Answer:
The solution is written using Python as it has a simple syntax.
- def getHighScores(gameScores, minScore):
- meetsThreshold = []
- for score in gameScores:
- if(score > minScore):
- meetsThreshold.append(score)
- return meetsThreshold
- gameScores = [2, 5, 7, 6, 1, 9, 1]
- minScore = 5
- highScores = getHighScores(gameScores, minScore)
- print(highScores)
Explanation:
Line 1-8
- Create a function and name it as <em>getHighScores</em> which accepts two values, <em>gameScores</em> and <em>minScore</em>. (Line 1)
- Create an empty list/array and assign it to variable <em>meetsThreshold</em>. (Line 2)
- Create a for loop to iterate through each of the score in the <em>gameScores</em> (Line 4)
- Set a condition if the current score is bigger than the <em>minScore</em>, add the score into the <em>meetsThreshold</em> list (Line 5-6)
- Return <em>meetsThreshold</em> list as the output
Line 11-12
- create a random list of <em>gameScores</em> (Line 11)
- Set the minimum score to 5 (Line 12)
Line 13-14
- Call the function <em>getHighScores()</em> and pass the<em> gameScores</em> and <em>minScore </em>as the arguments. The codes within the function <em>getHighScores()</em> will run and return the <em>meetsThreshold </em>list and assign it to <em>highScores.</em> (Line 13)
- Display <em>highScores</em> using built-in function print().
Answer:
See explaination
Explanation:
StackExample.java
public class StackExample<T> {
private final static int DEFAULT_CAPACITY = 100;
private int top;
private T[] stack = (T[])(new Object[DEFAULT_CAPACITY]);
/**
* Returns a reference to the element at the top of this stack.
* The element is not removed from the stack.
* atreturn element on top of stack
* atthrows EmptyCollectionException if stack is empty
*/
public T peek() throws EmptyCollectionException
{
if (isEmpty())
throw new EmptyCollectionException("stack");
return stack[top-1];
}
/**
* Returns true if this stack is empty and false otherwise.
* atreturn true if this stack is empty
*/
public boolean isEmpty()
{
return top < 0;
}
}
//please replace "at" with the at symbol
Note:
peek() method will always pick the first element from stack. While calling peek() method when stack is empty then it will throw stack underflow error. Since peek() method will always look for first element ffrom stack there is no chance for overflow of stack. So overflow error checking is not required. In above program we handled underflow error in peek() method by checking whether stack is an empty or not.
Answer:
Explanation:
temporal locality can be defined as: when a particular memory is referenced or accessed several times within a specific period of time. In the question, i think the variable that exhibit temporal locality are I, J and 0(all the variable). This is because the variable J and 0 are accessed several times within the loop. I would not have been part of it, but in the A[I][J]=B[I][0]+A[J][I], the variable "I" is also accessed in the addition. this is why it is part of the temporal locality.
An “iframe” maybe idk what is asking
