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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: JavaScript. What kind of script is used to run code on the client
Answer:
Advantages: computers don't make human error
It can be used for communication
Ease of access
Disadvantages: computers can be a distraction
Potential loss of privacy
It can limit learning and create a dependency
Answer:
B
Explanation:
When you initialize an instance of FunEvent(tags, year) and assign it to bc. The instance variables in this case are: self.tags = ["g", "ml"] and self.year = 2022. But then you alter tags, which will also change self.tags, since self.tags is a reference to the list you passed in as an argument. This is not the case when you do year=2023 because, first of all, integers are not mutable, and also because even if somehow integers were mutable, you're not changing the object in-place, you're simply changing the where the "variable" is pointing to. So for example if you did tags = ["g", "ml", "bc"] instead of tags.append("bc"), it would also not change the value of the instance variable "tags", because you wouldn't be changing the object in-place. So when you print(bc), the instance variables will be ["g", "ml", "bc"] and 2022. When you try to print an object, it call try to convert it into a string using the __str__ magic method. In this case it will return a string formatted as "Event(tags={self.tags}, year={self.year}) which will output "Event(tags=['g', 'ml', 'bc'], year=2022)" So the correct answer is B
