Create mentoring.md for Say (#2211)

* Create mentoring.md for Say

* Minor changes based on review

* Spelling error (Ternery -> Ternary)

* dict -> dictionary

Co-authored-by: Isaac Good <[email protected]>

* is -> are

Co-authored-by: Isaac Good <[email protected]>

* Update tracks/python/exercises/say/mentoring.md

Co-authored-by: Isaac Good <[email protected]>

* Update tracks/python/exercises/say/mentoring.md

Co-authored-by: Isaac Good <[email protected]>

* Consistent use of quotes

* Update to Parsing the number section

* Apply suggestions from code review

* Update tracks/python/exercises/say/mentoring.md

---------

Co-authored-by: Isaac Good <[email protected]>
Co-authored-by: Derk-Jan Karrenbeld <[email protected]>

tracks/python/exercises/say/mentoring.md

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+# Mentoring
+
+## Problems and Challenges
+
+This problem requires the student to transform an integer (0 <= n < 1,000,000,000,000) into its full English equivalent, and raise an exception for values outside that range.
+Take note of the tests - the use of 'and' as a separator is not required, and adding it will cause the tests to fail, but it is suggested as an extension exercise. 
+
+## Reasonable Solution
+
+```python
+digits = [""] + "one two three four five six seven eight nine".split()
+teens = "ten eleven twelve thirteen fourteen fifteen sixteen seventeen eighteen nineteen".split()
+tens = ["", ""] + "twenty thirty forty fifty sixty seventy eighty ninety".split()
+denominations = (9, "billion"), (6, "million"), (3, "thousand"), (2, "hundred")
+
+def say(number):
+    if 0 > number or number > 999999999999:
+        raise ValueError("input out of range")
+    if 10 <= number < 20:
+        return teens[number % 10]
+    if number < 100:
+        if number // 10 and number % 10:
+            return "-".join([tens[number // 10], digits[number % 10]])
+        else:
+            return tens[number // 10] or digits[number % 10] or "zero"
+    for k, denomination in denominations:
+        if number > 10**k - 1:
+            result = [say(number // 10**k), denomination]
+            if number % 10 ** k:
+                result.append(say(number % 10**k))
+            return " ".join(result)
+```
+Or without using recursion:
+
+```python
+denominations = (9, "billion"), (6, "million"), (3, "thousand"), (0, "")
+
+def say_three_digits(number):
+    hundreds_col, tens_col, ones_col = number // 100, number // 10 % 10, number % 10
+    result = [str(digits[hundreds_col]) + " hundred"] if hundreds_col else []
+    if tens_col == 1:
+        result.append(teens[ones_col])
+    elif tens_col and ones_col:
+        result.append("-".join([tens[tens_col], digits[ones_col]]))
+    else:
+        result.append(tens[tens_col] or digits[ones_col])
+    return result
+
+def say(number):
+    if 0 > number or number > 999999999999:
+        raise ValueError("input out of range")
+    result = []
+    for k, denomination in denominations:
+        if number > 10**k - 1:
+            result += say_three_digits(number // 10**k)
+            result += [denomination]
+            number %= 10**k
+    return " ".join(e for e in result if e) or "zero"
+```
+
+## Common Suggestions
+
+### Defining the literals
+
+Typically this problem will begin with defining the literals needed. 
+These can be done as either lists or dictionaries; many students will find dictionaries easier to work with for this problem. 
+Encourage the student to define these at the start rather than embedding them in their code.
+
+### Values below 1000
+
+Next, the instructions suggest beginning by creating a function which works for values of up to three digits.
+The return value of this function will typically be a string, but it can also be a list of strings for later assembly.
+Whether in this function or for the final return, encourage them to use a list and combine the elements using `str.join`.
+
+The main challenges are ensuring that the 'hundred' denominator is conditionally included and that the tens and ones columns are joined with a hyphen if they are both present.
+Simple `if` checks for each of these conditions is the most straightforward approach.
+
+Another approach is to only handle values below 100, and instead handle the hundreds alongside the other denominators.
+This typically requires the use of recursion. 
+
+### Larger numbers
+
+Encourage the student to use the first function they wrote for each set of three digits rather than duplicating code.
+Each set of three can be checked individually and conditionally added along with the appropriate denominator, or alternatively the denominators can be stored in a list and iterated over.
+
+### Zero or invalid input
+
+Finally, the student must add a guard that returns a `ValueError` if the input is out of range. 
+A guard is also the easiest way to handle the zero case, but this can also be handled through short circuit evaluation.
+
+### Parsing the number
+
+The number needs to be broken up into groups of three as well as isolating each digit. 
+Two common approaches are to use the `//` and `%` operators or to cast the integer to a string and access the digits by indexing or iterating.
+
+## Talking Points
+
+List comprehensions are very useful for this exercise.
+
+There are a wide variety of valid approaches to this problem, which gives students a lot of opportunity to experiment and encounter new ways of manipulating data.
+Some useful techniques that students might not have encountered are listed below.
+Students will not need or want to use all of these, but mentors should look for opportunities in the student's code to introduce one or more of them.
+* Casting between numeric types and strings
+* String indexing, iteration, formatting, or concatenation
+* String methods such as `zfill`, `isnumeric`, `split`, or `join`
+* List concatenation using the `+` operator
+* List methods such as `append` or `extend`
+* Iterator functions such as `zip`, `enumerate`, or `filter`
+* Assignment operators
+* Unpacking operator
+* Short circuit evaluation and the use of `or`/`and` for control flow
+* Ternary conditional operator
+* Integer division and modulo operators, and the divmod function
+* Recusion
+
+Students who wish to extend the exercise can rewrite the function to include 'and' as a separator as well as handle negative or larger numbers.
+This will require rewriting the tests, which might also serve as a learning opportunity for the student.
+
+This problem is a good opportunity to practice writing concise code and balancing readability. 
+It is also a good chance to code verbosely while the problem is solved, and then iterate on the solution to reduce length and complexity.
+A terse solution which demonstrates some techniques:
+
+```python
+digits = [""] + "one two three four five six seven eight nine".split()
+teens = "ten eleven twelve thir four fif six seven eigh nine".split()
+tens = ["", ""] + "twenty thirty forty fifty sixty seventy eighty ninety".split()
+denominations = (9, "billion"), (6, "million"), (3, "thousand"), (2, "hundred")
+
+def say(n):
+    if 0 > n or n > 999999999999:
+        raise ValueError("input out of range")
+    if 10 <= n < 20:
+        return teens[n % 10] + "teen" if n % 10 > 2 else ""
+    if n < 100:
+        return "-".join(filter(bool, [tens[n // 10], digits[n % 10]])) or "zero"
+    k, d = [t for t in denominations if n > 10**t[0] - 1][0]
+    return " ".join([say(n // 10**k), d] + ([say(n % 10**k)] if n % 10**k else []))
+```