Problem Description

You are given an array of meeting time intervals where each interval is represented as intervals[i] = [start_i, end_i]. Each interval indicates when a meeting starts and ends.

Your task is to find the minimum number of conference rooms required to schedule all the meetings without any conflicts. In other words, if two or more meetings overlap in time, they need separate conference rooms.

For example, if you have meetings [[0,30],[5,10],[15,20]], the first meeting runs from time 0 to 30. The second meeting (5 to 10) overlaps with the first, so you need a second room. The third meeting (15 to 20) also overlaps with the first, but since the second meeting has already ended by time 15, you can reuse that room. Therefore, you need a minimum of 2 conference rooms.

The solution uses a difference array technique. It tracks the changes in the number of ongoing meetings at each time point. By marking +1 when a meeting starts and -1 when it ends, then calculating the running sum, you can determine the maximum number of simultaneous meetings at any point in time, which equals the minimum number of rooms needed.

Intuition

To understand how many conference rooms we need, think about what happens at each moment in time. At any given moment, the number of rooms needed equals the number of meetings happening simultaneously.

Imagine a timeline where meetings start and end. When a meeting starts, we need one more room. When a meeting ends, we free up one room. The key insight is that we only care about these "change points" - the moments when meetings start or end.

We can track these changes using a difference array. For each meeting with interval [start, end], we mark +1 at the start time (one more room needed) and -1 at the end time (one room freed). This creates a record of all the changes in room demand throughout the day.

By calculating the running sum (prefix sum) of this difference array, we get the actual number of rooms in use at each time point. The running sum works because:

• Starting from 0 rooms
• Each +1 we encounter means a meeting started (add a room)
• Each -1 we encounter means a meeting ended (free a room)
• The sum at any point tells us exactly how many meetings are happening simultaneously

The maximum value in this running sum represents the peak moment when the most meetings overlap - this is our answer, the minimum number of conference rooms required.

For example, with meetings [[0,30],[5,10],[15,20]]:

• At time 0: +1 (1 room in use)
• At time 5: +1 (2 rooms in use)
• At time 10: -1 (1 room in use)
• At time 15: +1 (2 rooms in use)
• At time 20: -1 (1 room in use)
• At time 30: -1 (0 rooms in use)

The maximum is 2, so we need 2 conference rooms.

Solution Approach

The implementation uses a difference array to efficiently track room usage changes over time.

Step 1: Find the maximum end time

m = max(e[1] for e in intervals)

We first determine the maximum end time among all meetings to know the size of our difference array. This ensures our array covers the entire time range.

Step 2: Create and populate the difference array

d = [0] * (m + 1)
for l, r in intervals:
    d[l] += 1
    d[r] -= 1

We create a difference array d of size m + 1 initialized with zeros. For each meeting interval [l, r]:

• Increment d[l] by 1 to mark a meeting starting at time l
• Decrement d[r] by 1 to mark a meeting ending at time r

Note that multiple meetings can start or end at the same time, so we use += and -= to accumulate the changes.

Step 3: Calculate the prefix sum and find the maximum

ans = s = 0
for v in d:
    s += v
    ans = max(ans, s)
return ans

We iterate through the difference array, maintaining a running sum s. At each time point:

• Add the current value v to the running sum s
• The running sum represents the number of rooms in use at that moment
• Track the maximum value seen in ans

The maximum value of the running sum represents the peak number of simultaneous meetings, which equals the minimum number of conference rooms required.

Time Complexity: O(n + m) where n is the number of intervals and m is the maximum end time Space Complexity: O(m) for the difference array

Example Walkthrough

Let's walk through a small example with meetings [[0,4],[2,6],[3,5],[7,9]].

Step 1: Find maximum end time The end times are 4, 6, 5, and 9. Maximum is 9, so we'll create an array of size 10.

Step 2: Build the difference array Initialize: d = [0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

For each meeting, mark start with +1 and end with -1:

• Meeting [0,4]: Mark +1 at index 0, -1 at index 4 d = [1, 0, 0, 0, -1, 0, 0, 0, 0, 0]

• Meeting [2,6]: Mark +1 at index 2, -1 at index 6 d = [1, 0, 1, 0, -1, 0, -1, 0, 0, 0]

• Meeting [3,5]: Mark +1 at index 3, -1 at index 5 d = [1, 0, 1, 1, -1, -1, -1, 0, 0, 0]

• Meeting [7,9]: Mark +1 at index 7, -1 at index 9 d = [1, 0, 1, 1, -1, -1, -1, 1, 0, -1]

Step 3: Calculate running sum to find maximum rooms needed

The maximum running sum is 3, which occurs at time 3 when three meetings ([0,4], [2,6], and [3,5]) are all happening simultaneously. Therefore, we need a minimum of 3 conference rooms.

Solution Implementation

Time and Space Complexity

The time complexity is O(n + m) where n is the number of intervals (meetings) and m is the maximum end time among all intervals.

• Finding the maximum end time takes O(n) time as it iterates through all intervals once
• Creating the difference array d takes O(m) space initialization
• Updating the difference array with start and end points takes O(n) time (two operations per interval)
• The final sweep through the difference array to calculate the maximum overlapping meetings takes O(m) time
• Overall: O(n) + O(n) + O(m) = O(n + m)

The space complexity is O(m) where m is the maximum end time among all intervals.

• The difference array d has size m + 1, requiring O(m) space
• Other variables (ans, s, loop variables) use O(1) space
• Overall: O(m)

Common Pitfalls

1. Memory Inefficiency with Large Time Values

The current solution creates an array of size max_end_time + 1, which can be extremely memory-inefficient when dealing with large time values. For example, if you have meetings like [[1000000, 1000005], [1000002, 1000007]], the solution would create an array of over 1 million elements just to track a few meetings.

Solution: Use a sorted events approach instead of a difference array:

class Solution:
    def minMeetingRooms(self, intervals: List[List[int]]) -> int:
        # Create events for all start and end times
        events = []
        for start, end in intervals:
            events.append((start, 1))   # 1 for start
            events.append((end, -1))     # -1 for end
      
        # Sort events by time, with ends before starts at the same time
        events.sort(key=lambda x: (x[0], x[1]))
      
        max_rooms = 0
        current_rooms = 0
      
        for time, change in events:
            current_rooms += change
            max_rooms = max(max_rooms, current_rooms)
      
        return max_rooms

2. Handling Same Time Start and End

When a meeting ends at the exact same time another begins (e.g., [0, 10] and [10, 20]), the original difference array approach handles this correctly by processing both the -1 and +1 at time 10. However, if you modify the solution to use a dictionary or other data structure, you might accidentally overwrite values instead of accumulating them.

Incorrect approach:

# Wrong - overwrites instead of accumulates
difference = {}
for start, end in intervals:
    difference[start] = 1  # Should be +=
    difference[end] = -1   # Should be -=

Correct approach:

from collections import defaultdict

difference = defaultdict(int)
for start, end in intervals:
    difference[start] += 1
    difference[end] -= 1

3. Integer Overflow in Other Languages

While Python handles arbitrarily large integers, implementing this solution in languages like Java or C++ with fixed-size integers could cause overflow if the time values are near the maximum integer limit. Creating an array of size Integer.MAX_VALUE + 1 would cause an error.

Solution: Always validate input ranges or use the sorted events approach which doesn't depend on the absolute values of the times, only their relative ordering.