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amid [387]
3 years ago
9

)a___ is a complete binary tree such that each node in the tree contains a comparable object that us greater than or equal to th

e objects jn the nodes descendants
A. Binheap

B. Maxheap

C. Fullheap

D. Minheap
Computers and Technology
1 answer:
lesantik [10]3 years ago
7 0

Answer:

(B) Maxheap.

Explanation:

Complete binary tree:-It is a binary tree having every level filled completely excluding the last level .

Max heap:-It is a binary tree which is complete or complete binary tree in which the root node is greater than or equal to it's children or we can say descendants.So according to the question the answer is MaxHeap.

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OLga [1]
   Hey friends

The answer would be B

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3 years ago
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Which of the following is NOT a way the media communicates unrealistic body images?
Leokris [45]

The media communicates unrealistic body image through various ways, which are mentioned in the available options for the question. It definitely digitally alters images that they use for their various publications, and it also provides forums to discuss actor and actresses, including their looks.

The media also often criticizes high-profile individual’s current body condition – whether they are too underweight or too overweight. Thus, the best answer for the question would be (D) promoting all body types.

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3 years ago
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I have six nuts and six bolts. Exactly one nut goes with each bolt. The nuts are all different sizes, but it’s hard to compare t
juin [17]

Answer:

Explanation:

In order to arrange the corresponding nuts and bolts in order using quicksort algorithm, we need to first create two arrays , one for nuts and another for bolts namely nutsArr[6] and boltsArr[6]. Now, using one of the bolts as pivot, we can rearrange the nuts in the nuts array such that the nuts on left side of the element chosen (i.e, the ith element indexed as nutArr[i]) are smaller than the nut at ith position and nuts to the right side of nutsArr[i] are larger than the nut at position "I". We implement this strategy recursively to sort the nuts array. The reason that we need to use bolts for sorting nuts is that nuts are not comparable among themselves and bolts are not comparable among themselves(as mentioned in the question)

The pseudocode for the given problem goes as follows:

// method to quick sort the elements in the two arrays

quickSort(nutsArr[start...end], boltsArr[start...end]): if start < end: // choose a nut from nutsArr at random randElement = nutsArr[random(start, end+1)] // partition the boltsArr using the randElement random pivot pivot = partition(boltsArr[start...end], randElement) // partition nutsArr around the bolt at the pivot position partition(nutsArr[start...end], boltsArr[pivot]) // call quickSort by passing first partition quickSort(nutsArr[start...pivot-1], boltsArr[start...pivot-1]) // call quickSort by passing second partition quickSort(nutsArr[pivot+1...end], boltsArr[pivot+1...end])

// method to partition the array passed as parameter, it also takes pivot as parameter

partition(character array, integer start, integer end, character pivot)

{

       integer i = start;

loop from j = start to j < end

       {

check if array[j] < pivot

{

swap (array[i],array[j])

               increase i by 1;

           }

 else check if array[j] = pivot

{

               swap (array[end],array[j])

               decrease i by 1;

           }

       }

swap (array[i] , array[end])

       return partition index i;

}

7 0
3 years ago
Select the phrases that apply to Java classes or methods.
oee [108]

Answer:

graficar una lista de entradas

graficar una lista de entradas

crear un nuevo objeto

condiciones de prueba

romper un ciclo

romper un ciclo

Explanation:

7 0
2 years ago
Given two 2x3 (2 rows, 3 columns) arrays of integer , x1 and x2, both already initialized , two integer variables , i and j, and
Gnesinka [82]
JAVA programming was employed...

What we have so far:
* Two 2x3 (2 rows and 3 columns) arrays. x1[i][j] (first 2x3 array) and x2[i][j] (second 2x3 array) .
* Let i = row and j = coulumn.
* A boolean vaiable, x1rules

Solution:

for(int i=0; i<2; i++)
{
   for(int j=0; j<3; j++)
   {
      x1[i][j] = num.nextInt();
   }
}// End of Array 1, x1.

for(int i=0; i<2; i++)
{
   for(int j=0; j<3; j++)
   {
      x2[i][j] = num.nextInt();
   }
}//End of Array 2, x2
This should check if all the elements in x1 is greater than x2:

x1rules = false;
if(x1[0][0]>x2[0][0] && x1[0][1]>x2[0][1] && x1[0][2]>x2[0][2] && x1[1][0]>x2[1][0] && x1[1][1]>x2[1][1] && x1[1][2]>x2[1][2])
{
   x1rules = true;
   system.out.print(x1rules);
}
else
{
   system.out.print(x1rules);
}//Conditional Statement
7 0
3 years ago
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