The reason why this cost is going to force the optimal algorithm to explore the entire state space is the fact that large negative costs can cause concurrent automatically for optimal solutions.
<h3>The step to take with a negative cost</h3>
Now if there is the existence of a large negative cost for each of these actions, the optimal algorithm is going to try an exploration of the whole space state.
The reason for this would be the fact that the route that has the least consequences would be better of.
Read more on negative cost here: brainly.com/question/7143854
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:
prioritizing
Explanation:
The definition of prioritizing is determine the order for dealing with (a series of items or tasks) according to their relative importance.
So this applies to you.
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