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:
The statement about Multiprocessors that is FALSE is:
a. Asymmetric multiprocessors are a popular form of tightly coupled architecture
Explanation:
Symmetric multiprocessors house two or more identical processors sharing a single main memory. The multiprocessors are tightly coupled, and all of them can access all the connected devices without any preferential treatment of one over the others. This is unlike asymmetric multiprocessors that do not share a single main memory. Instead, they have distributed memories.
1.Extended border node (EBN)
2.Central directory server (CDS)
3.Branch extender (BEX or BrEx or BrNN)
Answer:
True
Explanation:
BSA (The Software Alliance), styled as "BSA | The Software Alliance," is a trade group founded by Microsoft in 1988 that attempts to eliminate software piracy of the software made by its members. Many major software makers are part of the BSA, including Adobe, Apple, Autodesk and Oracle, among others.
<span>Joel is a victim of a security breech. By leaving his computer unattended and not locked, he allowed for another person to use his credentials to access (or attempt) to access information that would otherwise be restricted to him. The fact that the unauthorized person did not find the information he was seeking does not minimize the risk.</span>
