Answer:
Answer explained below
Explanation:
I have given two approaches in implementing the solution.
1. Using the for loop, in which you have to iterate over all the elements in list 1 and check in list 2
2. Use the set intersection method. As intersection will give u the common elements. And we can get there length by using len method.
I have added the code along with the snapshot and inline comment for the ease of you to understand. Please check the methods below. You can use either of them.
METHOD-1:
********** CODE *****************
def matches(tickets,winner):
tickets = set(tickets)
winner = set(winner)
counter = 0 #To Count the common elements
for i in tickets: # Iterate over all the elements in tickets.
if i in winner: # Check the element in the winner list
counter = counter+1
return counter
METHOD -2:
********** CODE ********************
def matches(tickets, winner):
tickets = set(tickets)
winner = set(winner)
return len(tickets.intersection(winner))
Answer:
U can go to your profile, then go to your question to see what they have written.
Explanation: Hope this helps
<h2>Answer :</h2>
In case of Flash Sale, usually company uses Instant messages or SMS. As there are chances customer in most of the cases don’t have access to their emails due to internet unavailability. However SMS can be directed towards customers without the requirements of an internet. In case of SMS, there are companies available that can shoot out targeted SMS based on cities or even an entire country at a very minimal price. Company can provided them there contacts as well as they have a list of numbers from different sources to which they can send instant messages.
Answer:
There is also an attachment below
Explanation:
Since we are talking about binary search, let's assume that the items are sorted according to some criteria.
Time complexity of binary search is O(logN) in worst case, best case and average case as well. That means it can search for an item in Log N time where N is size of the input. Here problem talks about the item not getting found. So, this is a worst case scenario. Even in this case, binary search runs in O(logN) time.
N = 700000000.
So, number of comparisions can be log(N) = 29.3 = 29.
So, in the worst case it does comparisions 29 times
