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Answer:
The diagram of the ER and depreciation is in the attachments.
You need to get a certain amount of questions
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).
