Answer:
B.O(n).
Explanation:
In an ADT graph the method addEdge uses an Array of list.So in the worst case and the worst will be when the list already has n elements in it.
So to add an edge we have to iterate over the list upto nth element and then add the edge after that.So it has to travel over n elements.
So we can say that the answer is O(n).
Free version also know as the lite version that has less feature and but still allows the user to use the the core functionality
Answer:
SELECT
COUNT(SN), SUM(TaxAmount)
FROM ORDERS
or
SELECT
COUNT(SN) AS NumOrder, SUM(TaxAmount) As TotalTax
FROM ORDERS
Explanation:
Finding it difficult to add my explanation. So, I used an attachment instead
<em />
Answer:
Explanation:
The last name goes inside the main body of the mail specifically at the "closing block".
While the Address is inserted into the Subject Line specifically to the "To block"
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
