Example 1:
Input: events = [["u1","#ai"],["u1","#ai"],["u2","#ai"],["u2","#ml"],["u3","#ml"],["u4","#db"],["u4","#db"]], k = 2
Output: ["#ai","#ml"]
Explanation:
Example 2:
Input: events = [["alice","#x"],["bob","#y"],["alice","#y"],["alice","#z"]], k = 5
Output: ["#y","#x","#z"]
Explanation:
You are given a list of hashtag events events, where each event is [userId, hashtag].
A hashtag's popularity is the number of distinct users who used it.
If the same user uses the same hashtag multiple times, it should count only once for that hashtag.
Return the top k hashtags ranked by:
If k is larger than the number of distinct hashtags, return all hashtags in ranked order.
#ai and #ml both have popularity 2, so tie-break by lexicographic order.
Popularities are #y = 2, #x = 1, #z = 1.
0 <= events.length <= 10^5events[i].length == 21 <= userId.length, hashtag.length <= 501 <= k <= 10^4