Well think of it like this lets say this wasn't and security system and it was a video game console and remote.
Now the console or keypad or the main components the power on or shut off the devices and the remote or sensors are the secondary parts but they are needed to help the console or keypad figure out what they are doing.
Natural systems are systems that came into being by natural processes. Like the rain cycle.
Technological systems are those to whom people intervened or designed. Like diverting water for a rider for land irrigation.
<u>Answer:</u>
<em>using System;
</em>
<em>public class Program
</em>
<em>{
</em>
<em> public static void Main()
</em>
<em> {
</em>
<em> String input ;
</em>
<em> while(true)
</em>
<em> {
</em>
<em> input = Console.ReadLine();
</em>
<em> if(input.Equals(""quit""))
</em>
<em> break;
</em>
<em> Reverse_String(input);
</em>
<em> }
</em>
<em> return;
</em>
<em> }
</em>
<em>static void Reverse_String(string input_text)
</em>
<em>{
</em>
<em> char[] text = input_text.ToCharArray();
</em>
<em> Array.Reverse(text);
</em>
<em> Console.WriteLine(text);
</em>
<em>}
</em>
<u>Explanation:</u>
<em>In the above program a separate function is written to reverse the string.</em>
This method takes the string as an argument and place it in a array and then use the built-in function reverse and print the reversed string in the console.
<em>In the main(), the input is obtained from the console and it is passed to the reversestring().
</em>
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.
