435. Non-overlapping Intervals

Problem statement

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.

Solution

This is similar with 646. Maximum Length of Pair Chain.

The only difference is the return value.

/**
 * Definition for an interval.
 * struct Interval {
 *     int start;
 *     int end;
 *     Interval() : start(0), end(0) {}
 *     Interval(int s, int e) : start(s), end(e) {}
 * };
 */
class Solution {
public:
    int eraseOverlapIntervals(vector<Interval>& intervals) {
        if(intervals.empty()){
            return 0;
        }
        sort(intervals.begin(), intervals.end(), [](Interval l, Interval r) { return l.end < r.end ;});
        int len = 1;
        for(int i = 0, j = 1; j < intervals.size(); j++){
            if(intervals[i].end <= intervals[j].start){
                len++;
                i = j;
            }
        }
        return intervals.size() - len;
    }
};