Well knowing how the government is with security id go with A. as it makes a lot of sense unlike B and D and if they were looking for software the question would mention it
Answer:
The correct answer to the following question will be "Data-in-use".
Explanation:
- Data-in-use is an IT term referring to active information that is usually preserved in a semi-persistent physical state in RAM of computer, CPU registers or caches.
- It might be created, modified or changed, deleted or accessed via different endpoints of the interface. This is indeed a useful term for IT departments to pursue institutional defense.
Therefore, it's the right answer.
C. is the answer to that question
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:
Explanation:
There are many weaknesses/vulnerabilities, some of which are the following...
Default Network Hardware, many individuals will go out and buy a new router and install it in their home. These devices come with a preset configuration including a preset security password which is many times a default password used for every router of the same model. This can be easily obtained by anyone who can then access the network.
Access Point hacking, an experienced individual can use a packet sniffer to detect the SSID that is frequently sent from the router in order to create an access point to be able to access the network.
WEP encryption is another vulnerability. These are very low security passwords that can be cracked using different software in a short period of time.
These are some of many vulnerabilities that wireless local networks have, but there are also many precautions and security measures that can be taken to prevent them.
