25. Non-overlapping Intervals

Topic :

arrays

Difficulty :

medium

Problem Link :

problem statement

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.

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

solution

import java.io.*;
import java.util.*;
 class Overlapping_Intervals
{
    public static void main(String args[]){
       int [][] intervals = {{1,2},{2,3},{3,4},{1,3}};
       System.out.println(eraseOverlapIntervals(intervals));
    }
    static int eraseOverlapIntervals(int[][] intervals) {
        if (intervals == null || intervals.length == 0) 
         return 0;
        
        Arrays.sort(intervals, (a, b) -> Integer.compare(a[0], b[0])); //sort the intervals according to the start time
        
        int prevEnd = intervals[0][1];
        int count = 0;
        
        for (int i = 1; i < intervals.length; i++) {
            // if the start time of the curr interval is less than the end time of the prev then overlap exits
            if (prevEnd > intervals[i][0]) {
                count++;
                prevEnd = Math.min(intervals[i][1], prevEnd); // keep the interval with min end time
            } 
            // incase of no overlap update the prev end
            else {
                prevEnd = intervals[i][1];
            }
        }
        return count;
    }
}