= Python Challenge -- Level 2 -- Native Solution =

==== Code ====

{{{
#!d
import tango.io.Stdout;

void main () {
        char[] text = "";

        size_t[char[]] quantity;
        foreach (char letter; text)
                quantity["" ~ letter] += 1;
        foreach (char letter; text)
                if (quantity[""~letter] == 1)
                        Stdout.print(key);
        Stdout.flush();
}
}}}

==== Explanation ====

{{{
#!d
        char[] text = "";
}}}

The source text from the challenge page should be put between the quotes.

{{{
#!d
        size_t[char[]] quantity;
}}}

An associative array is the easiest way to store information on each individual character.

{{{
#!d
        foreach (inout char letter; text)
}}}

The code is executed for every character.

{{{
#!d
                quantity["" ~ letter]++;
}}}

Increment the counter for the number of occurences of the current character.  The concatenation is a workaround for the lack of implicit conversion between char and char![].

{{{
#!d
        foreach (char letter; text)
}}}

Go again through all the text.
The code is executed once for each character.

{{{
#!d
                if (quantity[letter] == 1)
}}}

We only want characters that only were in the source text once.

{{{
#!d
                        Stdout.print(key);
        Stdout.flush();
}}}

Print the character to the screen, flushing upon completion.