Adding Efficient Solution for Problem 435 - Non-overlapping Intervals

Author: Vanditg

Non-Overlapping Intervals/EfficientSolution.py

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+##==================================
+## Leetcode
+## Student: Vandit Jyotindra Gajjar
+## Year: 2020
+## Problem: 435
+## Problem Name: Non-overlapping Intervals      
+##===================================
+#
+#Given a collection of intervals, find the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping.
+#
+#Example 1:
+#
+#Input: [[1,2],[2,3],[3,4],[1,3]]
+#Output: 1
+#Explanation: [1,3] can be removed and the rest of intervals are non-overlapping.
+#Example 2:
+#
+#Input: [[1,2],[1,2],[1,2]]
+#Output: 2
+#Explanation: You need to remove two [1,2] to make the rest of intervals non-overlapping.
+#Example 3:
+#
+#Input: [[1,2],[2,3]]
+#Output: 0
+#Explanation: You don't need to remove any of the intervals since they're already non-overlapping.
+
+class Solution: 
+    def eraseOverlappingIntervals(self, intervals):
+        if intervals == []:	#Condition-check: If intervals is empty
+            return 0	#We'll return 0 as no overlapping interval presenet
+        intervals.sort(key = lambda x:x[1])	#Sort intervals list by 2nd element 
+        tmp = 0	#Initialize our tmp counter 
+        end = intervals[0][0]	#Initialize end which will be first element of the first sublist
+        for i in intervals:	#Loop through intervals
+            start = i[0]	#Initialize start which will be first element in every sublist
+            if start < end:	#Condition-check: If start is less than end 
+                tmp += 1	#Then we'll increase out tmp counter as we encounter the overlapping list
+            else:	#Condition-check: Else
+                end = i[1]	#We'll update the end by 2nd element of ith sublist
+        return tmp	#We'll return tmp counter