Why Meeting Rooms Problem Minimum Rooms Required in DSA Go?

The Big Idea

What if you could instantly know the exact number of rooms needed for any meeting schedule without confusion?

The Scenario

Imagine you are organizing meetings in a small office with limited rooms. You have a list of meeting times, and you need to figure out how many rooms to prepare. Doing this by checking each meeting against all others manually is confusing and slow.

The Problem

Manually comparing every meeting time with all others means lots of back-and-forth checking. It's easy to miss overlaps or double count rooms. This wastes time and causes mistakes, like booking too few or too many rooms.

The Solution

The Meeting Rooms Problem solution uses sorting and smart tracking to quickly find the minimum number of rooms needed. It checks meeting start and end times in order, so you only focus on when rooms become free or busy, avoiding repeated comparisons.

Before vs After

✗ Before

for i := 0; i < len(meetings); i++ {
  for j := i + 1; j < len(meetings); j++ {
    if meetings[i] overlaps meetings[j] {
      rooms++
    }
  }
}

✓ After

sort meetings by start time
use a min-heap to track end times
for each meeting {
  if meeting start >= earliest end in heap {
    reuse room
  } else {
    add room
  }
}

What It Enables

This method lets you quickly and accurately find the minimum number of rooms needed for any list of meetings, no matter how many or how overlapping they are.

Real Life Example

When planning conference schedules or booking classrooms, this approach helps organizers know exactly how many rooms to reserve without guesswork or last-minute changes.

Key Takeaways

Manual checking of meeting overlaps is slow and error-prone.

Sorting and tracking end times simplifies finding room needs.

Efficiently plans room usage for any meeting schedule.

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