Adding Efficient Solution for Problem - 57 - Insert Interval

Author: Vanditg

Insert_Interval/EfficientSolution.py

@@ -0,0 +1,33 @@
+##==================================
+## Leetcode
+## Student: Vandit Jyotindra Gajjar
+## Year: 2020
+## Problem: 57
+## Problem Name: Insert Interval   
+##===================================
+#Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).
+#
+#You may assume that the intervals were initially sorted according to their start times.
+#
+#Example 1:
+#
+#Input: intervals = [[1,3],[6,9]], newInterval = [2,5]
+#Output: [[1,5],[6,9]]
+#
+#Example 2:
+#
+#Input: intervals = [[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval = [4,8]
+#Output: [[1,2],[3,10],[12,16]]
+#Explanation: Because the new interval [4,8] overlaps with [3,5],[6,7],[8,10].
+class Solution:
+    def insert(self, intervals, newInterval):
+        l, r = [], []	#Initialize l and r empty list
+        intervals.sort(key = lambda x:x[0])	#Sort the intervals by its 1st element in sublists 
+        for i in intervals:	#Loop through intervals
+            if i[1] < newInterval[0]:	#Condition-check: If i's last element is less than newInterval's first element
+                l.append(i)	#We append i in l 
+            elif i[0] > newInterval[1]: #Condition-check: Elif i's first element is greater than newInterval's last element
+                r.append(i)	#We append i in r  
+            else:
+                newInterval = [min(newInterval[0], i[0]), max(newInterval[1], i[1])]	#Update the newInterval by taking min and max of newInterval and i's first and last element resepectively 
+        return l + [newInterval] + r	#Return the final list