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As part of my college class, I have to write a program that returns the "smallest" letter in the string. Input is assumed to be a non empty string. "Smallest" is defined as:

The smallest decimal value of the character, pertaining to this chart.

The range of acceptable values is 0 <= x <= 127

I would like to get feedback on the algorithm of the code. Other suggestions are accepted and welcome, but the main algorithm is the main focus.

def smallest_letter(string: str) -> str:
    """
    Returns the smallest letter in the string

    Input is restricted to ASCII characters in range 0 <= character <= 127

    :param string: A string containing only ASCII letters

    :return: A string length one of the smallest letter
    """

    # Ensure all characters are within the acceptable range #
    for character in string:
        assert ord(character) <= 127

    smallest_letter = 1000
    for letter in string:
        if ord(letter) < smallest_letter:
            smallest_letter = ord(letter)
    return chr(smallest_letter)

if __name__ == "__main__":

    # Test Cases #

    assert smallest_letter("abydsufaksjdf") == "a"
    assert smallest_letter("bfsufsjfbeywafbiu") == "a"
    assert smallest_letter("ABYUVGDuibfsafuofiw") == "A"
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  • 2
    \$\begingroup\$ Is there a reason you don't simply use min("abydsufaksjdf")? \$\endgroup\$
    – L. F.
    Commented Nov 3, 2019 at 7:28
  • \$\begingroup\$ There is a bug: what happens if you pass an empty string ("") to smallest_letter? \$\endgroup\$
    – Jan Kuiken
    Commented Nov 3, 2019 at 9:09
  • \$\begingroup\$ @L.F. I was completely unaware that the min method could do those computations. I'll tag this as reinventing-the-wheel. \$\endgroup\$
    – Linny
    Commented Nov 3, 2019 at 9:36
  • \$\begingroup\$ @JanKuiken This is assuming the input is a non empty string. I will update the question accordingly. \$\endgroup\$
    – Linny
    Commented Nov 3, 2019 at 9:36
  • \$\begingroup\$ I updated my answer, if you want to take a look :) \$\endgroup\$
    – AMC
    Commented Nov 26, 2019 at 0:00

3 Answers 3

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I'll second what Gloweye said in this answer, namely that assert should not be used for control flow.


This solution combines many of the other answers:

def smallest_character(str_in: str) -> str:
    min_ord = 128
    for curr_char_ord in (ord(c) for c in str_in):
        if curr_char_ord > 127:
            raise ValueError(f'Character {chr(curr_char_ord)} in smallest_letter() arg has ord value {curr_char_ord} '
                             f'which is above the allowed maximum of 127')
        else:
            min_ord = min(min_ord, curr_char_ord)
    return chr(min_ord)

This solution uses min() and max():

def smallest_character(str_in: str) -> str:
    min_val = min(str_in)
    max_val = max(str_in)
    if ord(max_val) > 127:
        raise ValueError(
            f'Character {max_val} in smallest_letter() arg has ord value {ord(max_val)} which is above the allowed '
            f'maximum of 127')
    else:
        return min_val
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Regarding min(string): Python exceeding your expectations. It happens a lot.

assert

All assert statements can be disabled with a switch to the interpreter, and sometimes are. Therefore, they're not suitable for flow control. I think you should replace:

for character in string:
    assert ord(character) <= 127

With:

for character in string:
    if ord(character) > 127:
        raise ValueError(f"Character {character} is out of range.")

or to optimize with a generator instead of the loop (requires Python 3.8+ if you want to report the character that violates the condition):

if any(ord(character := char) > 127 for char in string):
    raise ValueError(f"Character {character} out of range")

(Thanks to @Graipher for proper handling of the := token in 3.8+, which I hadn't worked with myself yet.)

That's all I can see being wrong here, though.

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  • \$\begingroup\$ Me neither, took me a while to find an online interpreter, but TIO has it: tio.run/#python38pr \$\endgroup\$
    – Graipher
    Commented Nov 4, 2019 at 16:03
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    \$\begingroup\$ Whoopsie. fixed. \$\endgroup\$
    – Gloweye
    Commented Nov 4, 2019 at 17:58
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Of course min("3sdsdf44ldfkTsdfnsnприветsdfa5É") (contains unicode chars) approach won't be suitable in case if validating/requiring only ASCII chars.

Issues of initial approach :

  • validating empty string. To ensure non-empty input string we'll add a simple assertion at start:
    assert string != "", "Empty string"
  • doubled traversals. On valid input strings like "3sdsdf44ldfkTe45456fghfgh678567sdfnsnsdfa23" where the char with minimal code would be at the end part of the string the former approach will make a double traversal though 2 for loops.
    To avoid that inefficiency we can combine validation and comparison/accumulation logic to be on a single iteration. (you may run time performance measurements to see the difference)

  • ord(letter). Duplicated calls can be eliminated through applying Extract variable technique: char_code = ord(char)


The final optimized version:

def smallest_letter(string: str) -> str:
    """
    Returns the smallest letter in the string
    Input is restricted to ASCII characters in range 0 <= character <= 127

    :param string: A string containing only ASCII letters
    :return: A string length one of the smallest letter
    """

    assert string != "", "Empty string"

    max_code, min_code = 128, 128

    for char in string:
        char_code = ord(char)
        assert char_code < max_code, f"{char} is not ASCII character"  # ensure the char is in acceptable code range
        if char_code < min_code:
            min_code = char_code

    return chr(min_code)
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