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:
Option (d) is the correct answer.
Explanation:
An Array is used to store multiple variables in the memory in the continuous memory allocation on which starting index value is starting from 0 and the last index value location is size-1.
In java programming language the array.length is used to tells the size of the array so when the user wants to get the value of the last element, he needs to print the value of (array.length-1) location so the correct statement for the java programming language is to print the last element in the array named ar is--
System.out.println(ar[ar.length-1]);
No option provides the above statement, so option d (None of these) is correct while the reason behind the other option is not correct is as follows--
- Option a will prints the size of the array.
- Option b also gives the error because length is an undeclared variable.
- Option c will give the error of array bound of an exception because it begs the value of the size+1 element of the array.
Answer:
d. This statement results in input failure.
Explanation:
We are taking input of 3 variables only and those are x,ch,and y. cin is a C++ keyword used take input entered by the user onto the screen cin is not a variable.
The input should of type int char int as written in the code. First is x and character variable ch and then the integer variable y and there should be only three inputs.
Answer:
Dependent sample t-test
Explanation:
The Dependent sample t-test compares the mean score of measurements in one group to that of another other group. It mainly used when analyzing comparable sample units as it pairs repeatable observations within a time frame.
For example, a researcher administered a measure of working memory to a group of subjects at 8am on Day 1 of the study and then again at 8am on Day 2 of the study, after keeping the subjects awake the entire night.
