Python 3   ||   7 lines,  w/example   ||   T/S: 92% / 91%
\nclass Solution:\n    def circularGameLosers(self, n, k):\n\n        remain, ball, nxt = set(N:=range(1,n)), k,0\n\n        for _ in N:                        
```\nclass Solution:\n    def circularGameLosers(self, n, k):\n\n        remain, ball, nxt = set(N:=range(1,n)), k,0\n\n        for _ in N:                         #  Example: n = 5   k = 2\n\n            nxt = (nxt + ball)%n            #  ball     nxt    remain\n            if nxt not in remain: break     #  -----   -----   -----\n                                            #    2       0     [1, 2, 3, 4]\n            remain.remove(nxt)              #    4       2     [1, 3, 4]\n            ball+= k                        #    6       1     [3, 4]\n                                            #    6       2     [3, 4] <-- 2 appears again\n\n        return [i+1 for i in remain]        #  return [3+1, 4+1] --> [4, 5]\n```\n[https://leetcode.com/problems/find-the-losers-of-the-circular-game/submissions/1297107321/](https://leetcode.com/problems/find-the-losers-of-the-circular-game/submissions/1297107321/)\n\nI could be wrong, but I think that time complexity is *O*(*N*) and space complexity is *O*(*N*).\n\n[Edited 7/28/23] See [@almostmonday](/almostmonday) in comments.
['Python3']