How does the non-overlapping intervals problem find the minimum removals?

Sort intervals by end time. Iterate through and greedily keep intervals that do not overlap with the last kept interval. When two intervals overlap, remove the one ending later because it is more likely to overlap with future intervals. Count the number of removals needed. The minimum number of intervals to remove equals the total count minus the maximum number of non-overlapping intervals you can keep. This is equivalent to the interval scheduling maximization problem. ---