Day15. Non-overlapping_Intervals.cpp

/*

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.

 

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.
 

Note:

You may assume the interval's end point is always bigger than its start point.
Intervals like [1,2] and [2,3] have borders "touching" but they don't overlap each other.

*/


static const auto fast = []() {
   ios_base::sync_with_stdio(false);
   cin.tie(0);
   cout.tie(0);
   return 0;
} ();
class Solution {
public:
    static bool comp(vector<int> a,vector<int> b)
    {
        return a[1]<b[1];
    }
    int eraseOverlapIntervals(vector<vector<int>>& intervals) 
    {
        int n=intervals.size();
        if(n==0) return 0;
        sort(intervals.begin(),intervals.end(),comp);
        int mx=INT_MIN;
        int count=0;
        for(int i=0;i<n;i++)
        {
            if(intervals[i][0]>=mx)
            {
                count++;
                mx=intervals[i][1];
            }
        }
        
        return n-count;
    }
};