See https://www.geeksforgeeks.org/find-the-smallest-positive-number-missing-from-an-unsorted-array/ This seems to be a popular coding interview problem.

The Python code shown on the site seems almost maliciously misleading.

The full problem is this:

You are given an unsorted array with both positive and negative elements. You have to find the smallest positive number missing from the array in O(n) time using constant extra space. You can modify the original array.

Here's a common-enough Python solution.

def smallest_missing_1(x):
"""
>>> smallest_missing_1([2, 3, 7, 6, 8, -1, -10, 15])
1
>>> smallest_missing_1([2, 3, -7, 6, 8, 1, -10, 15])
4
>>> smallest_missing_1([1, 1, 0, -1, -2])
2
"""
try:
return min(set(range(1, max(x)+2))-set(x))
except ValueError as ex:
# min() arg is an empty sequence
return 1

Some folks don't like the try/except to detect the edge case of all negative values. If max(x) <= 0, then the exception will be raised, and we could use an if statement for a LBYL solution.

What's more important is this solution violates the constant extra space constraint. It builds two sets, which isn't a simple constant size object; it depends on the size of the input object.

To avoid the sets, we'll use a generator.

def smallest_missing_2(x):
"""
>>> smallest_missing_2([2, 3, 7, 6, 8, -1, -10, 15])
1
>>> smallest_missing_2([2, 3, -7, 6, 8, 1, -10, 15])
4
>>> smallest_missing_2([1, 1, 0, -1, -2])
2
"""
try:
return next(n for n in range(1, max(x)+2) if n not in x)
except StopIteration as ex:
# next() found an empty sequence
return 1

This violates the **O**(*n*) constraint with the repeated use of the **in** operator.

To get to **O**(*n*) and no extra storage, we're forced to (a) demand the input is a mutable list, so we can (b) reuse the input list object to track which numbers are present and which are absent. This reuse of a mutable list is atypical for Python programming. Indeed, it seems like a bad idea.

Consistent with the spirit of the problem, we're constrained to making arithmetic changes to the values in the original list, x, to track the state of the computation. The idea is that the value x[i] will have *both* an original input value, and the presence (or absence) of some value, p+i, in the sequence.

One traditional approach is to use the sign as a way to carry this extra bit of information. That's why negative numbers are thrown in to the input data. They make the sign business super confusing. Also. That's why zero is excluded. Conventional integer math doesn't have a negative zero, confounding the problem with array slots that have numbers that make sign processing icky.

I'm not a fan of using the sign for this. I'd prefer to use Least Significant Bits (LSB's) because we have a fairly large number of available LSB's. And. We can trivially ignore zero values and their habit of not having useful signs. Unless the list has 2**62 elements, a little shifting won't hurt any.

Here's a solution that would *rarely* be used in normal Python work. Maybe on a Circuit Playground Express MicroPython. But not anywhere else.

from typing import List
def smallest_missing_3(x: List[int]) -> int:
"""
>>> smallest_missing_3([2, 3, 7, 6, 8, -1, -10, 15])
1
>>> smallest_missing_3([2, 3, -7, 6, 8, 1, -10, 15])
4
>>> smallest_missing_3([1, 1, 0, -1, -2])
2
"""
# Purge negative numbers. Scale the other numbers.
for i in range(len(x)):
if x[i] < 0:
x[i] = 0
else:
x[i] = x[i] << 1
# Set LSB on positions which are filled; ignoring None values.
# This can raise an index out-of-bounds exception, which we silence.
for v in filter(None, (scaled >> 1 for scaled in x)):
try:
x[v-1] = x[v-1] | 1
except IndexError:
pass
# Find first value with LSB not set.
for i in range(len(x)):
if x[i] & 1 == 0:
return i+1

This is pretty atypical Python code. I'm kind of shocked folks would use something like this as an interview question. It's rather complex and requires some very old-school programming tricks to make the whole thing remotely palatable.

The index out-of-bounds is particularly nasty. It means there's a number, n, that's greater than len(x). This is worrisome, but, it also means any gap MUST be less than this large number n. For this reason, we can silence array index errors.

I would not be able to simply stand up in a conference room and solve this without some additional direction. The "making arithmetic changes to the values in the original list" secret is something I knew about and did -- when I was younger -- but I haven't done that kind of thing in decades.