Answer:
E)II and III only
Step-by-step explanation:
This can be seen with examples. Say a[0]=1 and and a[1]=2.
for I , the first line of code would be:
a[0]=a[1];
thus, we would get a new value for a[0]=2.
The second line of code
a[1]=a[0]; uses the new value of a[0], so we would get a[1]=2.
The end result is a[0]=2, and a[1]=2 which doesn't exchange the values of the first two elements.
For II the first line of code
int t= a[0]; saves the original value of a[0] to t, so we get t=1.
the second line of code
a[0]=a[1]; changes the value of a[0] to that of a[1]. Thus, in our example a[0]=2.
the final line
a[1]=t; changes the value of a[1] to the original value of a[0], giving us a[1]=1 and a[0]=2, what we were looking for.
For III
the first line of code
a[0]=a[0]-a[1];
gives us
a[0]=1-2
the secon line
a[1]=a[1]+a[0];
takes the new value of a[0] and replaces it in the expression
a[1]= 2+(1-2)=1
the last line
a[0]=a[1]-a[0];
takes the new value of a[0] and a[1] and replaces the in the expression
a[0]=1-(1-2)=1-1+2=2
which exchanges the values needed.
So we can see that only II and III do what we require, giving us E as the answer.
