The answer is C.54.45.43.54
Claymation option A
Because Claymation is not the game design application.it is also called clay animation.
Answer:
Is there an early pay discount?
Explanation:
This determines and instructs what path the code should take,
if there is no early pay discount, it has different instructions to follow.
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:
An array.
Explanation:
An array can be defined as a structure that organizes data in a list that is commonly 1-dimensional or 2-dimensional.
Simply stated, an array refers to a set of memory locations (data structure) that comprises of a group of elements with each memory location sharing the same name. Therefore, the elements contained in array are all of the same data type e.g strings or integers.
Basically, in computer programming, arrays are typically used by software developers to organize data, in order to search or sort them.
Binary search is an efficient algorithm used to find an item from a sorted list of items by using the run-time complexity of Ο(log n), where n is total number of elements. Binary search applies the principles of divide and conquer.
In order to do a binary search on an array, the array must first be sorted in an ascending order.
Hence, array elements are mainly stored in contiguous memory locations on computer.
