Answer:
"NOT lookup
" is the correct approach.
Explanation:
- This methodology significantly reduces the quantity of congestion of DNS messages on a certain file. The application establishes that whenever a question reaches if it is processed. Unless the file is loaded, then perhaps the response is returned with the cached cache.
- Typically the name resolution occurs in something like a DNS File. The conversion usually occurs throughout this cycle from Username to IP, including IP via Username.
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:
Answer to the following question is CoC(chain of custody
).
Explanation:
- CoC(chain of custody
) is the legal document and it is the evidence that is handled during an investigation of the case.
- In the criminal trials, prosecution must be typically proved that all the evidences is handled according the documented and unbroken CoC.
- In the court, CoC(chain of custody) documentation has been presented by prosecutions to prove the evidence, related to an alleged crimes, and it has been in possession of defendant.
Void sort3(int& a, int& b, int& c)
{
if (a > b)
{
std::swap(a, b);
}
if (b > c)
{
std::swap(b, c);
}
if (a > b)
{
std::swap(a, b);
}
}
