Skip to main content
4 events
when toggle format what by license comment
Jul 17, 2021 at 0:13 vote accept EmmanuelMess
Jul 15, 2021 at 3:21 comment added Sam Westrick Yes, the whole algorithm runs in O(log n) parallel time, where n is number of input pairs. Finding the boundaries takes O(1) parallel time: in parallel for each i, if element at i is different from at i-1, then it is a boundary. And then you can filter out the non-boundaries in O(log n) parallel time.
Jul 15, 2021 at 0:43 comment added EmmanuelMess how do I find the boundry in $O(1)$ if I have to check every entry until I find the entry where the keys didn't match? In other words, if the entire list has a single key, does the algorithm run in parallel in O(log n)?
Jul 14, 2021 at 23:17 history answered Sam Westrick CC BY-SA 4.0