redis unsubscribe time complexity

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

1. There is a server side hashtable called server.pubsub_channels, it keeps track of all channels and its subscribed clients
2. 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;
}