It's either microsoft or office. Based on research I would say office.
I really hope this helps you! :-)
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.
The best possible solution for the technician to do is to go into the disk management and find out what exactly is going on. The technician should check whether there is partition that has unallocated space. It is 100% the case that the rest of the 500 GB is in the unallocated space.
The techie need to grow his partition. Possible option for a scenario like this is delete the unallocated 500 GB space using NTFS. He can then recreate the available 500 GB free space as 1TB partition.
U can see the printscr botton at the top right of ur keyboard and then u can type paint and the control v.
Answer:
Explanation:
Transitive dependency
In this case, we have three fields, where field 2 depends on field 1, and field three depends on field 2.
For example:
Date of birth --> age --> vote
Partial dependency
It is a partial functional dependency if the removal of any attribute Y from X, and the dependency always is valid
For example:
Course and student these tables have a partial dependency, but if we have the field registration date, this date will depend on the course and student completely, we must create another table with the field registration date to remove this complete dependency.
If we remove or update the table registration date, neither course nor student must not change.
