python 3 code for challenge 67

These turned out to be a lot simpler than my Perl versions, in large part
because you don't have to worry about references in Python.

2 files changed, 24 insertions, 0 deletions

diff --git a/challenge-067/walt-mankowski/python/ch-1.py b/challenge-067/walt-mankowski/python/ch-1.py
new file mode 100644
index 0000000000..83b9cd47dd
--- /dev/null
+++ b/challenge-067/walt-mankowski/python/ch-1.py
@@ -0,0 +1,7 @@
+from sys import argv
+from itertools import combinations
+
+m, n = [int(x) for x in argv[1:]]
+it = combinations(range(1,m+1), n)
+combs = [comb for comb in it]
+print(combs)
diff --git a/challenge-067/walt-mankowski/python/ch-2.py b/challenge-067/walt-mankowski/python/ch-2.py
new file mode 100644
index 0000000000..5cb94cfa10
--- /dev/null
+++ b/challenge-067/walt-mankowski/python/ch-2.py
@@ -0,0 +1,17 @@
+from sys import argv
+from itertools import product
+
+key = {'1': '_@',
+       '2': 'ABC',
+       '3': 'DEF',
+       '4': 'GHI',
+       '5': 'JKL',
+       '6': 'MNO',
+       '7': 'PQRS',
+       '8': 'TUV',
+       '9': 'WXYZ',
+       '0': ' ',
+      }
+
+s = argv[1]
+print([''.join(p) for p in product(*[key[c] for c in s])])