Answer:
The answer is YES.
Explanation:
<em>Things pertaining to weapons, destructive materials should be shared with precautions or much better if these things should not be put online. Since internet is widely available to almost all types of users and almost to all places, we cannot be sure of who will get hold of these information. Kids could easily see this and gain knowledge and out of curiosity will try these dangerous things. Terrorist groups will also have access to these information and who knows might happen to poor innocent souls around. </em>
<em>
</em>
<em>I don't think it would violate the right of First Amendment regarding freedom of speech. Freedom should be granted on matters that are most sensible. And this time, we are just protecting the rights of children and other innocent people who can be a victim including you. We are talking about weapons and destructive materials here which is non beneficial if one could get hold of this information, in my opinion. If there is really a need to construct these things, it is better to consult an expert or an authority to ensure that nobody gets harm.</em>
Shutter speed<span> is </span>generally measured<span> by the scientific symbol “s”. The </span>measurement means<span> that the </span>measurements<span> in "s" is the reciprocal of the number when the denominator is put on the numerator side instead. ... Aperture is </span>measured<span> by f's.</span>
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:
True
Explanation:
When using databases in a project, not everyone has the same access level, e.g the database admin may have the highest level of access (access to data on live mode), the software testers have their own level of access (access to data on test mode) and so on.
For Example,
If You Were Hosting Ark Survival And Suddenly The Server Was Off Then The Players Goes Off.The Server Is How We Connect To The Platform In Order To Do The Action Intended.
