Answer:
Sequential
Explanation:
Based on the information provided within the question it can be said that the search algorithm that is being described in this scenario is a Sequential algorithm. This is because sequential programming focuses on programming (or in this case searching for) a result by doing one step at a time as opposed to various functions running simultaneously.
Answer:
See explaination
Explanation:
Keep two iterators, i (for nuts array) and j (for bolts array).
while(i < n and j < n) {
if nuts[i] == bolts[j] {
We have a case where sizes match, output/return
}
else if nuts[i] < bolts[j] {
this means that size of nut is smaller than that of bolt and we should go to the next bigger nut, i.e., i+=1
}
else {
this means that size of bolt is smaller than that of nut and we should go to the next bigger bolt, i.e., j+=1
}
}
Since we go to each index in both the array only once, the algorithm take O(n) time.
Answer:
Java solution (because only major programming language that has public static methods)
(import java.io.* before hand)
public static boolean s2f(String fileName, String text){
try{
PrintWriter out = new PrintWriter(new File(fileName));
out.println(text);
out.close();
return true;
}
catch(Exception e){
return false;
}
}
Answer:
i think its translation complexity
Explanation:
