Given an array of meeting time intervals [[start, end]], determine if a person could attend all meetings.

Example 1:

• Input: intervals = [[0,30],[5,10],[15,20]]
• Output: false

Example 2:

• Input: intervals = [[7,10],[2,4]]
• Output: true

Constraints:

• 0 <= intervals.length <= 10^4

Sort by start time, check if intervals[i][0] < intervals[i-1][1].

Interval Sorting Overlap Check

graph TD
    Start["Input Data"] --> Process["Process Element by Element"]
    Process --> Lookup{"Hash Map / Set Lookup"}
    Lookup -- "Match Found" --> Return["Return Indices / Result"]
    Lookup -- "No Match" --> Store["Store in Map / Set"]
    Store --> Process

function canAttendMeetings(intervals) {
  intervals.sort((a, b) => a[0] - b[0]);
  for (let i = 1; i < intervals.length; i++) {
    if (intervals[i][0] < intervals[i-1][1]) return false;
  }
  return true;
}

• Time Complexity: O(n log n)
• Space Complexity: O(1)
• Explanation: Sort and adjacent check.

function canAttendMeetings(intervals) {
  intervals.sort((a, b) => a[0] - b[0]);
  for (let i = 1; i < intervals.length; i++) {
    if (intervals[i][0] < intervals[i-1][1]) return false;
  }
  return true;
}

• Time Complexity: O(n log n)
• Space Complexity: O(1)
• Explanation: Sort by start time.

1. Initialize State: Setup necessary pointers, dynamic programming arrays, or hash maps.
2. Iterate & Evaluate: Process the input according to the boundary conditions.
3. Update & Return: Compute the optimal answer and return early or at termination.

If any meeting starts before the previous one ends, attending all is impossible.

1. Sort intervals by start time.

👉 Solve this problem interactively in the DSA Lab