Answer:
- def check_subset(l1, l2):
- status = False
- count = 0
- if(len(l1) > len(l2)):
- for x in l2:
- for y in l1:
- if x == y:
- count += 1
-
- if(count == len(l2)):
- return True
- else:
- return False
-
- else:
- for x in l1:
- for y in l2:
- if x==y:
- count += 1
-
- if(count == len(l1)):
- return True
- else:
- return False
-
- print(check_subset([1,4,6], [1,2,3,4,5,6]))
- print(check_subset([2,5,7,9,8], [7,8]))
- print(check_subset([1, 5, 7], [1,4,6,78,12]))
Explanation:
The key idea of this solution is to create a count variable to track the number of the elements in a shorter list whose value can be found in another longer list.
Firstly, we need to check which list is shorter (Line 4). If the list 2 is shorter, we need to traverse through the list 2 in an outer loop (Line 5) and then create another inner loop to traverse through the longer list 1 (Line 6). If the current x value from list 2 is matched any value in list 1, increment the count variable by 1. After finishing the outer loop and inner loop, we shall be able to get the total count of elements in list 2 which can also be found in list 1. If the count is equal to the length of list 2, it means all elements in the list 2 are found in the list 1 and therefore it is a subset of list 1 and return true (Line 10-11) otherwise return false.
The similar process is applied to the situation where the list 1 is shorter than list 2 (Line 15-24)
If we test our function using three pairs of input lists (Line 26-28), we shall get the output as follows:
True
True
False
earliest = ""
while True:
month = int(input("Enter a date (month): "))
day = int(input("Enter a date (date): "))
year = int(input("Enter a date (year): "))
if month == 0 and day == 0 and year == 0:
break
if month < 10:
month = "0"+str(month)
if day < 10:
day = "0"+str(day)
string_date = str(month)+"/"+str(day)+"/"+str(year)
print(string_date)
if earliest == "":
earliest = string_date
else:
year,month,day=int(year),int(month),int(day)
lst = list(map(int,earliest.split("/")))
if year < lst[2]:
earliest = string_date
elif year == lst[2] and month < lst[0]:
earliest = string_date
elif year == lst[2] and lst[0] == month and day < lst[1]:
earliest = string_date
print(earliest,"is the earliest date")
I wrote my code in python 3.8. Best of luck.
Depends on what application - If it is a number crunching application, it will use a lot of CPU power
If it's a graphics intensive video game, it will take a lot of graphics card power
If it's a application that has a lot of things going on, like tabs in google chrome, then it will use a lot of RAM
