What is the brute force approach and complexity for Meeting Scheduler?

Try every pair of slots and keep the earliest overlap that is long enough with time complexity O(M · N) and space complexity O(1).

What is the optimal approach and complexity for Meeting Scheduler?

Sort both lists, walk them together, return at the first overlap that is long enough with optimal time complexity O(M log M + N log N) and space complexity O(1).

Meeting Scheduler

Q: How do you identify the pattern for Meeting Scheduler?

Look for key signals in the prompt: "two people, each with free slots" indicates Two interval lists → intersect them with two pointers.; "both are free" indicates We need the overlap: [max of starts, min of ends].; "at least duration long, earliest" indicates Check hi − lo ≥ duration, and stop at the first hit because the walk goes in time order.; "slots are not sorted" indicates Sort both lists first, or the walk is meaningless..

Medium·LeetCode #1229 Premium Pro OnlySlight Variation

LeetCode

Problem Statement

Two people want to meet. slots1 lists the times person 1 is free, and slots2 lists the times person 2 is free. Each slot is [start, end]. The slots of one person never overlap each other, but they are not given in any order.

Find the earliest time slot of length exactly duration when both are free. Return it as [start, start + duration]. If there is no such slot, return an empty array.

Input

Two lists of slots slots1 and slots2 (each slot is [start, end]), and an integer duration.

Output

[start, start + duration] for the earliest possible meeting, or [] if none fits.

Example 1

Input: slots1 = [[10,50],[60,120],[140,210]], slots2 = [[0,15],[60,70]], duration = 8

Output: [60,68]

[10,50] and [0,15] share [10,15], only 5 long. [60,120] and [60,70] share [60,70], 10 long. The meeting starts at 60 and lasts 8, so [60,68].

Example 2

Input: slots1 = [[10,50],[60,120],[140,210]], slots2 = [[0,15],[60,70]], duration = 12

Output: []

The longest shared time is [60,70], which is only 10 long.

Constraints

• 1 ≤ slots1.length, slots2.length ≤ 10^4
• slots1[i][0] < slots1[i][1], slots2[i][0] < slots2[i][1]
• 0 ≤ slots1[i][j], slots2[i][j] ≤ 10^9
• 1 ≤ duration ≤ 10^6
• One person's slots never overlap each other

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“two people, each with free slots”

→ Two interval lists → intersect them with two pointers.

“both are free”

→ We need the overlap: [max of starts, min of ends].

“at least duration long, earliest”

→ Check hi − lo ≥ duration, and stop at the first hit because the walk goes in time order.

“slots are not sorted”

→ Sort both lists first, or the walk is meaningless.

Approach Complexity Ladder

Brute Force — check every pair of slotsO(M · N)O(1)

Try every pair of slots and keep the earliest overlap that is long enough

Optimal — sort both, then two pointersO(M log M + N log N)O(1)

Sort both lists, walk them together, return at the first overlap that is long enough

Brute Force Intuition

For every slot of person 1 and every slot of person 2, compute the shared part [lo, hi]. If hi − lo ≥ duration, a meeting can start at lo. Keep the smallest such lo. Since the input isn't sorted, we must look at all pairs before we know which is earliest.

Recognition rule for this sub-pattern:

Two (or k) interval lists in time order — sort first if needed. Output their overlap, then advance whichever ends first; with k lists, use a min-heap.

Remember this · Intervals

Trigger

Input is [start, end] pairs; overlap, merge, insert, meeting rooms, free time, minimum removals.

Think

After sorting, do I only ever need to compare against the most recent interval?

Key question

Sort by start or by end — which one makes the greedy choice obvious?

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