435. Non-overlapping Intervals
Given an array of intervals intervals where intervals[i] = [starti, endi], return the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping.
Example 1:
Input: intervals = [[1,2],[2,3],[3,4],[1,3]] Output: 1 Explanation: [1,3] can be removed and the rest of the intervals are non-overlapping.
Example 2:
Input: intervals = [[1,2],[1,2],[1,2]] Output: 2 Explanation: You need to remove two [1,2] to make the rest of the intervals non-overlapping.
Example 3:
Input: intervals = [[1,2],[2,3]] Output: 0 Explanation: You don't need to remove any of the intervals since they're already non-overlapping.
Constraints:
1 <= intervals.length <= 105intervals[i].length == 2-5 * 104 <= starti < endi <= 5 * 104
Solution :
class Solution:
def eraseOverlapIntervals(self, intervals: List[List[int]]) -> int:
intervals.sort()
ans = 0
end_prev = intervals[0][1]
for s,e in intervals[1:]:
if s >= end_prev :
end_prev = e
else:
ans += 1
end_prev = min(e, end_prev)
return ans
Explaination :
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