435. Non-Overlapping Intervals

Problem Statement

Pattern:

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Solution

public int eraseOverlapIntervals(int[][] intervals) {
	Arrays.sort(intervals, Comparator.comparingInt(a -> a[1]));
	int prevEnd = Integer.MIN_VALUE, count = 0;
	for (int[] curr : intervals) {
		if(prevEnd > curr[0]){      //curr is overlapping
			count++;
		} else prevEnd = curr[1];   // curr is non-overlapping
	}
	return count;
}

Notes

bascically the inverse of 452. Minimum Number of Arrows to Burst Balloons prevEnd denotes the end of the last non-overlapping interval. if any interval overlaps with this interval remove it (increase count), until you find the next non-overlappping element