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kolezko [41]
3 years ago
8

A binary message consisting of four bits was sent to you by a friend. The message was supposed to be ABAB. Unfortunately, your f

riend set the bit on the wire once every 2 seconds, but you read the wire once every second. Assuming that the first bit was sent and read at the same time, what message did you receive instead?
A. ABABB. AABBC. AAAAD. BBBBE. ABBB
Computers and Technology
1 answer:
Ostrovityanka [42]3 years ago
5 0

Answer:

ABBB

Explanation:

The first bit is sent and read at the same time, and thus, it is "A".

The second bit is sent after 2 seconds, but the reader reads after each second. So, the second, third and fourth terms are all B.

Hence, the overall message is ABBB, and the above option is the correct answer. The others are showing five bits and thus, those options are not the correct ones.

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Suppose you have two arrays of ints, arr1 and arr2, each containing ints that are sorted in ascending order. Write a static meth
telo118 [61]
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]) {
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} else {
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boolean isArr1 = i+1 < arr1.length;
for(int index = isArr1 ? i : j; index < isArr1 ? arr1.length : arr2.length; index++) {
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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.


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4 years ago
Universal Windows Platform is designed for which Windows 10 version?
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How many Little mermaid movies are there? I know there is little mermaid 1 and 2, but I am not sure how many total movies there
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There have only been 3 Little Mermaid movies. 

<span>1. The Little Mermaid (1989) </span>
<span>2. The Little Mermaid 2 Return to the Sea (2000) </span>
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What is related to Gamut in Adobe illustrator?
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Answer:

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Explanation:

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