435. Non-overlapping Intervals Solution

Author: doh6077

insert-interval/doh6077.py

@@ -0,0 +1,48 @@
+class Solution:
+    def eraseOverlapIntervals(self, intervals: List[List[int]]) -> int:
+
+        # # first structure it as graph (dictionary)
+        # intervals = [[1,2],[2,3],[3,4],[1,3]]
+        # dict = {}
+        # for i in intervals:
+        #     dict[i[0]] = []
+        # for a,b in intervals:
+        #     dict[a].append(b)
+
+        # #Overlapping edge cases
+        # # 1. [1,4], [2,3] -> not sure how to figure out 
+        # # 2. [1,3], [1,2] , [3,4] -> check if the value has connection 
+        # # 3. [1,2], [1,2], [1,2] ( same values) use hashset 
+
+        # count = 0 
+        # visited = set()
+        # for key, value in dict.items():
+        #     if value not in visited:
+        #         visited.append(value)
+        #     else:
+        #         count += 1
+        # return count
+
+        # count = 0 
+        # visited = set()
+        # for a, b in intervals:
+        #     if b not in visited:
+        #         visited.add(b)
+        #     else:
+        #         count += 1
+        # return count 
+
+
+        # greedy Approach 
+        intervals.sort()
+
+        res = 0 
+        prevEnd = intervals[0][1]
+        for start, end in intervals[1:]:
+            if start >= prevEnd:
+                prevEnd = end
+            else:
+                res += 1
+                prevEnd = min(prevEnd, end)
+        return res
+