Meeting Rooms Problem Minimum Rooms Required in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to sort the intervals by their start time.

DSA C++

std::sort(intervals.begin(), intervals.end(), [](const Interval &a, const Interval &b) {
    return a.[1] < b.[1];
});

Drag options to blanks, or click blank then click option'

Alength

Bend

Cstart

Dindex

Attempts:

3 left

💡 Hint

Common Mistakes

Sorting by end time instead of start time.

Using an attribute that does not exist.

We sort intervals by their start time to process meetings in chronological order.

2fill in blank

medium

Complete the code to check if the current meeting overlaps with the earliest ending meeting.

DSA C++

if (intervals[i].start < minHeap.[1]()) {
    rooms++;
} else {
    minHeap.pop();
}

Drag options to blanks, or click blank then click option'

Amin

Bfront

Cpeek

Dtop

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'front()' which is for queues, not priority queues.

Using 'peek()' which is not a C++ priority_queue method.

The 'top' function returns the smallest end time in the min heap to check overlap.

3fill in blank

hard

Fix the error in the code to push the current meeting's end time into the min heap.

DSA C++

minHeap.[1](intervals[i].end);

Drag options to blanks, or click blank then click option'

Apush

Binsert

Cadd

Denqueue

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'insert' which is not a method of priority_queue.

Using 'enqueue' which is not a C++ priority_queue method.

The correct method to add an element to a priority_queue in C++ is 'push'.

4fill in blank

hard

Fill both blanks to create a min heap that stores meeting end times.

DSA C++

std::priority_queue<int, std::vector<int>, [1]> minHeap;

minHeap.push(intervals[0].[2]);

Drag options to blanks, or click blank then click option'

Astd::greater<int>

Bstd::less<int>

Cstart

Dend

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'std::less' which creates a max heap.

Pushing the 'start' time instead of 'end' time.

A min heap uses 'std::greater' as comparator, and we push the 'end' time of the interval.

5fill in blank

hard

Fill all three blanks to complete the loop that processes all meetings and counts minimum rooms.

DSA C++

for (int i = 1; i < intervals.size(); i++) {
    if (intervals[i].[1] < minHeap.[2]()) {
        rooms++;
    } else {
        minHeap.[3]();
    }
    minHeap.push(intervals[i].end);
}

Drag options to blanks, or click blank then click option'

Astart

Btop

Cpop

Dend

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing with 'end' instead of 'start'.

Using 'pop' incorrectly or missing it.

We compare the current meeting's start with the earliest end time (top), pop if no overlap, and push the current end.

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