Answer:
d)anotherFunc(myints);
Explanation:
When we call a function we just have to pass the names of the arguments to the function.In case of arrays we also just have to pass the name of the array.We don't have to worry about the square brackets.So the function call will be like this.
anotherFunc(myints);
Hence the answer is option d.
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.
Michael will use a Adobe Photoshop or CorelDraw tool to help format the text for creating an informative poster.
<h3>What application is used for graphic design?</h3>
They are:
- Adobe Photoshop
- Illustrator, GIMP
- CorelDraw
- Canva and others
Based on the American Institute of Graphic Arts (AIGA), graphic design is known to be a term that is described as “the art and method of planning and bringing forth ideas and experiences along with the use of visual and textual content.”
Therefore, Michael will use a Adobe Photoshop or CorelDraw tool to help format the text for creating an informative poster.
Learn more about graphic design from
brainly.com/question/27019704
#SPJ1
Answer:
<u>First test:</u> Give a list of disordered numbers to the sorting algorithm an examine if the output is correctly sorted.
<u>Second test:</u> Give a list of ordered numbers to the sorting algorithm an analyze if the output is still correctly ordered.
<u>Third test:</u> Give a list of ordered numbers and some non-numeric values to the sorting algorithm and check how is managed the exception in case of error or if the output is correctly ordered.