diff --git a/greedy_methods/n_meetings_in_one_room.py b/greedy_methods/n_meetings_in_one_room.py
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+++ b/greedy_methods/n_meetings_in_one_room.py
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+"""
+Question:
+There is one meeting room in a firm. There are N meetings in the form
+of (start[i], end[i]) where start[i] is start time of meeting i and end[i]
+is finish time of meeting i.
+What is the maximum number of meetings that can be accommodated in the
+meeting room when only one meeting can be held in the meeting room at a particular time?
+
+Note: Start time of one chosen meeting can't be equal to the end time of
+the other chosen meeting.
+
+Example 1:
+Input:
+N = 6
+start = [1,3,0,5,8,5]
+end =  [2,4,6,7,9,9]
+Output:
+4
+
+Explanation:
+Maximum four meetings can be held with
+given start and end timings.
+The meetings are - (1, 2),(3, 4), (5,7) and (8,9)
+
+Input:
+N = 3
+start= [10, 12, 20]
+end = [20, 25, 30]
+Output:
+1
+
+Explanation:
+Only one meetings can be held
+with given start and end timings.
+
+
+Approach:
+
+The idea is to solve problem using greedy way.
+i.e. sort the meetings by their finish time and the start
+selecting meetings, starting with one with least end time and the select
+other meetings such that the start time of current meeting is greater
+than the last meeting selected
+"""
+
+
+from typing import List
+
+
+def maximumMeetings(n: int, start: List[int], end: List[int]) -> int:
+    """
+    >>> maximumMeetings(6, [1,3,0,5,8,5], [2,4,6,7,9,9])
+    4
+
+    >>> maximumMeetings(3, [10, 12, 20], [20, 25, 30])
+    1
+    """
+    all_meetings = [(start_time,end_time) for start_time,end_time in zip(start, end)]
+    all_meetings.sort(key=lambda x:(x[1],x[0]))
+
+    possible_meet = 1
+    prev = all_meetings[0]
+    for each in range(1, n):
+        if all_meetings[each][0] > prev[1]:
+            possible_meet += 1
+            prev = all_meetings[each]
+    return possible_meet
+
+
+if __name__ == "__main__":
+    import doctest
+
+    doctest.testmod()