leetcode: 899. Ordered Queue [Thinking Question]

Analysis

k = 1 is nothing to say, go back and forth
If k > 1, we can hold down one at a time and connect its next one
This changes from 1 to 2 to 3, until all can be restored to any ordering we want
So when k is greater than 1 we can restore the smallest lexicographical order

ac code

class Solution:def orderlyQueue(self, s: str,k: int) -> str:# mathif k == 1:ans = sn = len(s)for i in range(n):s = s[1:] + s[0]ans = min(ans, s)return ans# k >= 2, do everythingreturn ''.join(sorted(s))

Summary

Is your head rusted today?