Here is a brief explanation of the difference between P$,$ NP$,$ NP$-$complete and NP$-$hard complexity classes:

P is the class of decision problems where the $"$P$"$ in the P class stands for Polynomial Time. It is a group of problems that can be resolved by a deterministic computer in polynomial time and require a "yes" or $"$no$"$ response.

NP is the class of decision problems whose solutions can be verified in polynomial time where An NP$-$hard problem is at least as hard as the hardest problem in NP and it is a class of problems such that every problem in NP reduces to NP$-$hard.

NP$-$hard is the class of decision problems that are at least as hard as any problem in NP$.$

NP$-$complete is the class of decision problems that are both NP$-$hard and NP$.$

One of the most critical unsolved problems in computer science is the P versus NP problem.

It asks whether every problem in NP can also be solved in P$.$ If P $=$ NP$,$ then it would mean that all NP$-$complete problems could be solved in polynomial time. This would have a profound impact on many areas of computer science, including cryptography and artificial intelligence.

However, if P $$ NP$,$ then there are some problems that are inherently difficult to solve. This would mean that there are some problems that we will never be able to solve in polynomial time, no matter how much computing power we have.

In other hand, The main difference between NP$-$complete and NP$-$hard complexity classes is that NP$-$complete problems are a subset of NP$-$hard problems. This means that all NP$-$complete problems are also NP$-$hard, but not all NP$-$hard problems are NP$-$complete.

NP$-$complete named complexity class is the class of decision problems that are both NP and NP$-$hard but NP$-$hard named complexity class is the class of decision problems that are at least as hard as any NP$-$complete problem.