Greedy — Non-overlapping Intervals

solve(intervals) — intervals[i] = [start, end]. Return the **minimum number of
intervals to remove** so that the rest do not overlap. Intervals that only touch
([1,2] and [2,3]) do not overlap.

Equivalently, keep the maximum number of compatible intervals. The classic
greedy choice: sort by end time and always keep the interval that ends earliest,
because it leaves the most room for the others.

Examples

solve([[1,2],[2,3],[3,4],[1,3]])           → 1
solve([[1,2],[1,2],[1,2]])                 → 2
solve([[1,2],[2,3]])                       → 0
solve([[1,100],[11,22],[1,11],[2,12]])     → 2
solve([])                                  → 0