Answer:
See explaination
Explanation:
Diverse Subset Problem is NP: When presented with a set of k customers, it can be checked in polynomial time that you wont at any time have two customers in the set have ever bought the same product.
Independent Set is known to be NP-complete.
Independent Set ?P Diverse Subset Problem: Suppose we have a black box for Diverse Subset Problem and want to solve an instance of Independent Set. For our Independent Set Problem, we have a graph G=(V,E) and a number k, and need to find out if G contains an independent set of size (at least) k. We need to reduce the Independent Set Problem to a Diverse Subset Problem. We do this by constructing an array where each v in V is a customer and each e in E is a product, and customer v purchased every product e for which the product edge e touches the customer node v. Then we ask the black box for the Diverse Subset Problem if there is a subset of k customers that is diverse.
The black box for the Diverse Subset Problem will return.
The service that can be used is the Database Migration Service.
<h3>What is the Database Migration Service?</h3>
A database migration service is a term used to describe a service that allows data contained in a database such as a SQL server database and the likes to a new cloud service account. The term database here denotes an organized collection of data stored electronically in a computer system.
For example, the amazon web service (AWS) database migration service falls under the category of a good tool that allows for such database transfers to a newly created AWS account. Hence, the company can use this tool to migrate the SQL server database to a newly-created AWS account.
You can learn more about database migration services from a related question here brainly.com/question/518894
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Answer:
Explanation:
The following code was written in Java and performs the exact requirements listed in the question. It has also been tested by the runner code and works perfectly.
public static boolean insert(String[] words, String newWord, int place) {
if (place > words.length) {
return false;
} else {
for (int x = words.length - 1; x >= 0; x--) {
if (place == x) {
words[x] = newWord;
break;
} else {
words[x] = words[x-1];
}
}
return true;
}
}
Answer:
I dont think it will help much
Explanation:
Well, just tracking it will not necessarily make you stop. I would recommend putting app limits or downtime (if you have a mac). Also make a log. Write down how much time you need for different things, like 1 hour of free time, 5 hours of schoolwork. Also, make a schedule based off the things you put on your log.
I found doing this helped me reduce my time on screen and made me more productive.
