435. Non-overlapping Intervals

1. Description

Given an array of intervals intervals where intervals[i] = [$start_i$, $end_i$], return the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping.
Note that intervals which only touch at a point are non-overlapping. For example, [1, 2] and [2, 3] are non-overlapping.

2. Example

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.

3. Constraints

• 1 <= intervals.length <= $10^5$
• intervals[i].length == 2
• -5 * $10^4$ <= starti < endi <= 5 * $10^4$

4. Solutions

Sort && Greedy

n = intervals.size()
Time complexity: O(nlogn)
Space complexity: O(logn)

// Greedy strategy:
// Sort intervals by their right endpoint in ascending order.
// Always pick the interval with the smallest right endpoint
// to minimize overlap and leave more room for subsequent intervals.
class Solution {
public:
    int eraseOverlapIntervals(vector<vector<int>> intervals) {
        sort(intervals.begin(), intervals.end(), [](const vector<int> &a, const vector<int> &b) {
            return a[1] < b[1];
        });

        const int n = intervals.size();
        int nonoverlap_right = intervals[0][1], nonoverlap_count = 1;
        for (int i = 1; i < n; ++i) {
            if (intervals[i][0] >= nonoverlap_right) {
                nonoverlap_right = intervals[i][1];
                ++nonoverlap_count;
            }
        }

        return n - nonoverlap_count;
    }
};