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 < b; });
int ans = 0, e = INT_MIN;
for (auto &v : A) {
if (v >= e) e = v;
else ++ans;
}
return ans;
}
};


Java

• 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 != interval2)
return interval1 - interval2;
else
return interval1 - interval2;
}
});
int eraseCount = 0;
int prev = 0;
int length = intervals.length;
for (int i = 1; i < length; i++) {
if (intervals[prev] > intervals[i]) {
if (intervals[prev] > intervals[i])
prev = i;
eraseCount++;
} else
prev = i;
}
return eraseCount;
}
}

• // 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 < b; });
int ans = 0, e = INT_MIN;
for (auto &v : A) {
if (v >= e) e = v;
else ++ans;
}
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