Merge pull request #142 from BrianLusina/feat/algorithms-intervals-can-attend-meetings

feat(algorithms, intervals): can attend meetings

Author: BrianLusina

DIRECTORY.md

@@ -131,6 +131,8 @@
     * [Decoding](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/huffman/decoding.py)
     * [Encoding](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/huffman/encoding.py)
   * Intervals
+    * Can Attend Meetings
+      * [Test Can Attend Meetings](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/intervals/can_attend_meetings/test_can_attend_meetings.py)
     * Car Pooling
       * [Test Car Pooling](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/intervals/car_pooling/test_car_pooling.py)
     * Count Days

algorithms/intervals/can_attend_meetings/README.md

@@ -0,0 +1,23 @@
+# Can Attend Meetings
+
+Write a function to check if a person can attend all the meetings scheduled without any time conflicts. Given an array 
+intervals, where each element [s1, e1] represents a meeting starting at time s1 and ending at time e1, determine if
+there are any overlapping meetings. If there is no overlap between any meetings, return true; otherwise, return false.
+
+Note that meetings ending and starting at the same time, such as (0,5) and (5,10), do not conflict.
+
+Examples:
+
+```text
+Input: intervals = [(1,5),(3,9),(6,8)]
+Output: False
+
+Explanation: The meetings (1,5) and (3,9) overlap.
+```
+
+```text
+Input: intervals = [(10,12),(6,9),(13,15)]
+Output: True
+
+Explanation: There are no overlapping meetings, so the person can attend all.
+```

algorithms/intervals/can_attend_meetings/__init__.py

@@ -0,0 +1,60 @@
+from typing import List
+
+
+def can_attend_meetings(intervals: List[List[int]]) -> bool:
+    """
+    Checks if an employee can attend all meetings given a list of intervals representing the start and end times of each
+    meeting.
+
+    A person can attend all meetings if and only if none of the meetings overlap. By sorting the intervals by start time,
+    we can easily check if any two consecutive intervals overlap.
+
+    We iterate over each interval, beginning with the second interval in the sorted list. We compare the start time of
+    the current interval with the end time of the previous interval. If the start time of the current interval is less
+    than the end time of the previous interval, then the two intervals overlap and the person cannot attend both meetings,
+    so we return false.
+
+    Otherwise, the person can attend both meetings, and we continue to the next interval. If we reach the end of the
+    list without finding any overlapping intervals, then the person can attend all meetings, and we return true.
+
+    Complexity Analysis
+
+    Time Complexity: O(n * logn) where n is the number of intervals. The time complexity is dominated by the sorting step.
+
+    Space Complexity:
+
+    Since we are sorting the intervals and creating a new sorted_intervals variable that has the sorted intervals by time
+    the space incurred is O(n). However, if sorting in place, then the space cost becomes O(1) and in that case no extra
+    extra space would be used beyond a few variables.
+
+    Args:
+        intervals(list): list of intervals where each entry is a list containing the start and end time of a meeting
+    Returns:
+        bool: True if an employee can attend all meetings, false otherwise
+    """
+    if len(intervals) == 0:
+        return True
+
+    # Sort the intervals by start time first to enable easier iteration through the intervals. Meetings with similar
+    # start times will be close to each other, allowing quick and early exit if they overlap.
+    sorted_intervals = sorted(intervals, key=lambda x: x[0])
+
+    # Keep track of the last seen end time.
+    # We initialize the first interval's end time to keep track of the last interval's end that we have seen so far
+    last_end_time = sorted_intervals[0][1]
+
+    # We then iterate through the list checking if there is any overlaps
+    # Start from the second interval in the list to check if it overlaps with the previous interval.
+    for current_interval in sorted_intervals[1:]:
+        # Get the start and end of the current interval
+        current_start, current_end = current_interval
+
+        if current_start < last_end_time:
+            # there is an overlap, we return here
+            return False
+
+        # Otherwise, we update the last end time we have seen with this interval's end time
+        last_end_time = current_end
+
+    # If no overlap is found, we return True
+    return True

algorithms/intervals/can_attend_meetings/test_can_attend_meetings.py

@@ -0,0 +1,29 @@
+import unittest
+from typing import List
+from parameterized import parameterized
+from algorithms.intervals.can_attend_meetings import can_attend_meetings
+
+CAN_ATTEND_MEETINGS_TEST_CASES = [
+    ([[1, 5], [3, 9], [6, 8]], False),
+    ([[10, 12], [6, 9], [13, 15]], True),
+    ([[0, 30], [5, 10], [15, 20]], False),
+    ([[7, 10], [2, 4]], True),
+    ([[1, 2], [2, 3], [3, 4]], True),
+    ([[1, 2], [2, 3], [3, 4]], True),
+    ([[1, 3], [2, 4], [4, 6]], False),
+    ([[0, 1], [3, 5], [6, 7]], True),
+    ([[10, 20], [20, 30], [30, 40]], True),
+    ([[1, 5], [6, 10], [11, 15]], True),
+    ([[5, 10], [15, 20], [10, 15]], True),
+]
+
+
+class CanAttendMeetingsTestCase(unittest.TestCase):
+    @parameterized.expand(CAN_ATTEND_MEETINGS_TEST_CASES)
+    def test_can_attend_meetings(self, intervals: List[List[int]], expected: bool):
+        actual = can_attend_meetings(intervals)
+        self.assertEqual(expected, actual)
+
+
+if __name__ == "__main__":
+    unittest.main()