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 grows when we find the minimum number of meeting rooms required for given meeting times.

How does the program's work increase as the number of meetings grows?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

#include <stdio.h>
#include <stdlib.h>

int compare(const void *a, const void *b) {
    return (*(int*)a - *(int*)b);
}

int minMeetingRooms(int* startTimes, int* endTimes, int n) {
    qsort(startTimes, n, sizeof(int), compare);
    qsort(endTimes, n, sizeof(int), compare);

    int rooms = 0, endPtr = 0;
    for (int i = 0; i < n; i++) {
        if (startTimes[i] < endTimes[endPtr]) {
            rooms++;
        } else {
            endPtr++;
        }
    }
    return rooms;
}

int main() {
    int starts[] = {0, 5, 15};
    int ends[] = {30, 10, 20};
    int n = 3;
    printf("Minimum rooms required: %d\n", minMeetingRooms(starts, ends, n));
    return 0;
}
    

This code finds the minimum number of meeting rooms needed by sorting start and end times and then comparing them.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Sorting the start and end time arrays.
• How many times: Each sorting runs once on n elements, and then a single loop runs n times to count rooms.

How Execution Grows With Input

Sorting takes more time as the number of meetings grows, and the loop grows linearly with the number of meetings.

Input Size (n)	Approx. Operations
10	About 10 * log 10 + 10 = 43 operations
100	About 100 * log 100 + 100 = 700 operations
1000	About 1000 * log 1000 + 1000 = 10,000 operations

Pattern observation: The work grows a bit faster than the number of meetings because of sorting, but the counting loop grows directly with the number of meetings.

Final Time Complexity

Time Complexity: O(n log n)

This means the time needed grows a little faster than the number of meetings because sorting takes more time as the list grows.

Common Mistake

[X] Wrong: "The time complexity is just O(n) because we only loop once through the meetings."

[OK] Correct: Sorting the start and end times takes more time than just looping, so the sorting step dominates the total time.

Interview Connect

Understanding how sorting and simple loops combine to affect time helps you explain your solution clearly and shows you know how to analyze real problems.

Self-Check

"What if the meeting times were already sorted? How would the time complexity change?"

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 minimum number of meeting rooms needed to hold all meetings without overlap

B. The maximum duration of a single meeting

C. The total number of meetings scheduled

D. The earliest start time among all meetings

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 asked

   We want to find how many rooms are needed to schedule all meetings without time conflicts.

3. Final Answer:

   The minimum number of meeting rooms needed to hold all meetings without overlap -> Option A

4. Quick Check:

   Minimum rooms = no overlapping meetings at the same time [OK]

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

Common Mistakes:

• Confusing total meetings with rooms needed
• Thinking about meeting durations instead of overlaps
• Ignoring overlapping intervals

2. Which of the following C code snippets correctly declares an array of meeting intervals where each interval has a start and end time?

easy

A. int intervals[2][3] = {{0, 5, 15}, {30, 10, 20}};

B. int intervals[3] = {0, 30, 5, 10, 15, 20};

C. int intervals[3][2] = {0, 30, 5, 10, 15, 20};

D. int intervals[3][2] = {{0, 30}, {5, 10}, {15, 20}};

Solution

1. Step 1: Check array dimensions and initialization

   We need a 2D array with 3 intervals, each having 2 values (start and end).

2. Step 2: Validate each option

   int intervals[3][2] = {{0, 30}, {5, 10}, {15, 20}}; correctly declares int intervals[3][2] with proper nested braces for each interval.

3. Final Answer:

   int intervals[3][2] = {{0, 30}, {5, 10}, {15, 20}}; -> Option D

4. Quick Check:

   Correct 2D array syntax for intervals [OK]

Hint: Use int arr[n][2] for intervals with start and end [OK]

Common Mistakes:

• Using 1D array for 2D interval data
• Incorrect brace nesting in initialization
• Swapping rows and columns in declaration

3. Given the following C code snippet, what is the output printed?

int intervals[3][2] = {{0, 30}, {5, 10}, {15, 20}};
// Assume a function minMeetingRooms(intervals, 3) returns minimum rooms needed
int rooms = minMeetingRooms(intervals, 3);
printf("%d\n", rooms);

Assuming minMeetingRooms correctly implements the meeting rooms logic, what is printed?

medium

A. 1

B. 3

C. 2

D. 0

Solution

1. Step 1: Analyze meeting intervals for overlaps

   Intervals: [0,30], [5,10], [15,20]. The first and second overlap (5 < 30), so need at least 2 rooms.

2. Step 2: Check if any three overlap simultaneously

   Third interval [15,20] overlaps with first [0,30], but second ends at 10, so max overlap is 2 meetings at once.

3. Final Answer:

   2 -> Option C

4. Quick Check:

   Max overlapping meetings = 2 [OK]

Hint: Count max overlaps to find rooms needed [OK]

Common Mistakes:

• Assuming all meetings overlap simultaneously
• Counting total meetings instead of overlaps
• Ignoring partial overlaps

4. The following C function aims to find the minimum number of meeting rooms required but has a bug. What is the bug?

int minMeetingRooms(int intervals[][2], int n) {
    int rooms = 0;
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            if (intervals[i][0] < intervals[j][1] && intervals[j][0] < intervals[i][1]) {
                rooms++;
            }
        }
    }
    return rooms;
}

medium

A. It counts overlapping pairs but does not track maximum overlaps for rooms

B. It uses wrong indices in the loops causing out-of-bounds access

C. It returns zero always because rooms is never incremented

D. It sorts intervals incorrectly before counting overlaps

Solution

1. Step 1: Understand the nested loops and condition

   The code counts each overlapping pair by checking if intervals overlap and increments rooms.

2. Step 2: Identify why counting pairs is incorrect

   Counting pairs does not give the minimum rooms needed; we need the maximum number of simultaneous overlaps.

3. Final Answer:

   It counts overlapping pairs but does not track maximum overlaps for rooms -> Option A

4. Quick Check:

   Counting pairs ≠ max simultaneous overlaps [OK]

Hint: Track max overlaps, not just count pairs [OK]

Common Mistakes:

• Confusing pair counts with room count
• Not sorting intervals before processing
• Ignoring simultaneous overlaps

5. You have the following meeting intervals: [[1, 4], [2, 5], [7, 9], [3, 6], [10, 11]]. Using the minimum rooms required approach, what is the minimum number of meeting rooms needed?

hard

A. 2

B. 3

C. 4

D. 5

Solution

1. Step 1: Sort intervals by start time

   Sorted intervals: [1,4], [2,5], [3,6], [7,9], [10,11]

2. Step 2: Find maximum overlapping meetings

   Between 1-6, three meetings overlap: [1,4], [2,5], [3,6]. After 6, meetings do not overlap.

3. Final Answer:

   3 -> Option B

4. Quick Check:

   Max overlaps = 3 rooms needed [OK]

Hint: Sort intervals, count max overlaps for rooms [OK]

Common Mistakes:

• Not sorting intervals before counting
• Counting total meetings instead of overlaps
• Missing overlapping intervals in middle