Find the Survivor of Clockwise Elimination

### Problem People labeled 1 through `n` sit clockwise. Start counting at person 1, counting that person as one. Remove the kth counted person, then begin again at the next surviving person clockwise. Repeat until one remains and return the survivor’s label. ```cpp class Solution { public: int findTheWinner(int n, int k); }; ``` ### Examples **Example 1:** ```text n = 4, k = 2 Output: 1 Explanation: People 2, 4, and 3 leave in that order. ``` **Example 2:** ```text n = 3, k = 1 Output: 3 Explanation: People 1 and 2 leave before counting reaches the final survivor. ``` ### Constraints - `1 <= k <= n <= 500`. ### Clarifying Questions Each round includes its starting person in the count. Labels are one-based even if an internal recurrence uses zero-based positions. ```hint Inspect the structure Which person is counted as one after a removal, and how are surviving labels reported? ``` ### Reasoning checkpoints - Preserve clockwise counting, restarting rules, and one-based output labels. - Meet the source follow-up of linear time and constant space. ### Optional extensions The source asks whether this can be solved in linear time and constant space; compare the recurrence with explicit queue simulation.