Added doctest and more explanation about Dijkstra execution. (#1014)

* Added doctest and more explanation about Dijkstra execution.

* tests were not passing with python2 due to missing __init__.py file at number_theory folder

* Removed the dot at the beginning of the imported modules names because 'python3 -m doctest -v data_structures/hashing/*.py' and 'python3 -m doctest -v data_structures/stacks/*.py' were failing not finding hash_table.py and stack.py modules.

* Moved global code to main scope and added doctest for project euler problems 1 to 14.

* Added test case for negative input.

* Changed N variable to do not use end of line scape because in case there is a space after it the script will break making it much more error prone.

* Added problems description and doctests to the ones that were missing. Limited line length to 79 and executed python black over all scripts.

* Changed the way files are loaded to support pytest call.

* Added __init__.py to problems to make them modules and allow pytest execution.

* Added project_euler folder to test units execution

* Changed 'os.path.split(os.path.realpath(__file__))' to 'os.path.dirname()'

project_euler/problem_17/sol1.py

@@ -1,35 +1,63 @@
-from __future__ import print_function
-'''
+"""
 Number letter counts
 Problem 17
 
-If the numbers 1 to 5 are written out in words: one, two, three, four, five, then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total.
+If the numbers 1 to 5 are written out in words: one, two, three, four, five,
+then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total.
 
-If all the numbers from 1 to 1000 (one thousand) inclusive were written out in words, how many letters would be used?
+If all the numbers from 1 to 1000 (one thousand) inclusive were written out in
+words, how many letters would be used?
 
 
-NOTE: Do not count spaces or hyphens. For example, 342 (three hundred and forty-two) contains 23 letters and 115 (one hundred and fifteen) 
-contains 20 letters. The use of "and" when writing out numbers is in compliance with British usage.
-'''
+NOTE: Do not count spaces or hyphens. For example, 342 (three hundred and
+forty-two) contains 23 letters and 115 (one hundred and fifteen) contains 20
+letters. The use of "and" when writing out numbers is in compliance withBritish
+usage.
+"""
 
-ones_counts = [0, 3, 3, 5, 4, 4, 3, 5, 5, 4, 3, 6, 6, 8, 8, 7, 7, 9, 8, 8] #number of letters in zero, one, two, ..., nineteen (0 for zero since it's never said aloud)
-tens_counts = [0, 0, 6, 6, 5, 5, 5, 7, 6, 6] #number of letters in twenty, thirty, ..., ninety (0 for numbers less than 20 due to inconsistency in teens)
 
-count = 0
+def solution(n):
+    """Returns the number of letters used to write all numbers from 1 to n.
+    where n is lower or equals to 1000.
+    >>> solution(1000)
+    21124
+    >>> solution(5)
+    19
+    """
+    # number of letters in zero, one, two, ..., nineteen (0 for zero since it's
+    # never said aloud)
+    ones_counts = [0, 3, 3, 5, 4, 4, 3, 5, 5, 4, 3, 6, 6, 8, 8, 7, 7, 9, 8, 8]
+    # number of letters in twenty, thirty, ..., ninety (0 for numbers less than
+    # 20 due to inconsistency in teens)
+    tens_counts = [0, 0, 6, 6, 5, 5, 5, 7, 6, 6]
 
-for i in range(1, 1001):
-	if i < 1000:
-		if i >= 100:
-			count += ones_counts[i/100] + 7 #add number of letters for "n hundred"
+    count = 0
 
-			if i%100 != 0:
-				count += 3 #add number of letters for "and" if number is not multiple of 100
+    for i in range(1, n + 1):
+        if i < 1000:
+            if i >= 100:
+                # add number of letters for "n hundred"
+                count += ones_counts[i // 100] + 7
 
-		if 0 < i%100 < 20:
-			count += ones_counts[i%100] #add number of letters for one, two, three, ..., nineteen (could be combined with below if not for inconsistency in teens)
-		else:
-			count += ones_counts[i%10] + tens_counts[(i%100-i%10)/10] #add number of letters for twenty, twenty one, ..., ninety nine
-	else:
-		count += ones_counts[i/1000] + 8
+                if i % 100 != 0:
+                    # add number of letters for "and" if number is not multiple
+                    # of 100
+                    count += 3
 
-print(count)
+            if 0 < i % 100 < 20:
+                # add number of letters for one, two, three, ..., nineteen
+                # (could be combined with below if not for inconsistency in
+                # teens)
+                count += ones_counts[i % 100]
+            else:
+                # add number of letters for twenty, twenty one, ..., ninety
+                # nine
+                count += ones_counts[i % 10]
+                count += tens_counts[(i % 100 - i % 10) // 10]
+        else:
+            count += ones_counts[i // 1000] + 8
+    return count
+
+
+if __name__ == "__main__":
+    print(solution(int(input().strip())))