Answer: without satellites there would be no way for the world to communicate
Explanation:
I'm assuming hardware.
Remember, Hard = Physical.
If that's not it, try peripherals.
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
