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Studentka2010 [4]
2 years ago
15

Array A is not a heap. Clearly explain why does above tree not a heap? b) Using build heap procedure discussed in the class, con

struct the heap data structure from the array A above. Represent your heap in the array A as well as using a binary tree. Clearly show all the steps c) Show how heap sort work in the heap you have constructed in part (b) above. Clearly show all the step in the heap sort

Computers and Technology
1 answer:
Aleksandr [31]2 years ago
6 0

Answer:

Sorted Array A { } = { 1, 4, 23, 32, 34, 34, 67, 78, 89, 100 }

Explanation:

Binary tree is drawn given that the binary tree do not follow both minimum heap and maximum heap property, therefore, it is not a heap.

See attached picture.

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A two-dimensional array can have elements of ________ data type(s).
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The answer is "One". The number of dimensions is the representation of the array and has nothing to do with the data types the array can contain.

All array's can only hold one data type, the type it was initialized with.
4 0
2 years ago
Rideshare companies like Uber or Lyft track the x,y coordinates of drivers and customers on a map. If a customer requests a ride
TiliK225 [7]

Answer:

The program in Java is as follows:

import java.util.*;

public class Main{

 public static void main (String[]args){

   Scanner input = new Scanner(System.in);

   double user[] = new double[2];    double dr1[] = new double[2];

   double dr2[] = new double[2];    double dr3[] = new double[2];    

   System.out.print("Enter user coordinates: ");

   for(int i =0;i<2;i++){        user[i] = input.nextDouble();    }

   System.out.print("Enter driver 1 coordinates: ");

   for(int i =0;i<2;i++){        dr1[i] = input.nextDouble();    }

   System.out.print("Enter driver 2 coordinates: ");

   for(int i =0;i<2;i++){        dr2[i] = input.nextDouble();    }

   System.out.print("Enter driver 3 coordinates: ");

   for(int i =0;i<2;i++){        dr3[i] = input.nextDouble();    }

   double dr1dist = Math.abs(user[0] - dr1[0]) + Math.abs(user[1] - dr1[1]);

   double dr2dist = Math.abs(user[0] - dr2[0]) + Math.abs(user[1] - dr2[1]);

   double dr3dist = Math.abs(user[0] - dr3[0]) + Math.abs(user[1] - dr3[1]);

   System.out.println("Estimated pickup time of driver 1 "+(3.5 * dr1dist)+" minutes");

   System.out.println("Estimated pickup time of driver 2 "+(3.5 * dr2dist)+" minutes");

   System.out.println("Estimated pickup time of driver 3 "+(3.5 * dr3dist)+" minutes");

 }

}

Explanation:

The following array declarations are for the customer and the three drivers

<em>    double user[] = new double[2];    double dr1[] = new double[2];</em>

<em>    double dr2[] = new double[2];    double dr3[] = new double[2];    </em>

This prompts the user for the customer's coordinates

   System.out.print("Enter user coordinates: ");

This gets the customer's coordinate

   for(int i =0;i<2;i++){        user[i] = input.nextDouble();    }

This prompts the user for the driver 1 coordinates

   System.out.print("Enter driver 1 coordinates: ");

This gets the driver 1's coordinate

   for(int i =0;i<2;i++){        dr1[i] = input.nextDouble();    }

This prompts the user for the driver 2 coordinates

   System.out.print("Enter driver 2 coordinates: ");

This gets the driver 2's coordinate

   for(int i =0;i<2;i++){        dr2[i] = input.nextDouble();    }

This prompts the user for the driver 3 coordinates

   System.out.print("Enter driver 3 coordinates: ");

This gets the driver 3's coordinate

   for(int i =0;i<2;i++){        dr3[i] = input.nextDouble();    }

This calculates the distance between driver 1 and the customer

   double dr1dist = Math.abs(user[0] - dr1[0]) + Math.abs(user[1] - dr1[1]);

This calculates the distance between driver 2 and the customer

   double dr2dist = Math.abs(user[0] - dr2[0]) + Math.abs(user[1] - dr2[1]);

This calculates the distance between driver 3 and the customer

   double dr3dist = Math.abs(user[0] - dr3[0]) + Math.abs(user[1] - dr3[1]);

The following print statements print the estimated pickup time of each driver

<em>   System.out.println("Estimated pickup time of driver 1 "+(3.5 * dr1dist)+" minutes");</em>

<em>    System.out.println("Estimated pickup time of driver 2 "+(3.5 * dr2dist)+" minutes");</em>

<em>    System.out.println("Estimated pickup time of driver 3 "+(3.5 * dr3dist)+" minutes");</em>

7 0
2 years ago
Consider an application that transmits data at a steady rate (for example, the sender generates an N-bit unit of data every k ti
brilliants [131]

Answer:  For this type of application, it should be more appropriate a circuit-switched network.

Explanation: As the data are transmitted at a steady rate (x bits per second, for instance), and it's not transmitted in bursts, but during a relatively long period of time, it makes more sense to use a circuit-switched network instead of a packet-switched network, more useful when data shows in bursts (short intervals of data, followed by idle time, that can be filled with other data).

5 0
2 years ago
Create a program to deteate a program to determine whether a user-specified altitude [meters] is in the troposphere, lower strat
allsm [11]

Answer:

#include <iostream>

using namespace std;

int main()

{

   float altitude;

   cout<<"Enter alttitude in meter";

cin >>altitude;

if(altitude<0 || altitude > 50000)

{

   cout<<"Invalid value entered";

   return 0;

}

else{

   if(altitude<= 10000)

   {cout <<"In troposphere "<<endl;}

   

   else if(altitude>10000 && altitude<=30000)

   {cout <<"In lower stratosphere"<<endl;}

   

   else if(altitude>30000 && altitude<=50000)

   {cout <<"In upper stratosphere"<<endl;}

   

}

   return 0;

}

Explanation:

Define a float type variable. Ask user to enter altitude in meters. Store value to altitude variable.

Now check if value entered is positive and less than 5000,if its not in the range of 0-50,000 display a termination message and exit else check if it's in less than 10000, in between 10000 and 30000 or in between 30000 and 50000.

10,000 is above sea level is troposphere.

10,000-30,000 is lower stratosphere.

30,000-50,000 is upper stratosphere.

8 0
2 years ago
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