Eased the mobilization and transportation of troops
Answer:
A HUB
Explanation:
A hub can described as an hardware or network device use for multi purpose functions like connecting series of internet devices together there by making it possible for all devices to work or function together on a single network.
Hubs are classified into active hub, passive hub, and intelligent hub.
ACTIVE HUB: This type of hub boost the signals but can not perform calculations.
PASSIVE HUB: this type only send the signals but can not boost a signal.
INTELLIGENT HUB: it brings about the boosting of signals and also bring about packets to be analyzed.
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:
Worm malware
Explanation:
The worm operates destructively as it replicates itself and keeps spreading within a computer or network. It takes advantage of software and security vulnerabilities.
Once a computer or network is infected, worms replicate themselves, therefore using up the computer and network resources. Among others, one of the major missions of worms is to create a backdoor into a network, to be able to enable the attacker to carry out a more devastating attack.
Since worms take advantage of weak security in the Operating system, it is very important to constantly update the security features of the Operating system. Users must be careful of the emails they open and attachments they download.
