The answer is mobile communication, this consists of emails, texts, and parts of social media (Ex. DM's)
Answer:
Explanation:
Other advantages of object-oriented programming languages are you can use it to kinds of web applications for thorough data analysis, less development time, accurate coding, easy testing, reusability, debugging, less data corruption, and maintenance.
Answer:
item based for loop is used to iterate items in a collection .it is useful to apply some operations on item. Index based for loop is used to execute some logic repetitively. While loop also useful to execute a logic repetitively
Explanation:
item based for loop is used to iterate items in a collection .it is useful to apply some operations on item. Index based for loop is used to execute some logic repetitively. While loop also useful to execute a logic repetitively
in c#.net , following example explains this
using system;
void main(){
String[] names=new names[20];
int counter=0;
//index based for loop
for(int i=0;i<20;i++){
console.read(names[i]);
}
//item based for loop
foreach(string s in names){
console.writeline(s);
}
//while loop
while(counter<20)
{
console.read(names[counter];
counter++;
}
}
Answer:
Follows are the explanation of the choices:
Explanation:
Following are the Pseudocode for selection sort:
for j = 0 to k-1 do:
SS = i
For l = i + 1 to k-1 do:
If X(l) < X(SS)
SS= l
End-If
End-For
T = X(j)
X(j) = X(SS)
X(SS) = T
End-For
Following are the description of Loop invariants:
The subarray A[1..j−1] includes the lowest of the j−1 components, ordered into a non-decreasing order, only at beginning of the iteration of its outer for loop.
A[min] is the least amount in subarray A[j.. l−1] only at beginning of the each loop-inner iterations.
Following are the explanation for third question:
Throughout the final step, two elements were left to evaluate their algorithm. Its smaller in A[k-1] would be placed as well as the larger in A[k]. One last is the large and medium component of its sequence because most and the last two components an outer loop invariant has been filtered by the previous version. When we do this n times, its end is a repetitive, one element-sorting phase.
Following is the description of choosing best-case and worst-case in run- time:
The body the if has never been activated whenever the best case time is the list is resolved. This number of transactions are especially in comparison also as a procedure, that will be (n-1)(((n+2)/2)+4).
A structure iterator at every point in the worst case that array is reversed, that doubles its sequence of iterations in the inner loop, that is:(n−1)(n+6) Since both of them take timeΘ(n2).
