My personal favorite is Metal gear solid 3: <em>Snake eater</em>.
Answer:
D: The protocols of the Internet are open and used by all devices connected to the network
Explanation:
There are billions of devices connected to the Internet, and hundreds of different kinds of devices: laptops, tablets, phones, refrigerators, handheld credit card readers, and so on. Protocols (standards) ensure that the variety of devices interact with each other smoothly. There are a lot of protocols! The Internet was designed with several layers of abstraction that sort the protocols according to what part of the process they support.
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:
space alignment
Explanation:
I'm not sure just taking a guess cuz I'm bored lol
Answers
- OS(The Operating System) sends <em>interrupts to the processor</em> to stop whatever is being processing at that moment and computer architecture send <em>data bus</em>. This bus sends<u> data between the processor,the memory and the input/output unit.</u>
- The operating system is a low-level software that supports a <em>computer’s basic functions</em>, such as <u>scheduling tasks and controlling peripherals</u> while the computer architecture has the <em>address bus bar</em>. This bus carries <u>signals related to addresses between the processor and the memory.
</u>
- The interface between <em>a computer’s hardware and its software</em> is its Architecture while An operating system (OS) is<u> system software that manages computer hardware and software resources and provides common services for computer programs.</u>
Explanation:
In short explanation,the Computer Architecture specifically <em>deals with whatever that's going on in the hardware part of the computer system </em>while the Operating System is the computer program <em>which has been program to execute at some instances depending on the programming instructions embedded in it</em>. An example is the MS Office.
