import java.util.Scanner;
public class U2_L3_Activity_Four {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
System.out.println("Enter a sentence.");
String sent = scan.nextLine();
int count = 0;
for (int i = 0; i < sent.length(); i++){
char c = sent.charAt(i);
if (c != ' '){
count++;
}
else{
break;
}
}
System.out.println("The first word is " + count +" letters long");
}
}
We check to see when the first space occurs in our string and we add one to our count variable for every letter before that. I hope this helps!
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)
the answer is Java.util.scanner
Answer:
Sensitivity Levels
Explanation:
Sensitivity Level is option use in email to inform the recipient that they should exercise discretion in accordance with sharing the content of the message.