A) <span>unauthorized duplication of software </span>
Since both arrays are already sorted, that means that the first int of one of the arrays will be smaller than all the ints that come after it in the same array. We also know that if the first int of arr1 is smaller than the first int of arr2, then by the same logic, the first int of arr1 is smaller than all the ints in arr2 since arr2 is also sorted.
public static int[] merge(int[] arr1, int[] arr2) {
int i = 0; //current index of arr1
int j = 0; //current index of arr2
int[] result = new int[arr1.length+arr2.length]
while(i < arr1.length && j < arr2.length) {
result[i+j] = Math.min(arr1[i], arr2[j]);
if(arr1[i] < arr2[j]) {
i++;
} else {
j++;
}
}
boolean isArr1 = i+1 < arr1.length;
for(int index = isArr1 ? i : j; index < isArr1 ? arr1.length : arr2.length; index++) {
result[i+j+index] = isArr1 ? arr1[index] : arr2[index]
}
return result;
}
So this implementation is kind of confusing, but it's the first way I thought to do it so I ran with it. There is probably an easier way, but that's the beauty of programming.
A quick explanation:
We first loop through the arrays comparing the first elements of each array, adding whichever is the smallest to the result array. Each time we do so, we increment the index value (i or j) for the array that had the smaller number. Now the next time we are comparing the NEXT element in that array to the PREVIOUS element of the other array. We do this until we reach the end of either arr1 or arr2 so that we don't get an out of bounds exception.
The second step in our method is to tack on the remaining integers to the resulting array. We need to do this because when we reach the end of one array, there will still be at least one more integer in the other array. The boolean isArr1 is telling us whether arr1 is the array with leftovers. If so, we loop through the remaining indices of arr1 and add them to the result. Otherwise, we do the same for arr2. All of this is done using ternary operations to determine which array to use, but if we wanted to we could split the code into two for loops using an if statement.
Answer:
% here x and y is given which we can take as
x = 2:2:10;
y = 2:2:10;
% creating a matrix of the points
point_matrix = [x;y];
% center point of rotation which is 2,2 here
x_center_pt = x(2);
y_center_pt = y(2);
% creating a matrix of the center point
center_matrix = repmat([x_center_pt; y_center_pt], 1, length(x));
% rotation matrix with rotation degree which is 45 degree
rot_degree = pi/4;
Rotate_matrix = [cos(rot_degree) -sin(rot_degree); sin(rot_degree) cos(rot_degree)];
% shifting points for the center of rotation to be at the origin
new_matrix = point_matrix - center_matrix;
% appling rotation
new_matrix1 = Rotate_matrix*new_matrix;
Explanation:
We start the program by taking vector of the point given to us and create a matrix by adding a scaler to each units with repmat at te center point which is (2,2). Then we find the rotation matrix by taking the roatational degree which is 45 given to us. After that we shift the points to the origin and then apply rotation ans store it in a new matrix called new_matrix1.
Answer:
The oldest malware vector was emails.
Explanation:
Websites were not widespread early on into the internet, however it was very easy to have a virus that uses an email contact list, where it can send itself to other email addresses and spread.
Liquidity Effect. When the Fed pursues a tight monetary policy, it takes money out of the system by selling Treasury securities and raising the reserve requirement at banks. This raises interest rates because the demand for credit is so high that lenders price their loans higher to take advantage of the demand.
