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The shortest correct test is len(set(s)) == len(s). It returns True when no character repeats and False otherwise. The rest of this article covers when to use a different approach, and what “character” means for Unicode text.
The one-line answer
def all_unique(s: str) -> bool:
return len(set(s)) == len(s)
all_unique("python") # True
all_unique("hello") # False (two "l")
all_unique("") # True
Python’s tutorial defines a set as “an unordered collection with no duplicate elements.” Building a set from a string therefore discards repeats. If the set is as long as the string, nothing was discarded, so every character was unique. An empty string has nothing to repeat, so it returns True.
The expected running time is O(n) with O(k) extra memory, where n is the string length and k is the number of distinct characters. This is an average-case figure. CPython’s time-complexity reference lists average O(1) set insertion and membership, but warns that the worst case for hash tables can degrade to linear. So describe the bound as expected, not guaranteed.
Choosing an approach
Seen-set loop: stop at the first duplicate
def all_unique_early_exit(s: str) -> bool:
seen = set()
for char in s:
if char in seen:
return False
seen.add(char)
return True
This has the same expected O(n) time and O(k) memory. However, it returns as soon as it meets a repeat, so it can do far less work on a long string whose duplicate appears early. The one-liner always builds the full set first. Choose the loop when early exit matters, when you need custom handling (such as reporting the first repeated character), or when you are explaining the algorithm.
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Counter: when you need to know which characters repeat
from collections import Counter
counts = Counter(s)
is_unique = all(n == 1 for n in counts.values())
repeated = [ch for ch, n in counts.items() if n > 1]
The collections documentation describes Counter as a tallying tool. It gives more information than a yes/no answer, but it is more machinery than the set comparison for a boolean-only task. Use it for questions like “how can I detect duplicate characters in a string?”, where the duplicates themselves matter.
Quick comparison
| Approach | Returns | Early exit | Gives counts | Best for |
|---|---|---|---|---|
len(set(s)) == len(s) |
Boolean | No | No | Compact default check |
| Seen-set loop | Boolean | Yes | No | Long inputs, custom handling, teaching |
Counter |
Counts, from which you derive anything | No | Yes | Finding or reporting duplicates |
What “character” means for Unicode text
A Python str is, per the language’s data model documentation, a sequence of values representing characters, more formally Unicode code points. So set(s) tests the uniqueness of code points, which is not always what a person sees as a character. Two consequences:
- No normalization. An accented “é” can be one precomposed code point or an “e” followed by a combining accent. A set treats these as different, so two visually identical spellings are not detected as duplicates of each other.
- Visible units can span several code points. A user-perceived character (a grapheme cluster) may be made of multiple code points, and iterating a
strdoes not group them.
If canonically equivalent spellings should count as the same, normalize first:
import unicodedata
def all_unique_normalized(s: str) -> bool:
s = unicodedata.normalize("NFC", s)
return len(set(s)) == len(s)
If the requirement is uniqueness of visible characters, you must define and segment grapheme clusters explicitly, because plain str iteration won’t do it. For most exercises and interview questions, checking code points as shown above is the intended meaning.
Case sensitivity
The checks above treat “A” and “a” as different. If they should match, convert first, for example s.casefold(), and then run the test.
Quick Recap
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