There is a extension from google chrome webstore that allows to take a screnshot
Answer:
The change in complex systems can be explained according to the relationship of the environment where the system is implemented.
The system environment is dynamic, which consequently leads to adaptation to the system, which generates new requirements inherent to changes in business objectives and policies. Therefore, changing systems is necessary for tuning and usefulness so that the system correctly supports business requirements.
An example is the registration of the justification of the requirements, which is a process activity that supports changes in the system so that the reason for including a requirement is understood, which helps in future changes
Explanation:
Answer:
public class SimpleSquare{
public int num;
private int square;
public SimpleSquare(int number){
num = number;
square = number * number;
}
public int getSquare(){
return square;
}
}
Explanation:
*The code is in Java.
Create a class called SimpleSquare
Declare two fields, num and square
Create a constructor that takes an integer number as a parameter, sets the num and sets the square as number * number.
Since the square is a private field, I also added the getSquare() method which returns the value of the square.
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.
