Answer:
O(n) which is a linear space complexity
Explanation:
Space complexity is the amount of memory space needed for a program code to be executed and return results. Space complexity depends on the input space and the auxiliary space used by the algorithm.
The list or array is an integer array of 'n' items, with the memory size 4*n, which is the memory size of an integer multiplied by the number of items in the list. The listSize, i, and arithmeticSum are all integers, the memory space is 4(3) = 12. The return statement passes the content of the arithmetic variable to another variable of space 4.
The total space complexity of the algorithm is "4n + 16" which is a linear space complexity.
Answer:
The program to this question can be given as follows:
Program:
public class data //defining class data
{
//main method
public static void main (String [] as) //declaring main method
{
int userNum = 4; //declare variable userNum and assign value
for (userNum = 1; userNum <= 4; userNum++) //loop for print values
{
System.out.println(userNum); //print values in new lines.
}
}
}
Output:
1
2
3
4
Explanation:
In the above java program code firstly a class "data" is defined, inside this class the main method is defined in which an integer variable userNum is defined that holds a value that is "4".
- In this method, a for loop is declare that uses variable userNum which starts from 1 and ends with a given value that is equal to 4.
- Inside a for loop, the print function is used that print userNum variable each values in the newline.
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Answer:
arson, kidnapping, gun fighting
Explanation:
Answer:
boolean isEven = false;
if (x.length % 2 == 0)
isEven = true;
Comparable currentMax;
int currentMaxIndex;
for (int i = x.length - 1; i >= 1; i--)
{
currentMax = x[i];
currentMaxIndex = i;
for (int j = i - 1; j >= 0; j--)
{
if (((Comparable)currentMax).compareTo(x[j]) < 0)
{
currentMax = x[j];
currentMaxIndex = j;
}
}
x[currentMaxIndex] = x[i];
x[i] = currentMax;
}
Comparable a = null;
Comparable b = null;
if (isEven == true)
{
a = x[x.length/2];
b = x[(x.length/2) - 1];
if ((a).compareTo(b) > 0)
m = a;
else
m = b;
}
else
m = x[x.length/2];
