Create non-overlapping-intervals.py

Author: kamyu104

Python/non-overlapping-intervals.py

@@ -0,0 +1,50 @@
+# Time:  O(nlogn)
+# Space: O(1)
+
+# Given a collection of intervals, find the minimum number of intervals
+# you need to remove to make the rest of the intervals non-overlapping.
+#
+# Note:
+# You may assume the interval's end point is always bigger than its start point.
+# Intervals like [1,2] and [2,3] have borders "touching" but they don't overlap each other.
+# 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.
+
+# Definition for an interval.
+# class Interval(object):
+#     def __init__(self, s=0, e=0):
+#         self.start = s
+#         self.end = e
+
+class Solution(object):
+    def eraseOverlapIntervals(self, intervals):
+        """
+        :type intervals: List[Interval]
+        :rtype: int
+        """
+        intervals.sort(key=lambda interval: interval.start)
+        result, prev = 0, 0
+        for i in xrange(1, len(intervals)):
+            if intervals[i].start < intervals[prev].end:
+                if intervals[i].end < intervals[prev].end:
+                    prev = i
+                result += 1
+            else:
+                prev = i
+        return result