Answer:
Sound card
Explanation:
In computer or laptop music or sound files are stored in digital form. Sound card is the the device that is used to convert the digital form of music into analog to produce the signal that is compatible with speaker or microphone.
If the speakers are not working on the laptop, means there is no sound produced on speakers. It means that sound card is not working properly.
Answer:
ExplanKelly arrives for work at a restaurant at 5:00 p.m. Once there, she washes and chops vegetables, then sets aside ingredients and organizes them for the cook. Kelly is most likely a
✔ Food Preparation Worker
.
Mark's work hours change every week, depending on when he is needed. He wears a suit and tie to work at an elegant restaurant. He greets guests and seats them in a waiting area. He then guides them to their tables. Mark is most likely a
✔ Host
You would need to select a theme
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:
I did this in C# & Java
Explanation:
C#:
public static void Main(string[] args)
{
int input = Convert.ToInt32(Console.ReadLine());
Multiply(input);
}
public static int Multiply(int input)
{
int ans = 0;
for(int i =1; i<=10; i++)
{
ans = i*input;
Console.WriteLine(i + "*" + input + "=" + ans);
}
return ans;
}
Java:
public static void main(String[] args)
{
Scanner myObj = new Scanner(System.in);
int input = Integer.parseInt(myObj.nextLine());
Multiply(input);
}
public static int Multiply(int input)
{
int ans = 0;
for(int i =1; i<=10; i++)
{
ans = i*input;
System.out.println(i + "*" + input + "=" + ans);
}
return ans;
}
