Runtime complexity of permutation function

Question

I am trying to find the asymptotic run time complexity of the following function which will return a list of all permutations of nums.

    def permute(nums):
        res = []
        dfs(nums, [], res)
        return res
def dfs(nums, curr, res):
    if not nums:
        res.append(curr)
        return
    for i in range(len(nums)):
        dfs(nums[:i]+nums[i+1:], curr+[nums[i]], res)

I think that the run time on an input of size $n$ is $T(n)=nT(n-1)+n$ because the function will make $n$ recursive calls on an input of size $n-1$ and it loops over $n$ terms. This gives $T(n)\in \mathcal O(n!)$ but some people say that it is $\mathcal O(n\cdot n!)$. Is the runtime of this $\mathcal O(n!)$ or $\mathcal O(n\cdot n!)$?

score -1 · Answer 1 · answered Jun 23 '22 at 23:41

Consider this piece of code:

        for i in range(len(nums)):
            dfs(nums[:i]+nums[i+1:], curr+[nums[i]], res)

Note that curr+[nums[i]] creates a new list, and on the bottom level, each of the $n!$ new lists will have size $n$. Hence $\mathcal{O} (n! \cdot n)$ total complexity.

On the other side, note that the complexity nums[:i]+nums[i+1:] is also linear in terms of $n$, but it does not present a problem, since its length at the bottom levels is $\mathcal{O} (1)$.

To look at it another way, the output has a size of $n! \cdot n$, so, barring any clever tricks in storing the output, the whole algorithm can't take less than that to produce the whole output.

In terms of recurrence relation, it gets a bit tricky, as the deeper we go, the larger the second argument (curr) gets. So, perhaps we can write $T (n, \ell) = n \cdot T (n - 1, \ell + 1) + n \cdot (n + \ell)$, where $n$ is the length of nums, and $\ell$ is the length of curr. We are interested in $T (n, 0)$, and the base is $T (0, \ell) = \ell$.

Runtime complexity of permutation function

1 Answers1