Frequency Queries

  • + 0 comments

    your solution is 98% correct, but when frequency reaches 0 then that element should be removed from the map.

    so in

    elif(queries[i][0] == 2):
                 if queries[i][1] in num:
                      num[queries[i][1]] = num[queries[i][1]] - 1
     								 if num[queries[i][1]] == 0:		 
     											 del num[queries[i][1]]