435. Non-overlapping Intervals

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

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.

// Sort by start
int eraseOverlapIntervals(vector<vector<int>>& intervals) { // time: O(nlogn); space: O(1)
    if (intervals.size() <= 1) return 0;
    sort(intervals.begin(), intervals.end());
    int res = 0, minEnd = intervals[0][1];
    for (int i = 1; i < intervals.size(); ++i) {
        if (minEnd <= intervals[i][0]) {
            minEnd = intervals[i][1];
        } else { // overlapping
            minEnd = min(minEnd, intervals[i][1]);
            ++res;
        }
    }
    return res;
}

// Sort by end
int eraseOverlapIntervals(vector<vector<int>>& intervals) { // time: O(nlogn); space: O(1)
    if (intervals.size() <= 1) return 0;
    sort(intervals.begin(), intervals.end(), [](vector<int>& a, vector<int>& b) {
        return a[1] < b[1];
    });
    int res = 0, minEnd = intervals[0][1];
    for (int i = 1; i < intervals.size(); ++i) {
        if (minEnd <= intervals[i][0]) {
            minEnd = intervals[i][1];
        } else {
            ++res;
        }
    }
    return res;
}

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