Answer:
O(n^2)
Explanation:
The number of elements in the array X is proportional to the algorithm E runs time:
For one element (i=1) -> O(1)
For two elements (i=2) -> O(2)
.
.
.
For n elements (i=n) -> O(n)
If the array has n elements the algorithm D will call the algorithm E n times, so we have a maximum time of n times n, therefore the worst-case running time of D is O(n^2)
Answer:
I am going to use the Python programming language to answer this. The source code is given below:
print("Enter your tweet here")
user_tweet = input()
decoded_tweet = user_tweet.replace('TTYL', 'talk to you later')
print("This is the decoded tweet: ")
print(decoded_tweet)
Explanation:
In the program the replace() module was used to replace 'TTYL' with 'talk to you later.'
Attached is the screenshot of the output.
Answer:
I will write the code in C++ and JAVA
Explanation:
<h2>
JAVA CODE</h2>
public class Main
{ public static void main(String[] args) {
// displays Gershwin,George
System.out.println("Gershwin,George"); } }
<h2>
C++ Code:</h2>
#include <iostream>
using namespace std;
int main()
{ cout<<"Gershwin,George";
}
// displays last name Gershwin followed by , followed by first name George
//displays Gershwin,George as output.
Answer:
Pseudocode is as follows:
// below is a function that takes two parameters:1. An array of items 2. An integer for weight W
// it returns an array of selected items which satisfy the given condition of sum <= max sum.
function findSubset( array items[], integer W)
{
initialize:
maxSum = 0;
ansArray = [];
// take each "item" from array to create all possible combinations of arrays by comparing with "W" and // "maxSum"
start the loop:
// include item in the ansArray[]
ansArray.push(item);
// remove the item from the items[]
items.pop(item);
ansArray.push(item1);
start the while loop(sum(ansArray[]) <= W):
// exclude the element already included and start including till
if (sum(ansArray[]) > maxSum)
// if true then include item in ansArray[]
ansArray.push(item);
// update the maxSum
maxSum = sum(ansArray[items]);
else
// move to next element
continue;
end the loop;
// again make the item[] same by pushing the popped element
items.push(item);
end the loop;
return the ansArray[]
}
Explanation:
You can find example to implement the algorithm.
Answer:
Explanation:
public class Temperature
{
double ftemp;
public int Constructor(double fahrenheit)
{
ftemp = fahrenheit;
return Convert.ToInt32(ftemp);
}
public void setFahrenheit(double fahrenheit)
{
ftemp = fahrenheit;
}
public void getFahrenheit()
{
ftemp = Constructor(ftemp);
}
public void getCelcius()
{
ftemp = (ftemp - 32) * 5 / 9;
}
public void getKelvin()
{
ftemp = (ftemp - 32) * 5 / 9 + 273.15;
}
}
