End Time
To manage the end times of the meetings efficiently, a min-heap (also known as a priority queue) is used. A min-heap is a data structure that always gives you access to the smallest element. In this case, the earliest ending meeting. It is essential because if you know the meeting that ends the earliest, you can quickly determine if the next meeting can start after it or if there is an overlap.
Initially, the end time of the first meeting is added to the min-heap. This way, we begin by tracking when the first meeting will finish.
Step 4: Processing Each Meeting
Now, for each subsequent meeting:
- You compare the start time of the current meeting with the earliest end time of any ongoing meeting (the one at the top of the min-heap).
- If the current meeting starts after or at the same time the earliest meeting ends, there's no conflict, and the earliest-ended meeting can be removed from the heap. This signifies that the earlier meeting is over, and the current meeting can proceed without any overlap.
- After that, you add the current meeting's end time to the heap, which signify that this meeting is now in progress and will potentially overlap with future meetings.
Step 5: Checking for Overlaps
The heap is designed to hold the end times of ongoing meetings. If the heap grows too large or if multiple meetings overlap in time, it means that the person will not be able to attend all meetings. Specifically, if a meeting starts before the earliest meeting ends, this creates a conflict.
If, by the end of processing all the meetings, the heap contains only one meeting, it means that all the meetings were handled without any conflicts, and the person can attend all of them.
Complexity analysis:
Time Complexity
Sorting the Meetings:
Sorting the list of meetings based on their start times is the most time-consuming operation in this approach. The sorting algorithm typically used (like std::sort) has a time complexity of O(nlogn), where n is the number of meetings.
Processing Meetings with the Min-Heap:
- After sorting, the algorithm processes each meeting once. For each meeting, it performs two primary operations: Inserting an end time into the min-heap: This operation takes O(logk) time, where k is the number of elements currently in the heap. In the worst case, k can be up to n, making the insertion operation O(logn). Removing the smallest end time from the min-heap: This operation also takes O(logk) time, which, in the worst case, is O(logn).
- As each meeting is processed once and we perform these operations at most n times, the total time complexity for processing all meetings with the heap is O(nlogn).
- Inserting an end time into the min-heap: This operation takes O(logk) time, where k is the number of elements currently in the heap. In the worst case, k can be up to n, making the insertion operation O(logn).
- Removing the smallest end time from the min-heap: This operation also takes O(logk) time, which, in the worst case, is O(logn).
The overall time complexity is O(nlogn) This is because sorting the meetings dominates the time complexity, and heap operations add a logarithmic factor.
Space Complexity
Space for Sorting:
Sorting is performed in place, which means it uses only a small, constant amount of additional space beyond the input list. Thus, the space complexity for sorting itself is O(1) if we ignore the space required for the input list.
Space for the Min-Heap:
- The min-heap stores the end times of ongoing meetings. In the worst case, the heap might need to store all meetings' end times simultaneously, which requires O(n) space.
Space for Input Storage:
- The input list of meetings itself requires O(n) space to store the intervals.
The overall space complexity is O(n). This accounts for the space required for the input list and the heap.
Approach 3: Using sweep line algorithm
The sweep line algorithm involves processing a series of events as if a vertical line sweeps from left to right across the timeline. In the context of meeting rooms, the goal is to determine if a person can attend all meetings without any overlap.
Implementation:
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
enum EventType { START, END };
struct Event {
int time;
EventType type;
bool operator<(const Event& other) const {
if (time == other.time) {
return type < other.type; // Ensure that end events are processed before start events if they have the same time
}
return time < other.time;
}
};
bool canAttendMeetings(vector<pair<int, int>>& meetings) {
vector<Event> events;
// Convert meetings to events
for (const auto& meeting : meetings) {
events.push_back({meeting.first, START});
events.push_back({meeting.second, END});
}
// Sort events by time and type
sort(events.begin(), events.end());
int ongoingMeetings = 0;
// Process events
for (const auto& event : events) {
if (event.type == START) {
ongoingMeetings++;
if (ongoingMeetings > 1) {
return false; // Overlap detected
}
} else {
ongoingMeetings--;
}
}
return true; // No overlaps detected
}
int main() {
vector<pair<int, int>> meetings = {{1, 3}, {2, 4}, {5, 6}};
if (canAttendMeetings(meetings)) {
cout << "Yes, the person can attend all meetings." << endl;
} else {
cout << "No, the person cannot attend all meetings." << endl;
}
return 0;
}
Output:
No, the person cannot attend all meetings.
Explanation:
Convert Meetings into Events:
- Each meeting is represented by two key moments in time: the start time and the end time. To apply the sweep line algorithm, these moments are converted into discrete events.
- For each meeting, you create two events:
- Start Event: Occurs when the meeting begins.
- End Event: Occurs when the meeting ends.
Sort Events:
- To handle these events efficiently, they are sorted based on their timestamps. The primary criterion for sorting is the time of the event.
- When two events have the same timestamp, end events are given priority over start events. This is crucial because if a meeting ends and another starts at the same moment, we need to handle the end before the start to avoid incorrect overlap detection.
Process the Events:
- As you process each event in the sorted order, you maintain a counter to keep track of the number of ongoing meetings.
- Handling Start Events: Each time a start event is encountered, increment the counter. This signifies that a new meeting has begun and is now overlapping with ongoing meetings.
- Handling End Events: Each time an end event is encountered, decrement the counter. This signifies that a meeting has concluded, reducing the count of ongoing meetings.
Detect Overlaps:
- During event processing, continuously monitor the counter:
- If the counter exceeds one at any point, it means there are overlapping meetings. Consequently, the person cannot attend all meetings without a conflict.
- If the counter remains at one or less throughout the process, it indicates that all meetings are scheduled in such a way that they do not overlap, allowing attendance at all meetings.
Complexity analysis:
Time Complexity
Event Generation:
- Each meeting generates two events (start and end). Therefore if there are n meetings, there will be 2n events.
- Generating these events is O(n) because each meeting contributes a constant amount of work.
Sorting Events:
- Sorting the events takes O(mlogm) time, where m is the number of events. Since m=2n (two events per meeting), sorting is O(nlogn). Sorting is the most time-consuming step in this approach.
Processing Events:
- Processing each event involves updating a counter and checking conditions. Since each event is processed exactly once, this step is O(m), where m is the number of events, leading to O(n) time in this case.
Combining these steps, the overall time complexity is dominated by the sorting step: O(nlogn)
Space Complexity
Event Storage:
- You need to store all the events. With n meetings generating 2n events, this requires O(n) space.
Additional Variables:
- Space is needed for the counter and a few other variables , but these require constant space.
Input Storage:
- The original list of meetings requires O(n) space.
Combining these, the overall space complexity is O(n).