Using e-mail to send messages is the best choice to convey urgent and highly sensitive information. E-mail is just a conversation between you and the recipient. So it is the best when it comes to when you are sending a highly sensitive information. While telephone fax letter and dispatch radio may need to use a mediator to transfer messages which violates the confidentiality of the information.
to review the most important supporting points from the original information
Answer:
The Difference is because of the salt added to the password before hashing or encrypting it.
Explanation:
The Difference is because of the salt added to the password before hashing or encrypting it.
Salts are random data used to mention uniqueness. Suppose if two users have same password for the same system. By adding Salt to the password makes it unique in the crypto system. Its a added security layer which can safeguard the passwords or hashed data in the keystore or storage.
Salt can be prepended to the password or appended to the password.
Suppose the password is “stuti” then its SHA256 HASH value is <u><em>“4beb1eb6f438495eede7b14ac0d2b955636a49412cd4eb5714341f5716144954”</em></u>
So, if we add a salt with random string <em>brainlysalttest</em> then its SHA256 HASH value is <u><em>“5429e85778f1b9f493da637848f253dedf3edbfbb72782d43eb7337cee45ab0c”</em></u>
If we append the salt <em>brainlytestsalt</em> to the same password and hash it using SHA256 then its hash value is : <u><em>“3919b677d80ae0da2b58f70b464f6492670ccffee78cf52972c6983995ed8f52”</em></u>
So clearly Salting to the password and Hashing the salted password will make the difference and also unique.
Answer:
- public static String bothStart(String text1, String text2){
- String s = "";
-
- if(text1.length() > text2.length()) {
- for (int i = 0; i < text2.length(); i++) {
- if (text1.charAt(i) == text2.charAt(i)) {
- s += text1.charAt(i);
- }else{
- break;
- }
- }
- return s;
- }else{
- for (int i = 0; i < text1.length(); i++) {
- if (text1.charAt(i) == text2.charAt(i)) {
- s += text1.charAt(i);
- }else{
- break;
- }
- }
- return s;
- }
- }
Explanation:
Let's start with creating a static method <em>bothStart()</em> with two String type parameters, <em>text1 </em>&<em> text2</em> (Line 1).
<em />
Create a String type variable, <em>s,</em> which will hold the value of the longest substring that both inputs start with the same character (Line 2).
There are two possible situation here: either <em>text1 </em>longer than<em> text2 </em>or vice versa. Hence, we need to create if-else statements to handle these two position conditions (Line 4 & Line 13).
If the length of<em> text1</em> is longer than <em>text2</em>, the for-loop should only traverse both of strings up to the length of the <em>text2 </em>(Line 5). Within the for-loop, we can use<em> charAt()</em> method to extract individual character from the<em> text1</em> & <em>text2 </em>and compare with each other (Line 15). If they are matched, the character should be joined with the string s (Line 16). If not, break the loop.
The program logic from (Line 14 - 20) is similar to the code segment above (Line 4 -12) except for-loop traverse up to the length of <em>text1 .</em>
<em />
At the end, return the s as output (Line 21).
Answer:
It depends on what the company will be willing to use and it also depends on their budget
Explanation:
