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Andrej [43]
3 years ago
5

You have to sort 1 GB of data with only 100 MB of available main memory. Which sorting technique will be most appropriate?

Computers and Technology
1 answer:
Arte-miy333 [17]3 years ago
8 0
I’m guessing that what’s being looked at here moreso is the space complexity of these algorithms. Heap sort and insertion sort I believe have the lowest of these, but insertion sort is also known to not be the best with time complexity. Therefore heap sort should take the cake
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Which of the following transferable skills are generally the most look for in the it <br> field
kherson [118]

Answer:

mechanical, encryptions, communication, network admin, leadership, and teambuilding

Explanation:

6 0
3 years ago
1. Print out the string length of s1 2. Loop through characters in s2 with charAt() and display reversed string 3. Compare s2, s
konstantin123 [22]

Answer:

See explaination

Explanation:

public class StringLab9 {

public static void main(String args[]) {

char charArray[] = { 'C', 'O', 'S', 'C', ' ', '3', '3', '1', '7', ' ', 'O', 'O', ' ', 'C', 'l', 'a', 's', 's' };

String s1 = new String("Objected oriented programming language!");

String s2 = "COSC 3317 OO class class";

String s3 = new String(charArray);

// To do 1: print out the string length of s1

System.out.println(s1.length());

// To do 2: loop through characters in s2 with charAt and display reversed

// string

for (int i = s2.length() - 1; i >= 0; --i)

System.out.print(s2.charAt(i));

System.out.println();

// To do 3: compare s2, s3 with compareTo(), print out which string (s2 or s3)

// is

// greater than which string (s2 or s3), or equal, print the result out

if (s2.compareTo(s3) == 0)

System.out.println("They are equal");

else if (s2.compareTo(s3) > 0)

System.out.println("s2 is greater");

else

System.out.println("s3 is greater");

// To do 4: Use the regionMatches to compare s2 and s3 with case sensitivity of

// the first 8 characters.

// and print out the result (match or not) .

if (s2.substring(0, 8).compareTo(s3.substring(0, 8)) == 0)

System.out.println("They matched");

else

System.out.println("They DONT match");

// To do 5: Find the location of the first character 'g' in s1, print it out

int i;

for (i = 0; i < s2.length(); ++i)

if(s2.charAt(i)=='g')

break;

System.out.println("'g' is present at index " + i);

// To do 6: Find the last location of the substring "class" from s2, print it

// out

int index = 0, ans = 0;

String test = s2;

while (index != -1) {

ans = ans + index;

index = test.indexOf("class");

test = test.substring(index + 1, test.length());

}

System.out.println("Last location of class in s2 is: " + (ans + 1));

// To do 7: Extract a substring from index 4 up to, but not including 8 from

// s3, print it out

System.out.println(s3.substring(4, 8));

} // end main

} // end class StringLab9

7 0
3 years ago
How many fnaf games did Scott Cawthon make?
MrMuchimi

Answer:

8 but he is making FNaF security breach

Explanation:

there is FNaF VR help wanted

7 0
3 years ago
What provision of the Government Paperwork Elimination Act was designed to encourage a paperless society?
Ratling [72]
<span>Validation of electronic signatures was designed to encourage a paperless society.</span>
8 0
3 years ago
Read 2 more answers
How to convert 23.125 to binary (Hex) using the double - precision representation
Svet_ta [14]

Answer:-

(10111.001)₂

Explanation:

To convert a decimal number to a binary number we have to constantly divide the decimal number by 2 till the decimal number becomes zero and the binary number is writing the remainders in reverse order of obtaining them on each division.

Hence the binary number is 10111.001

To convert binary to hexa decimal we to make a group 4 binary bits starting from the decimal and moving outwards if the last group is not of 4 then add respective 0's and write the corresponding hexa decimal number.

<u>0001</u>  <u>0111</u> . <u>0010</u>

   1        7         2

Hence the hexadecimal number is 17.2

6 0
3 years ago
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