# 435. Non-overlapping Intervals

## Description

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 <= 105
• intervals[i].length == 2
• -5 * 104 <= starti < endi <= 5 * 104

## Solutions

Greedy.

• 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;
}
}

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


• 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;
}