Answer:
The answer to this question is a. Event log
Explanation:
Event log in Windows is a comprehensive record of security,system and application notifications stored by the Windows and it is used by administrators to determine problems in the system and foretell issues that going to happen in future.
Operating system and the Apps use event logs to keep account of important software and hardware activity that administrator can use to correct issues with the operating system.
Answer:
The answer to this question is given below in the explanation section. the correct option is C.
Explanation:
This is Java code statement:
System.out.print("Computing\nisInfun");
The output of this code statement is
Computing
isInfun
However, it is noted that the C option is not written correctly, but it is guessed that it will match to option C.
This Java code statement first prints "Computing" and then on the next line it will print "isInfun" because after the word "Computing" there is a line terminator i.e. \n. when \n will appear, the compiler prints the remaining text in the statement on the next line.
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
Explanation:
If you find something embarrassing, its more likely that other people with also think its embarrassing. All the other points are on the internet right now. Amazon sells things, News show information and there are political campaigns to show you there offers
