Non-overlapping Intervals: the idea to remember.

The core insight

Sort by ending time and greedily keep each interval whose start is at least the previous kept end.

Why it works

Replacing a kept interval with an earlier finishing compatible interval leaves at least as much room for every future interval.

The mistake to avoid

Touching intervals are allowed here, unlike merging closed intervals. Empty input needs no removals.

Commented Java solution

Use java.util.* imports and a surrounding solution class. Tree snippets assume the standard TreeNode definition; linked lists use ListNode and Clone Graph uses a Node with val and neighbors; Copy Random List uses Node with val, next and random. Read the input assumptions above.

int eraseOverlapIntervals(int[][] intervals) {
    Arrays.sort(intervals, (a, b) -> Integer.compare(a[1], b[1]));
    int kept = 0, end = Integer.MIN_VALUE;
    for (int[] interval : intervals) if (interval[0] >= end) { kept++; end = interval[1]; }
    return intervals.length - kept;
}

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