Hey there! So, another day, another problem! This time, it is called Odd Man Out, and it is quite simple to “get”:

You are hosting a party with G guests and notice that there is an odd number of guests! When planning the party you deliberately invited only couples and gave each couple a unique number C on their invitation. You would like to single out whoever came alone by asking all of the guests for their invitation numbers.

Well, just get the number that’s not duplicated! There are many ways to do this, be it counting the occurrences of every member of the list and finding the “odd man”, or the way I did, which is basically some push and pop into a new list. Python code follows:

for case in range(1,int(raw_input())+1):        # For all test cases
    raw_input()                                 # Discard the guest list size
    guests = raw_input().split(" ")             # Get the list of guest IDs

    guestList = []                              # Create an empty guest list

    for guest in guests:                        # For all guests
        if guestList.count(guest) > 0:          # If it's already on the final list
            guestList.remove(guest)             #    if is a couple, remove
        else:                                   # If it's not in the list yet
            guestList.append(guest)             #    Add its ID

    print "Case #%d: %s" % (case, guestList[0]) # Reports results

Needless to say, I bet there are many ways to improve this algorithm, but for this problem it performs decently, so no worries (unless you’re a hardcore bit banger). 🙂 See you tomorrow!