2026-09-01 18:08 UTC
Seems straightforward enough. For j values of 1 to i it will not do anything because the largest element in the array has already been moved to position i in some earlier iteration in the i loop. For j values greater than i it then proceeds to find the largest remaining element place in position i.
Replies (1)
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@queerlilhayseed@piefed.blahaj.zone 2026-09-01 18:22
For the j > i case I think you’re right, it sorts largest to smallest (or, backwards), but for the j i part of the algorithm. Sort of a “two wrongs that accidentally make a right” maneuver.