Answer:
PST
Resource:
This question has already been answered here: brainly.com/question/18415952
Explanation:
PST is the personal storage table, which is a file built by Microsoft to store data from outlook and other services too. Really quick, this may or may not be an option that you are looking for but, what I would do in my opinion would be to use the PST and export it on a removable USB, since USBs are removeable media.
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:
A blog
Explanation:
A blog is a regularly updated website or web page, typically one run by an individual or small group, that is organised by posts.
Answer:
The offset
Explanation:
In the question, we understand that the original text is HELLO and the encrypted text is OLSSV.
It should be noted that O replaced H, L replaced E, S replaced L and V replacement O.
This is made possible by a term refered to as offset.
The offset is used to determine the character that will replace another when it is encrypted.
7 characters after H is O; this is same for every other characters in the text.
Answer:
trading as a whole
Explanation:
pls give brainlest almost lvled up.
