<u>Answer:</u>
<em>A summary sentence should brief the whole content “what so ever the length of the original content” may be. </em>
For example, if you take a story, <em>moral will be the good example of summary. </em>One another example is when the teacher taught concept in the classroom, in the last few minutes of the class the teacher <em>would brief the whole into smaller points. </em>
Even nowadays, people go and visit movies only after seeing the review online. So once again the review is a small brief about the movie in one or two lines. <em>It should be crisp, use cherry-picked words, etc.</em>
Ash is the answer. Hope this helps
Answer:
public void trimToSize() {
modCount++;
if (size < elementData.length) {
elementData = (size == 0)
? EMPTY_ELEMENTDATA
: Arrays.copyOf(elementData, size);
}
}
Now, the running time for copyOf() function is O(N) where N is the size/capacity of the ArrayList. Hence, the time complexity of trimToSize() function is O(N).
Hence, the running time of building an N-item ArrayList is O(N^2).
Please find the sample code below.
CODE
==================
import java.util.ArrayList;
public class Driver {
public static void main(String[] args) throws Exception{
int N = 100000;
ArrayList<Integer> arr = new ArrayList<>(N);
long startTime = System.currentTimeMillis();
for(int i=0; i<N; i++) {
arr.add(i);
arr.trimToSize();
}
long endTime = System.currentTimeMillis();
double time = (endTime - startTime)/1000;
System.out.println("Total time taken = " + time + " seconds.");
}
}
Explanation:
Answer:
// code in C++
#include <bits/stdc++.h>
using namespace std;
// main function
int main()
{
// variables
int sum_even=0,sum_odd=0,eve_count=0,odd_count=0;
int largest=INT_MIN;
int smallest=INT_MAX;
int n;
cout<<"Enter 10 Integers:";
// read 10 Integers
for(int a=0;a<10;a++)
{
cin>>n;
// find largest
if(n>largest)
largest=n;
// find smallest
if(n<smallest)
smallest=n;
// if input is even
if(n%2==0)
{
// sum of even
sum_even+=n;
// even count
eve_count++;
}
else
{
// sum of odd
sum_odd+=n;
// odd count
odd_count++;
}
}
// print sum of even
cout<<"Sum of all even numbers is: "<<sum_even<<endl;
// print sum of odd
cout<<"Sum of all odd numbers is: "<<sum_odd<<endl;
// print largest
cout<<"largest Integer is: "<<largest<<endl;
// print smallest
cout<<"smallest Integer is: "<<smallest<<endl;
// print even count
cout<<"count of even number is: "<<eve_count<<endl;
// print odd cout
cout<<"count of odd number is: "<<odd_count<<endl;
return 0;
}
Explanation:
Read an integer from user.If the input is greater that largest then update the largest.If the input is smaller than smallest then update the smallest.Then check if input is even then add it to sum_even and increment the eve_count.If the input is odd then add it to sum_odd and increment the odd_count.Repeat this for 10 inputs. Then print sum of all even inputs, sum of all odd inputs, largest among all, smallest among all, count of even inputs and count of odd inputs.
Output:
Enter 10 Integers:1 3 4 2 10 11 12 44 5 20
Sum of all even numbers is: 92
Sum of all odd numbers is: 20
largest Integer is: 44
smallest Integer is: 1
count of even number is: 6
count of odd number is: 4
Answer:
B. Migration Assistant
Explanation:
Migration assistant also referred to as the Data Migration Assistant (DMA) is a Microsoft tool used during SQL server upgrades to detect any compatibility issues that could impact the functionality of the new version of SQL server being installed. The Data Migration Assistant will also recommend performance and reliability improvements as well as allow you to move your SQL server data and schema from the source server to the destination server.
