Upload: codedecks-in#56 Merge_Intervals (Python) (codedecks-in#4)

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* Upload: "Flower Planting with No adjacent"-Python

* Upload: codedecks-in#56 Merge Intervals (Python)

Author: lokendra1704

Python/56_MergeIntervals.py

@@ -0,0 +1,29 @@
+#Difficluty - Medium
+#Speed = 33.94%
+'''
+First we will sort the intervals (with start time of the interval as the key), so that the (overlapping) intervals that could be merged comes adjacent. For Ex:
+Original Array -> [[0,2], [5,9], [1,3], [11,12]]
+Sorted Array -> [[0,2], [1,3], [5,9], [11,12]]
+Note => Time Complexity of Sorting is O(nlogn)
+Overlapping Condition: Two intervals (X and Y) are overlapping if end Time of first interval (X) is greater than start time of second interval (Y)
+that is,
+            X.end > Y.start
+In case of [0,2] and [1,3], since 2 > 1, these intervals are overlapping.
+In case of [5,9] and [11,12], since 9 < 11, these intervals are *NOT* overlapping.
+So we will start with pushing the first interval in merged array.
+Then traverse the array, checking the overlapping condition of every interval with the last interval in merged array,
+1) If overlapping condition satisfy, We will update the end time of last interval in merged array as maximum of end time of both intervals.
+2) If overlapping condition doesnt satisy, We will push the interval in merged array
+'''
+#Time Complexity: O(nlogn)
+#Space Complexity: O(n)
+class Solution:
+    def merge(self, intervals: List[List[int]]) -> List[List[int]]:
+        intervals.sort(key=lambda x: x[0])                          #sort the intervals with starting time as the key
+        merged = []
+        for interval in intervals:
+            if not merged or merged[-1][1] < interval[0]:           #If merged list is empty (we need to push the first interval) or Overlapping condn doesnt satisfy
+                merged.append(interval)                             #push the interval in merged array
+            else:                                                   #else
+                merged[-1][1] = max(merged[-1][1], interval[1])     #update the end time of last interval in merged array as maximum of end time of both intervals
+        return merged

README.md

@@ -67,6 +67,7 @@
 ## Array
 |  #  | Title           |  Solution       |  Time           | Space           | Difficulty    | Tag          | Note| 
 |-----|---------------- | --------------- | --------------- | --------------- | ------------- |--------------|-----|
+56| [Merge Intervals](https://leetcode.com/problems/merge-intervals)|[Python](./Python/56_MergeIntervals.py)|33.94%|_O(nlogn)_|_O(n)_|Medium|Intervals||
 
 
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