Meeting Rooms Problem Minimum Rooms Required in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to sort the start times of meetings.

DSA Go

sort.Slice(intervals[1], func(i, j int) bool { return intervals[i].Start < intervals[j].Start })

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

A[0]

B(0)

C[:]

D(1)

Attempts:

3 left

💡 Hint

Common Mistakes

Using (0) or [0] which are invalid for sorting the whole slice.

Trying to sort intervals without slicing causes errors.

Using [:] slices the entire intervals slice to sort by start times.

2fill in blank

medium

Complete the code to push the end time of the first meeting into the heap.

DSA Go

heap.Push(&minHeap, intervals[0][1])

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

A[0]

B.Start

C[1]

D.End

Attempts:

3 left

💡 Hint

Common Mistakes

Pushing start time instead of end time.

Using incorrect indexing like [1] or [0] without field.

We push the end time to track when the room becomes free.

3fill in blank

hard

Fix the error in comparing the current meeting start time with the earliest end time in the heap.

DSA Go

if intervals[i].Start [1] minHeap[0].(int) {

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

A<

B>=

C<=

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<=' causes wrong logic for room reuse.

Using '==' misses cases where start is after end.

If current start is greater or equal, room is free to reuse.

4fill in blank

hard

Fill the blank to pop the earliest end time and push the current meeting's end time.

DSA Go

heap.Pop(&minHeap)
heap.Push(&minHeap, intervals[i][1])

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

A.End

B.Start

C[1]

D[0]

Attempts:

3 left

💡 Hint

Common Mistakes

Pushing start time instead of end time.

Using wrong indexing like [1] or [0].

Pop frees a room, then push current meeting's end time to track usage.

5fill in blank

hard

Fill all three blanks to return the minimum number of rooms needed after processing all meetings.

DSA Go

return [1](minHeap)

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

Alen

Bcap

Ccopy

Dappend

Attempts:

3 left

💡 Hint

Common Mistakes

Using cap() returns capacity, not current size.

Using copy() or append() are unrelated here.

The heap size shows how many rooms are used simultaneously.

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 earliest meeting start time

B. The maximum length of a single meeting

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

D. The total number of meetings scheduled

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 meeting rooms needed so no meetings overlap -> Option C

4. Quick Check:

   Minimum rooms = max overlapping meetings [OK]

Hint: Count max overlapping intervals to find rooms needed [OK]

Common Mistakes:

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

2. Which Go data structure is best to store meeting intervals for solving the minimum rooms problem?

easy

A. [][]int

B. string

C. map[int]int

D. chan int

Solution

1. Step 1: Identify suitable data structure for intervals

   Meeting intervals have start and end times, so a slice of slices ([][]int) is ideal to store pairs.

2. Step 2: Check other options

   map[int]int is less natural for intervals, string can't hold numeric ranges, chan int is for concurrency, not intervals.

3. Final Answer:

   [][]int -> Option A

4. Quick Check:

   Intervals stored as pairs in [][]int [OK]

Hint: Use [][]int to hold start and end times pairs [OK]

Common Mistakes:

• Using map which doesn't keep order
• Trying to store intervals as strings
• Using channels which are unrelated here

3. What is the output of the following Go code snippet that calculates minimum meeting rooms needed?

intervals := [][]int{{0, 30}, {5, 10}, {15, 20}}
rooms := minMeetingRooms(intervals)
fmt.Println(rooms)

func minMeetingRooms(intervals [][]int) int {
    starts := []int{}
    ends := []int{}
    for _, interval := range intervals {
        starts = append(starts, interval[0])
        ends = append(ends, interval[1])
    }
    sort.Ints(starts)
    sort.Ints(ends)
    rooms, endIndex := 0, 0
    for i := 0; i < len(starts); i++ {
        if starts[i] < ends[endIndex] {
            rooms++
        } else {
            endIndex++
        }
    }
    return rooms
}

medium

A. 1

B. 2

C. 3

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

   At start=0 < end=10, rooms=1; start=5 < end=10, rooms=2; start=15 >= end=10, endIndex++, rooms=2

3. Final Answer:

   2 -> Option B

4. Quick Check:

   Max overlapping meetings = 2 [OK]

Hint: Sort starts and ends, count overlaps to find rooms [OK]

Common Mistakes:

• Not sorting start and end times
• Incrementing rooms incorrectly
• Confusing start and end comparisons

4. Find the bug in this Go function for minimum meeting rooms:

func minMeetingRooms(intervals [][]int) int {
    starts := []int{}
    ends := []int{}
    for _, interval := range intervals {
        starts = append(starts, interval[0])
        ends = append(ends, interval[1])
    }
    sort.Ints(starts)
    sort.Ints(ends)
    rooms := 0
    endIndex := 0
    for i := 0; i <= len(starts); i++ {
        if starts[i] < ends[endIndex] {
            rooms++
        } else {
            endIndex++
        }
    }
    return rooms
}

medium

A. Loop condition should be i < len(starts), not <= to avoid out-of-range

B. Should sort starts and ends in descending order

C. Should compare starts[i] > ends[endIndex] instead

D. Should initialize rooms to 1 instead of 0

Solution

1. Step 1: Check loop boundary

   The loop uses i <= len(starts), which causes index out-of-range error because valid indices are 0 to len(starts)-1.

2. Step 2: Correct loop condition

   Change to i < len(starts) to safely iterate over all elements.

3. Final Answer:

   Loop condition should be i < len(starts), not <= to avoid out-of-range -> Option A

4. Quick Check:

   Index error fixed by proper loop bounds [OK]

Hint: Use < not <= in loops to avoid index errors [OK]

Common Mistakes:

• Using <= instead of < in for loops
• Sorting in wrong order
• Wrong comparison operators

5. Given meeting intervals [[1,4],[2,5],[7,9],[6,8],[9,10]], what is the minimum number of meeting rooms required?

hard

A. 3

B. 1

C. 4

D. 2

Solution

1. Step 1: Sort start and end times

   Starts = [1,2,6,7,9], Ends = [4,5,8,9,10]

2. Step 2: Count overlaps to find rooms

   i=0 start=1 < end=4 rooms=1 i=1 start=2 < end=4 rooms=2 i=2 start=6 >= end=4 endIndex=1 rooms=2 i=3 start=7 >= end=5 endIndex=2 rooms=2 i=4 start=9 >= end=8 endIndex=3 rooms=2

3. Final Answer:

   2 -> Option D

4. Quick Check:

   Max overlaps = 2 rooms needed [OK]

Hint: Sort starts and ends, increment rooms on overlap [OK]

Common Mistakes:

• Not sorting intervals before processing
• Miscounting overlaps at boundaries
• Assuming one room per meeting always