Meeting Rooms Problem Minimum Rooms Required in DSA C++ - Time & Space Complexity

Time Complexity: Meeting Rooms Problem Minimum Rooms Required

O(n log n)

Understanding Time Complexity

We want to find out how the time needed to find the minimum number of meeting rooms grows as the number of meetings increases.

Specifically, how does the code handle many meetings and overlapping times?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

#include <vector>
#include <algorithm>
#include <queue>
using namespace std;

int minMeetingRooms(vector<pair<int,int>>& intervals) {
    sort(intervals.begin(), intervals.end(), [](const pair<int,int>& a, const pair<int,int>& b) { return a.first < b.first; });
    priority_queue<int, vector<int>, greater<int>> minHeap;
    for (auto &interval : intervals) {
        if (!minHeap.empty() && minHeap.top() <= interval.first) {
            minHeap.pop();
        }
        minHeap.push(interval.second);
    }
    return minHeap.size();
}
    

This code sorts meetings by start time, then uses a min-heap to track end times to find the minimum rooms needed.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Looping through all meetings once after sorting.
• How many times: Exactly once for each meeting (n times).
• Additional operations: Sorting all meetings once before the loop (n log n), and heap operations inside the loop (push/pop) which take log n time each.

How Execution Grows With Input

As the number of meetings grows, sorting takes more time, and each meeting causes heap operations.

Input Size (n)	Approx. Operations
10	Sort ~30 ops, Loop ~10 * log 10 ≈ 33 ops
100	Sort ~700 ops, Loop ~100 * log 100 ≈ 664 ops
1000	Sort ~10,000 ops, Loop ~1000 * log 1000 ≈ 10,000 ops

Pattern observation: The total operations grow a bit faster than the number of meetings because of sorting and heap operations, roughly like n log n.

Final Time Complexity

Time Complexity: O(n log n)

This means as the number of meetings grows, the time needed grows a bit faster than just counting them, mainly due to sorting and managing the heap.

Common Mistake

[X] Wrong: "Since we just loop through meetings once, the time is O(n)."

[OK] Correct: Sorting the meetings first and using a heap inside the loop both add extra time, making the total time grow faster than just n.

Interview Connect

Understanding this time complexity helps you explain how your solution handles many meetings efficiently, showing you know how sorting and heaps affect performance.

Self-Check

"What if the meetings were already sorted by start time? How would that change the time complexity?"

Practice

1. What does the "Meeting Rooms Problem Minimum Rooms Required" ask you to find given a list of meeting time intervals?

easy

A. The maximum length of a single meeting

B. The total number of meetings scheduled

C. The minimum number of rooms needed so no meetings overlap

D. The earliest meeting start time

Solution

1. Step 1: Understand the problem goal

   The problem asks for the minimum number of rooms needed so that no two meetings overlap in the same room.

2. Step 2: Clarify what is being counted

   We count how many meetings happen at the same time to find the minimum rooms required.

3. Final Answer:

   The minimum number of rooms needed so no meetings overlap -> Option C

4. Quick Check:

   Minimum rooms = max overlapping meetings [OK]

Hint: Count max overlapping meetings to find minimum rooms [OK]

Common Mistakes:

• Confusing total meetings with rooms needed
• Thinking earliest start time is the answer
• Ignoring overlapping intervals

2. Which of the following C++ code snippets correctly sorts a vector of meeting intervals by their start time? Assume intervals is a vector of pairs where first is start and second is end time.

easy

A. std::sort(intervals.begin(), intervals.end(), [](auto &a, auto &b) { return a.first > b.first; });

B. std::sort(intervals.begin(), intervals.end(), [](auto &a, auto &b) { return a.first < b.first; });

C. std::sort(intervals.rbegin(), intervals.rend());

D. std::sort(intervals.begin(), intervals.end(), [](auto &a, auto &b) { return a.second < b.second; });

Solution

1. Step 1: Identify sorting by start time

   We want to sort intervals by their start time, which is stored in a.first.

2. Step 2: Check lambda comparator correctness

   std::sort(intervals.begin(), intervals.end(), [](auto &a, auto &b) { return a.first < b.first; }); uses a lambda comparing a.first < b.first, which sorts ascending by start time correctly.

3. Final Answer:

   std::sort(intervals.begin(), intervals.end(), [](auto &a, auto &b) { return a.first < b.first; }); -> Option B

4. Quick Check:

   Sort by start time ascending [OK]

Hint: Sort intervals by start time using lambda with a.first < b.first [OK]

Common Mistakes:

• Sorting by end time instead of start time
• Using default sort without comparator (may sort by first but not guaranteed)
• Sorting descending when ascending is needed

3. What is the output of the following C++ code snippet that calculates the minimum number of meeting rooms required?

std::vector> intervals = {{0,30}, {5,10}, {15,20}};
std::vector<int> starts, ends;
for(auto &i : intervals) {
    starts.push_back(i.first);
    ends.push_back(i.second);
}
std::sort(starts.begin(), starts.end());
std::sort(ends.begin(), ends.end());
int rooms = 0, end_ptr = 0;
for(int i = 0; i < starts.size(); i++) {
    if(starts[i] < ends[end_ptr]) rooms++;
    else end_ptr++;
}
std::cout << rooms << std::endl;

medium

A. 2

B. 3

C. 1

D. 0

Solution

1. Step 1: Sort start and end times

   Starts = [0,5,15], Ends = [10,20,30]

2. Step 2: Iterate and count rooms

   i=0: 0<10 true rooms=1
   i=1: 5<10 true rooms=2
   i=2: 15<10 false end_ptr=1 rooms=2

3. Final Answer:

   2 -> Option A

4. Quick Check:

   Max overlapping meetings = 2 [OK]

Hint: Compare starts and ends with two pointers to count rooms [OK]

Common Mistakes:

• Not sorting start and end arrays
• Incrementing rooms incorrectly
• Confusing when to move end pointer

4. The following code attempts to find the minimum number of meeting rooms needed but has a bug. What is the bug?

std::vector> intervals = {{1,5}, {2,6}, {8,10}};
std::sort(intervals.begin(), intervals.end());
int rooms = 1;
for(int i = 1; i < intervals.size(); i++) {
    if(intervals[i].first < intervals[i-1].second) rooms++;
}
std::cout << rooms << std::endl;

medium

A. The loop should start from i = 0, not i = 1

B. The intervals vector is not sorted correctly

C. The rooms variable should start at 0, not 1

D. It only compares each meeting with the previous one, missing overlaps with earlier meetings

Solution

1. Step 1: Analyze overlap check logic

   The code only checks if current meeting starts before previous meeting ends, ignoring overlaps with other meetings.

2. Step 2: Understand why this causes error

   This misses cases where a meeting overlaps with a meeting before the previous one, undercounting rooms needed.

3. Final Answer:

   It only compares each meeting with the previous one, missing overlaps with earlier meetings -> Option D

4. Quick Check:

   Check all overlaps, not just adjacent meetings [OK]

Hint: Check overlaps with all active meetings, not just previous [OK]

Common Mistakes:

• Assuming sorting fixes all overlap checks
• Starting loop at wrong index
• Incorrect initial rooms count

5. Given the following meeting intervals: {{0, 10}, {5, 15}, {10, 20}, {15, 25}, {20, 30}}, what is the minimum number of meeting rooms required to schedule all meetings without overlap?

hard

A. 2

B. 4

C. 3

D. 1

Solution

1. Step 1: Sort start and end times

   Starts = [0,5,10,15,20], Ends = [10,15,20,25,30]

2. Step 2: Use two pointers to count rooms

   i=0: 0<10 rooms=1
   i=1: 5<10 rooms=2
   i=2: 10<10 false end_ptr=1 rooms=2
   i=3: 15<15 false end_ptr=2 rooms=2
   i=4: 20<20 false end_ptr=3 rooms=2

3. Final Answer:

   2 -> Option A

4. Quick Check:

   Max overlapping meetings = 2 [OK]

Hint: Track starts and ends with pointers to find max overlaps [OK]

Common Mistakes:

• Counting all meetings as overlapping
• Not moving end pointer correctly
• Misreading intervals causing wrong overlaps