Is this dealing with the computer because I can help you then
Answer:
Should no such operating system be found it will show an error something like “Boot sector error” or “operating system missing”.
Explanation:
:)
Answer:
what the heck? what kind of question is that
Answer:In the table, right-click in the row that you want to repeat, and then click Table Properties. In the Table Properties dialog box, on the Row tab, select the Repeat as header row at the top of each page check box.
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