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Lunna [17]
3 years ago
14

Over the last few years, security cameras and other devices have become more common. Some people argue that these measures viola

te the right to personal privacy. Others argue that they are necessary to reduce crime. What do you think? Why?
Computers and Technology
2 answers:
jarptica [38.1K]3 years ago
7 0

Answer:

I think that privacy is an ever-growing concern for the population as more of our daily lives are documented by cameras. I believe that as long as cameras are not placed in areas without the publics knowledge or consent, then they are fine. Though people think that security cameras violate the right to personal privacy, it's also necessary for them to be set in place. An example would be a robbery, a murder, a hit and run, a shootout, and many more events. I think I am most scared of cameras being hidden in things that aren't supposed to have cameras.

Explanation:

Tomtit [17]3 years ago
3 0
Though people think that Security cameras violate the right to personal privacy, it's also necessary for them to be set in place. A few examples of this would be store robbery, School Fights, ect. Some could argue as well that people recording everyday things are invading privacy. Take SnapChat for example, people get self-conscious or upset that they're being recorded or having their picture taken. I hope this helps you !!
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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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boolean isArr1 = i+1 < arr1.length;
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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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Answer:

Answers explained with appropriate comments

Explanation:

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