How does the greedy approach solve interval scheduling and meeting rooms?

For the maximum non-overlapping intervals problem, sort intervals by end time and greedily select the earliest-ending interval that does not overlap with the last selected one. This works because finishing early leaves maximum room for future intervals. For the minimum meeting rooms problem, separate start and end times into two sorted arrays, then sweep through them chronologically: each start increments active meetings and each end decrements, tracking the maximum concurrent meetings as the answer. Both run in O of n log n due to sorting.