If every value one adds to a sorted set (redis) is one with the highest score, will the time complexity be `O(log(N))`

for each `zadd`

?

OR, for such edge cases, redis performs optimizations (e.g. an exception that in such cases where `score`

is higher than the highest `score`

in the set, simply add the value at the highest spot)?

Practically, I ask because I keep a global sorted set in my app where values are `zadded`

with time since epoch as the score. And I'm wondering whether this will still be `O(log(N))`

, or would it be faster?