Remove Covered Intervals - LeetCode
Given an array intervals where intervals[i] = [li, ri] represent the interval [li, ri), remove all intervals that are covered by another interval in the list. The interval [a, b) is covered by the interval [c, d) if and only if c <= a and b <= d. Return the number of remaining intervals.
Example 1: Input: intervals = [[1,4],[3,6],[2,8]] Output: 2 Explanation: Interval [3,6] is covered by [2,8], therefore it is removed. Example 2: Input: intervals = [[1,4],[2,3]] Output: 1
Constraints:
1 <= intervals.length <= 1000
intervals[i].length == 2
0 <= li <= ri <= 105
All the given intervals are unique.
- code
class Solution:
def removeCoveredIntervals(self, intervals: List[List[int]]) -> int:
intervals.sort()
last_start = intervals[0][0]
last_end = intervals[0][1]
ct = 1
for i in range(1, len(intervals)):
if intervals[i][1] <= last_end:
continue # removed cur
else:
last_end = intervals[i][1]
if last_start < intervals[i][0]:
ct += 1
last_start = intervals[i][0]
return ct
- code
class Solution:
def removeCoveredIntervals(self, intervals: List[List[int]]) -> int:
# Sort by start point.
# If two intervals share the same start point
# put the longer one to be the first.
intervals.sort(key = lambda x: (x[0], -x[1]))
count = 0
prev_end = 0
for _, end in intervals:
# if current interval is not covered
# by the previous one
if end > prev_end:
count += 1
prev_end = end
return count