Formatted question description: https://leetcode.ca/all/435.html

435. Non-overlapping Intervals (Medium)

Given a collection of intervals, find the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping.

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.


Note:

1. You may assume the interval's end point is always bigger than its start point.
2. Intervals like [1,2] and [2,3] have borders "touching" but they don't overlap each other.

Related Topics:
Greedy

Similar Questions:

Solution 1. Interval Scheduling Maximization Problem

// OJ: https://leetcode.com/problems/non-overlapping-intervals/
// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
int eraseOverlapIntervals(vector<vector<int>>& A) {
sort(begin(A), end(A), [](auto &a, auto &b) { return a[1] < b[1]; });
int ans = 0, e = INT_MIN;
for (auto &v : A) {
if (v[0] >= e) e = v[1];
else ++ans;
}
return ans;
}
};

• class Solution {
public int eraseOverlapIntervals(int[][] intervals) {
if (intervals == null || intervals.length == 0)
return 0;
Arrays.sort(intervals, new Comparator<int[]>() {
public int compare(int[] interval1, int[] interval2) {
if (interval1[0] != interval2[0])
return interval1[0] - interval2[0];
else
return interval1[1] - interval2[1];
}
});
int eraseCount = 0;
int prev = 0;
int length = intervals.length;
for (int i = 1; i < length; i++) {
if (intervals[prev][1] > intervals[i][0]) {
if (intervals[prev][1] > intervals[i][1])
prev = i;
eraseCount++;
} else
prev = i;
}
return eraseCount;
}
}

############

class Solution {
public int eraseOverlapIntervals(int[][] intervals) {
Arrays.sort(intervals, Comparator.comparingInt(a -> a[1]));
int t = intervals[0][1], ans = 0;
for (int i = 1; i < intervals.length; ++i) {
if (intervals[i][0] >= t) {
t = intervals[i][1];
} else {
++ans;
}
}
return ans;
}
}

• // OJ: https://leetcode.com/problems/non-overlapping-intervals/
// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
int eraseOverlapIntervals(vector<vector<int>>& A) {
sort(begin(A), end(A), [](auto &a, auto &b) { return a[1] < b[1]; });
int ans = 0, e = INT_MIN;
for (auto &v : A) {
if (v[0] >= e) e = v[1];
else ++ans;
}
return ans;
}
};

• class Solution:
def eraseOverlapIntervals(self, intervals: List[List[int]]) -> int:
intervals.sort(key=lambda x: x[1])
ans, t = 0, intervals[0][1]
for s, e in intervals[1:]:
if s >= t:
t = e
else:
ans += 1
return ans

############

# 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 i: i.end)
ans = 0
end = float("-inf")
for interval in intervals:
# print interval.start, interval.end
if interval.start >= end:
ans += 1
end = interval.end
return len(intervals) - ans


• func eraseOverlapIntervals(intervals [][]int) int {
sort.Slice(intervals, func(i, j int) bool {
return intervals[i][1] < intervals[j][1]
})
t, ans := intervals[0][1], 0
for i := 1; i < len(intervals); i++ {
if intervals[i][0] >= t {
t = intervals[i][1]
} else {
ans++
}
}
return ans
}

• function eraseOverlapIntervals(intervals: number[][]): number {
intervals.sort((a, b) => a[1] - b[1]);
let end = intervals[0][1],
ans = 0;
for (let i = 1; i < intervals.length; ++i) {
let cur = intervals[i];
if (end > cur[0]) {
ans++;
} else {
end = cur[1];
}
}
return ans;
}