Given an alphanumeric string, find the shortest substring that occurs
exactly once as a (contiguous) substring in it. Overlapping occurrences
are counted as distinct. If there are several candidates of the same
length, you must output all of them in the order of occurrence. The
*space* is NOT considered as a valid non-repeating substring.

### Examples

Consider the following cases:

**Case 1:** If the given string is `asdfsasa`

, the answer should be
`['d', 'f']`

**Case 2:** If the given string is `sadanands`

,

the answer should be `['sa', 'ad', 'da', 'na', 'nd', 'ds']`

**Case 3:** If the given string is `wwwwwwww`

, the answer should be
`['wwwwwwww']`

# My Solution

Here is my solution in *Python*.

It is quite brute force. I am not sure about the order of `find()`

and
`rfind()`

built-in methods in *Python*. Assuming these are $O(n)$, my
algorithm is in $O(n^3)$. Please put your answers in comments below,
if your answer has a better scaling.

The function definition that I use for finding non-empty non-repeating
strings is *recursive*.

```
def findNsubString(s, n):
subS = []
for index in range(len(s)+1):
x = s[index:index+n]
if s.find(x) == s.rfind(x):
subS.append(x)
if subS:
return subS
else:
return findNsubString(s, n+1)
```

I call this method as follows to get the desired results:

```
#! /usr/bin/python
import argparse
# Parse Command Line Arguments
parser = argparse.ArgumentParser()
parser.add_argument("-s", "--string", default = "asda", help="Input")
args = parser.parse_args()
s = args.string
# Call Method to find smallest non-repeating sub-string
ans = findNsubString(s, 1)
print(ans)
```

A similar solution can also be written in `JAVA`

or `C++`

. The corresponding
`find()`

and `rfind()`

methods in JAVA are called `indexOf()`

and
`lastIndexOf()`

, respectively. In C++, these methods are called same as
in Python.

Please feel free to put your method definition in any programming language.