Python Multiple Assignment Is a Puzzle

I think it&#x27;s easier to see what is going on without the nested subscripting:<p><pre><code> &gt;&gt;&gt; i = 0 &gt;&gt;&gt; a = [0, 0] &gt;&gt;&gt; i, a[i] = 1, 10 &gt;&gt;&gt; a [0, 10] </code></pre> versus<p><pre><code> &gt;&gt;&gt; i = 0 &gt;&gt;&gt; a = [0, 0] &gt;&gt;&gt; a[i], i = 10, 1 &gt;&gt;&gt; a [10, 0]</code></pre>

While it may require an extra set, the following is a lot more pythonic, and also finds the first missing positive in the same two passes that your code did (which is actually O(2n), and honestly made my eyes bleed trying to follow.)<p><pre><code> def missPos(a): b={x for x in a} for x in range(1,len(b)+2): if not x in b: return x</code></pre>

This reminded me of gotchas one can fall to when defining some Lisp macro from scratch. So I decided to test whether rotatef works exactly like Python&#x27;s multiple assignment:<p><pre><code> (let ((A (vector 2 1))) (rotatef (elt A 0) (elt A (1- (elt A 0)))) A) </code></pre> It returns a changed vector as if indexes were saved.<p>I am not sure which should be considered the right behavior.

Please note, that the solution from the post has O(n) space complexity (you modify the input and this counts as using the extra memory).

<p><pre><code> A = [1, 2, 4, 5, 7] A_s = sorted(A) b = A_s[0] for a in A_s[1:]: if a - b &gt; 1: print b + 1 break else: b = a else: print &#x27;No missing element&#x27; </code></pre> Trying to be more Pythonic:<p><pre><code> def test(l): l_s = sorted(l) print [l0 + 1 for l1, l0 in zip(l_s[1:], l_s[:-1]) if l1 - l0 &gt; 1]</code></pre>

Interesting post regarding python, but the algo is incorrect, consider test case firstMissingPositive([100,101,103,104]) will return 1 instead of 102

def findMissingPositive(A): A.sort() return min([x for x in range(A[0], A[-1]+1) if x not in A])