Leetcode 435. Non-overlapping Intervals
1 min readJul 2, 2022
Array
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
Solution
class Solution {
public:
struct ascending {
bool operator()(const vector<int>& a, const vector<int>& b) {
return a[1] < b[1];
}
};
int eraseOverlapIntervals(vector<vector<int>>& intervals) {
sort(intervals.begin(), intervals.end(), ascending());
int end = intervals[0][1];
int count = 1;
for (auto& interval: intervals) {
if (interval[0] >= end) {
end = interval[1];
count++;
}
}
return intervals.size() - count;
}
};
Runtime: 702 ms, faster than 43.94% of C++ online submissions for Non-overlapping Intervals.
Memory Usage: 89.8 MB, less than 82.03% of C++ online submissions for Non-overlapping Intervals.