Merge pull request #914 from dsrao711/issue_913

Python solution for activity selection in DSA 450

Author: smv1999

DSA 450 GFG/ActivitySelection.py

@@ -0,0 +1,85 @@
+# Link : https://practice.geeksforgeeks.org/problems/n-meetings-in-one-room-1587115620/1
+
+# Input:
+# N = 6
+# start[] = {1,3,0,5,8,5}
+# end[] =  {2,4,6,7,9,9}
+# Output: 
+# 4
+
+# Meetings = [(1,2) , (3,4) , (0,6) , (5,7) , (8,9) , (5,9)]
+# Required Output - > Get the max number of meetings that can be scheduled from the timings given 
+
+# Get the maximum number of short meetings from the list of meetings
+
+# Sorting The meetings wrt Start time -> [(0,6),(1,2),(3,4),(5,7),(5,9)]
+# O/p = (0,6) => 1 meeting can be scheduled ad the end time of the meeting is 6 and there is no meeting
+# starting after 6 , so Op is 1
+
+# Sorting the meetings wrt End Time , Start time->  [(1,2) , (3,4) , (0,6) , (5,7) , (5,9) , (8,9) ]
+
+# Op = [(1,2) , (3,4) , (5,7) , (8,9) ] => 4
+
+# Condition for selecting a meeting - > Start time of the meeting should be greater than
+# the end time of the previous meeting
+
+# Approach :
+
+# Variables  :
+# i = Track the previous meetings -> i = 0
+# j = Track the current meetings  -> j = 1
+# counter = Count of number of meetings that can be scheduled  -> counter = 1
+# By default the first meeting will be scheduled , so counter initialized with value 1
+
+# 1. Make a list of Meetings with start time and end time 
+# 2. Sort the list wrt start time and then sort it wrt to end  time
+# 3. Iterate through the meetings and check the condition that whether the start time of the current 
+# meeting is greater than the end time of the previous meeting 
+
+# for j in range(1, n):
+#   if(Meetings[j][0] > Meetings[i][1]):
+#       counter ++
+#       i = j
+
+# 4. Return counter
+
+
+import sys
+import io
+
+class Solution:
+    
+    #Function to find the maximum number of meetings that can
+    #be performed in a meeting room.
+    def maximumMeetings(self,n,start,end):
+        # code here
+        Meetings = []
+        for i in range(n):
+            Meetings.append([start[i] , end[i]])
+            
+        Meetings.sort(key = lambda x : x[0])
+        Meetings.sort(key = lambda x : x[1])
+        # To track the previous meeting
+        i = 0 
+        # To track the current meeting
+        j = 1
+        # First meeting always gets selected
+        counter = 1
+        
+        for j in range(1,n):
+            # Condition for scheduling
+            if(Meetings[j][0] > Meetings[i][1]):
+                counter += 1
+                i = j
+                
+        # print(Meetings)
+        return counter
+    
+if __name__ == '__main__':
+    test_cases = int(input())
+    for cases in range(test_cases):
+        n = int(input())
+        start_time = list(map(int , input().strip().split()))
+        end_time = list(map(int , input().strip().split()))
+        ob = Solution()
+        print(ob.maximumMeetings(n,start_time , end_time))