Answer:
your on unit 9 of edhesive?
Explanation:
You should follow a set of directions
Answer:
public static void swapPairs(int[] a){
int len=a.length;
if(len%2 ==0){
for(int i=0; i<len; i=i+2){
a[i]=a[i+1];
a[i+1]=a[i];
int[] b={a[i]+a[i+1]};
}
}
if(len%2 !=0){
for(int j=0; j<len; j=j+2){
a[j]=a[j+1];
a[j+1]=a[j];
a[len-1]=a[len-1];
int[] b={a[j]+a[j+1]+a[len-1]};
}
}
}
public static void printArray(int[] a){
System.out.println(a);
}
Big-O notation is a way to describe a function that represents the n amount of times a program/function needs to be executed.
(I'm assuming that := is a typo and you mean just =, by the way)
In your case, you have two loops, nested within each other, and both loop to n (inclusive, meaning, that you loop for when i or j is equal to n), and both loops iterate by 1 each loop.
This means that both loops will therefore execute an n amount of times. Now, if the loops were NOT nested, our big-O would be O(2n), because 2 loops would run an n amount of times.
HOWEVER, since the j-loop is nested within i-loop, the j-loop executes every time the i-loop <span>ITERATES.
</span>
As previously mentioned, for every i-loop, there would be an n amount of executions. So if the i-loop is called an n amount of times by the j loop (which executes n times), the big-O notation would be O(n*n), or O(n^2).
(tl;dr) In basic, it is O(n^2) because the loops are nested, meaning that the i-loop would be called n times, and for each iteration, it would call the j-loop n times, resulting in n*n runs.
A way to verify this is to write and test program the above. I sometimes find it easier to wrap my head around concepts after testing them myself.
