Cubic

def triplets_with_sum(n):
    triplets = []
    for a in range(1, n + 1):
        for b in range(a + 1, n + 1):
            for c in range(b + 1, n + 1):
                if a**2 + b**2 == c**2 and a + b + c == n:
                    triplets.append([a, b, c])
    return triplets

The strategy in this case is to scan through all three variables and test them in the innermost loop. But the innermost loop will test all variables every time the enclosing loop iterates. And the enclosing loop up will iterate through all of its range every time the outermost loop iterates.

So the 'work' this code has to do for each additional value in range(1, n+1) is n**3.

This gives code that is simple, clear, ...and so slow as to be useless for all but the smallest values of n.

We could tighten up the bounds on loop variables: for example, a is the smallest integer of a triplet that sums to n, so inevitably a < n //3.

However, this is not nearly enough to rescue an inappropriate algorithm.