# 1288. Remove Covered Intervals

## Description

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.

## Solutions

• Sort intervals by increasing of startTime and decreasing of endTime.
• cnt = 1: cnt is the result.
• pre = intervals[0]: pre is the last interval
• For each interval in intervals
• if pre.endTime < interval.endTime, means interval is not overlapped then we count cnt, and update pre = interval
• else we do nothing
• Return cnt
• class Solution {
public int removeCoveredIntervals(int[][] intervals) {
Arrays.sort(intervals, (a, b) -> a[0] - b[0] == 0 ? b[1] - a[1] : a[0] - b[0]);
int[] pre = intervals[0];
int cnt = 1;
for (int i = 1; i < intervals.length; ++i) {
if (pre[1] < intervals[i][1]) {
++cnt;
pre = intervals[i];
}
}
return cnt;
}
}

• class Solution {
public:
int removeCoveredIntervals(vector<vector<int>>& intervals) {
sort(intervals.begin(), intervals.end(), [](const vector<int>& a, const vector<int>& b) { return a[0] == b[0] ? b[1] < a[1] : a[0] < b[0]; });
int cnt = 1;
vector<int> pre = intervals[0];
for (int i = 1; i < intervals.size(); ++i) {
if (pre[1] < intervals[i][1]) {
++cnt;
pre = intervals[i];
}
}
return cnt;
}
};

• class Solution:
def removeCoveredIntervals(self, intervals: List[List[int]]) -> int:
intervals.sort(key=lambda x: (x[0], -x[1]))
cnt, pre = 1, intervals[0]
for e in intervals[1:]:
if pre[1] < e[1]:
cnt += 1
pre = e
return cnt


• func removeCoveredIntervals(intervals [][]int) int {
sort.Slice(intervals, func(i, j int) bool {
if intervals[i][0] == intervals[j][0] {
return intervals[j][1] < intervals[i][1]
}
return intervals[i][0] < intervals[j][0]
})
cnt := 1
pre := intervals[0]
for i := 1; i < len(intervals); i++ {
if pre[1] < intervals[i][1] {
cnt++
pre = intervals[i]
}
}
return cnt
}