#133 Easy Time O(n log n) Space O(1) Animated

Meeting Rooms

Determine if a person can attend all meetings.

LeetCode ↗

Tutor Mode Build the mental model before you memorize the code.

1 Predict
2 Trace
3 Explain
4 Implement

Pattern Sort intervals, then maintain one boundary

After sorting, overlap decisions depend only on the active interval and the next start.

Approach: Sorting

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Core Insight

After sorting, overlap decisions depend only on the active interval and the next start.

Common Pitfall

Compare the next start against the current end, not against the original unsorted input.

Description

Determine if a person can attend all meetings.

Constraints

Time Complexity: O(n log n)
Space Complexity: O(1)

Solution — Sorting

TimeO(n log n) SpaceO(1)

Implementation Plan

Translate the invariant into these small, testable moves.

Sort by the relevant endpoint.
Initialize the active interval.
Merge or select based on overlap.
Update the active boundary.

def canAttendMeetings(intervals):
    intervals.sort(key=lambda i: i[0])
    
    for i in range(1, len(intervals)):
        i1 = intervals[i - 1]
        i2 = intervals[i]
        
        if i1[1] > i2[0]:
            return False
            
    return True

Non-overlapping Intervals

Meeting Rooms II

Meeting Rooms

Steps

Description

Constraints

Tags

Solution — Sorting