Since you provide no choices, here are several things that should cover every part of the worksite :
- Emergency exit
- Safety measures
Hope this helps
Answer:
Option c is correct.
Example :
public class Subject
{
public static void main(String [] args)
{
int ar[]={5,4,6,7,8};
System.out.println("the number of array elements are: ",ar.length);
}
}
Explanation:
The above program is created in java language in which Subject is a class name.Inside Subject class , there is main method.
Inside the main method, there is An array named 'ar' which data type is an integer and we have assigned the value to this array.
in the next line, we are printing the total no. of the element in the array with the length function which displays the length of an array or variable.
Answer:
n := length(A)
repeat
swapped := false
for i := 1 to n-1 inclusive do
<em> /* if this pair is out of order */</em>
if A[i-1] > A[i] then
<em> /* swap them and remember something changed */</em>
swap(A[i-1], A[i])
swapped := true
end if ⇒
end for
until not swapped
end procedure
Explanation:
This is pseudocode
Answer:
Check the explanation
Explanation:
Player 1 Coin
Player 2 Coin
Player 1
Player 2
Round
Count at
Count at
Player
Player
Coin
Coin
Number
Beginning of
Beginning of
1
2
Outcome
Count at
Count at
Round
Round
Spends
Spends
End of
End of
Round
Round
Off-by-
one,
10 - 1
10
10
2
player 2
10 - 2
+
= 8
1
gains 1
= 10
coin
Same,
10 - 2
2
8
10
2
2
player 2
-
gains 1
6
+
=
1
=
9
coin
Off-by-
two,
6 - 1
+
3
6
9
3
player 1
2
9 - 3
gains 2
=
=
7
6
coins
Same,
4
7
6
2
2
player 2
7 - 2
6 - 2+
gains 1
=
5
=
coin
5
Kindly check the attached image below to see the well arranged table to solve the above question.
Answer:
The main benefit of the ordered list is that you can apply Binary Search( O( n log n) ) to search the elements. Instead of an unordered list, you need to go through the entire list to do the search( O(n) ).
The main cost of the ordered list is that every time you insert into a sorted list, you need to do comparisons to find where to place the element( O( n log n) ). But, every time you insert into an unsorted, you don't need to find where to place the element in the list ( O(1) ). Another cost for an ordered list is where you need to delete an element, you have an extra cost rearranging the list to maintain the order.
