non-overlapping intervals.cpp

// 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
// -5 * 104 <= starti < endi <= 5 * 104

//   CODE:

  class Solution {
public:
    int eraseOverlapIntervals(vector<vector<int>>& intervals) {
        sort(intervals.begin(),intervals.end());
        int previous = 0;
        int n = intervals.size();
        int ans = 0;
        for(int current = 1;current<n; current++){
            if(intervals[current][0]<intervals[previous][1]){
                ans++;
                if(intervals[current][1]<=intervals[previous][1])
                previous = current;
            }
           
            else{
                previous = current;

            }

        }
        return ans;

        
    }
};