Answer:
Permanent storage, also called persistent storage, is any computer data storage device that retains its data when the device is unpowered. A common example of permanent storage is the computer's hard drive or SSD.
Explanation:
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.
D) Alt Tags thats the answer i think
Answer:
Issue the Dcdiag on all domain controller servers.
Explanation:
Active directory is a database that holds information of client's authentication for service access in a server. Domain controllers in servers holds these active directories, to replicate client's credentials for quick response to access to the server.
When a server domain can not contact the active directory domain controller, the Dcdiag command is issued on the server to check for connectivity issue to the DNS, AD replication, SYSVOL replication etc.
Answer:
public class Digits
{
public static boolean allDigitsOdd(int num)
{
boolean flag=true;
int rem;
while(num>0)
{
rem=num%10;
num=num/10;
if(rem%2==0) // if a even digit found immediately breaks out of loop
{
flag=false;
break;
}
}
return flag; //returns result
}
public static void main(String args[])
{
System.out.println(allDigitsOdd(1375)); //returns true as all are odd digits
}
}
OUTPUT :
true
Explanation:
Above program has 2 static methods inside a class Digits. Logic behind above function is that a number is divided by 10 until it is less than 0. Each time its remainder by 0 is checked if even immediately breaks out of the loop.
