Answer:
enhanced for loop
Explanation:
Enhanced for loop is an improve concept about loops, this features was implemented in Java SE 5.0 version, this method simplify the For structure. For example:
for (int i = 0; i <array.length; i ++) {
System.out.print (array [i]);
}
Enhanced for loop
for (String element : array) {
System.out.print(element);
}
The part of a computing device that is the biggest power consumer is;
Central Processing Unit
<h3>Central Processing Unit</h3>
The part of a computing device that cosnumes the most power is called the Central Processing Unit (CPU). This is due to the following reasons;
- The CPU consists of motherboard and battery that draws power from the socket.
- The Mother board is equipped with logic gates that are used to implement logic in the computer.
- The Logic gates are voltage signals that signify the binary information that are fed into the system.
Read more about Central Processing Unit (CPU) at; brainly.com/question/4558917
A useful advantage of Asymmetric encryption over symmetric
encryption is that there is no secret channel necessary for the
exchange of the public key, unlike in the symmetric encryption which requires a
secret channel to send the secret key.
Another
advantage of Asymmetric encryption is that is has increased security. Asymmetric
uses two different keys (Public and private) for both encryption and decryption
of data while symmetric uses one.
Three inputs have 8 possible states.
Here they are:
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
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.
