What does it mean for a problem to be considered NP-Hard?
When a problem is classified as NP-Hard, it means that it is at least as difficult as the hardest problems in the class of problems known as NP (Non-deterministic Polynomial time). NP problems are decision problems that can be solved in polynomial time by a non-deterministic Turing machine, a hypothetical machine that can explore multiple solutions simultaneously.

An NP-Hard problem, on the other hand, is a problem that is at least as hard as the hardest problems in NP. In other words, if an NP-Hard problem could be solved in polynomial time, then all problems in NP could also be solved in polynomial time.

The concept of NP-Hardness is important in computer science and mathematics because it helps to identify problems that are difficult to solve. Many optimization problems, such as the traveling salesman problem and the knapsack problem, are NP-Hard. While it is not currently known whether these problems can be solved efficiently, researchers have developed algorithms that can approximate solutions to within a certain degree of accuracy.
Explain to me what polynomial time means
Before explaining polynomial time, I'll briefly explain Big O notation. Just note that for brevity it is a simplified explanation.
Any algorithm takes some number of steps to compute a result.
We also have a notion of input size, often denoted as *n*.
We say that an algorithm is *O(f(n))* for a function f, if it can be shown that the algorithm takes at most *c\*f(n)* steps
to compute a result with input size *n*. Where *c* is some constant.

To give a concrete example. Imagine we have a list of n integers, and we want to find which pair of integers gives the largest sum.
Let's suppose our algorithm is to check every possible pair, and update the best pair we found as soon as we find a better one.
The Big O of this algorithm would be *O(n^2)*, because we would have to check *n\*(n-1)/2* pairs, and do a finite number of steps
for each pair to consider. Say the number of steps is for instance 7. Then the number of steps the algorithm does for an input of
size n is *3.5n^2 - 3.5n*, we only really look at the dominant term: *3.5n^2*, the value of *c* here is 3.5, f(n) is n^2, thus this algorithm is O(n^2).

Okay, with that out of the way. We say that an algorithm is polynomial time, if it's Big O function is a polynomial.
where is kith treats at