I don't think you can check if array of n elements is sorted in O(1), if you skip the check though and just assume it is sorted now (have faith), then the time would be constant, depending on how long you're willing to wait until the miracle happens. As long as MTM (Mean Time to Miracle) is constant, the faithfull miracle sort has O(1) time complexity, even if MTM is infinite. Faithless miracle sort has at best the complexity of the algorithm that checks if the array is sorted.

Technically you can to down to O(0) if you assume all array are always sorted.

[–] 1boiledpotato@sh.itjust.works 1 points 2 weeks ago

Oh yeah, I didn't think about the time that it takes to check if it's sorted. The sorting time is constant though