redis unsubscribe time complexity - redis

Question

Document says that "O(N) where N is the number of clients already subscribed to a channel"
Does N means the number of clients who subscribing the channel which I want to unsubscribe, or who subscribing any channel?

Answer 1

It means the sum of all clients subscribed to all the channels you unsubscribe in the batch operation.
UNSUBSCRIBE [channel [channel ...]]

Answer 2

Good question, like #JeanJacquesGourdin said, it is sum of clients of all channels that you are going to unsubscribe. I look into source code of redis 5.0.5, here is what I found, please correct me if I am wrong, thanks in advance
There is a server side hashtable called server.pubsub_channels, it keeps track of all channels and its subscribed clients
when you want to unsubscribe one channel, the server will try to remove your client from list of the server.pubsub_channels[your_channel], and it is a O(N) time complexity operation to iter through this list. ln = listSearchKey(clients,c);
int pubsubUnsubscribeChannel(client *c, robj *channel, int notify) {
dictEntry *de;
list *clients;
listNode *ln;
int retval = 0;
/* Remove the channel from the client -> channels hash table */
incrRefCount(channel); /* channel may be just a pointer to the same object
we have in the hash tables. Protect it... */
if (dictDelete(c->pubsub_channels,channel) == DICT_OK) {
retval = 1;
/* Remove the client from the channel -> clients list hash table */
de = dictFind(server.pubsub_channels,channel);
serverAssertWithInfo(c,NULL,de != NULL);
clients = dictGetVal(de);
ln = listSearchKey(clients,c); /** the iteration occurs here **/
serverAssertWithInfo(c,NULL,ln != NULL);
listDelNode(clients,ln);
if (listLength(clients) == 0) {
/* Free the list and associated hash entry at all if this was
* the latest client, so that it will be possible to abuse
* Redis PUBSUB creating millions of channels. */
dictDelete(server.pubsub_channels,channel);
}
}
/* Notify the client */
if (notify) {
addReply(c,shared.mbulkhdr[3]);
...
}
decrRefCount(channel); /* it is finally safe to release it */
return retval;
}
listNode *listSearchKey(list *list, void *key)
{
listIter iter;
listNode *node;
listRewind(list, &iter);
while((node = listNext(&iter)) != NULL) {
if (list->match) {
if (list->match(node->value, key)) {
return node;
}
} else {
if (key == node->value) {
return node;
}
}
}
return NULL;
}