Challenge 234 solutions by Packy Anderson

Solutions in:
  * Perl
  * Raku
  * Python

One blog post

2 files changed, 101 insertions, 0 deletions

diff --git a/challenge-234/packy-anderson/python/ch-1.py b/challenge-234/packy-anderson/python/ch-1.py
new file mode 100755
index 0000000000..ddfc986b12
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+++ b/challenge-234/packy-anderson/python/ch-1.py
@@ -0,0 +1,66 @@
+#!/usr/bin/env python
+
+from collections import Counter
+
+def charFrequency(word):
+    # https://docs.python.org/3/library/collections.html#counter-objects
+    freq = Counter()
+    for c in word:
+        freq[c] += 1
+    return freq
+
+def commonCharacters(words):
+    # get the character freqencies for each word
+    freq = list(map(charFrequency, words))
+
+    # grab the character frequency map for the first word
+    first = freq.pop(0)
+
+    # make a copy of the dictionary since we'll
+    # be modifying it in the loop
+    first_orig = dict(first)
+
+    # now check the characters in the first word against
+    # the characters in all the subsequent words
+    for subsequent in freq:
+        for c in first_orig:
+            if c not in subsequent:
+                # this character isn't in subsequent words,
+                # so let's remove it from the frequency map
+                # of the first word
+                first.pop(c)
+            else:
+                # the character IS in subsequent words,
+                # so let's set the frequency count to be
+                # the minimum count found in those words
+                first[c] = min(first[c], subsequent[c])
+
+    # now we generate a list of characters in the order they
+    # appear in the first word
+    output = []
+    # once again, loop over the characters in the first word
+    for c in words[0]:
+        if c not in first:
+            continue
+        if first[c] > 1:
+            first[c] -= 1
+        else:
+            first.pop(c)
+        output.append(c)
+    return output
+
+def solution(words):
+    quoted = '"' + '", "'.join(words) + '"'
+    print(f'Input: @words = ({quoted})')
+    output = commonCharacters(words)
+    quoted = '"' + '", "'.join(output) + '"'
+    print(f'Output: ({quoted})')
+
+print("Example 1:")
+solution(["java", "javascript", "julia"])
+
+print("\nExample 2:")
+solution(["bella", "label", "roller"])
+
+print("\nExample 3:")
+solution(["cool", "lock", "cook"])
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diff --git a/challenge-234/packy-anderson/python/ch-2.py b/challenge-234/packy-anderson/python/ch-2.py
new file mode 100755
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+++ b/challenge-234/packy-anderson/python/ch-2.py
@@ -0,0 +1,35 @@
+#!/usr/bin/env python
+
+def findTriplets(ints):
+    solutions = []
+    for i in range(0, len(ints) - 3 ):
+        for j in range(i + 1, len(ints) - 2):
+            for k in range(j + 1, len(ints) - 1):
+                if (ints[i] != ints[j] and
+                    ints[j] != ints[k] and
+                    ints[i] != ints[k]):
+                    solutions.append([i, j, k])
+    return solutions
+
+def solution(ints):
+    intlist = ", ".join([ str(i) for i in ints ])
+    print(f'Input: @ints = ({intlist})')
+    solutions = findTriplets(ints)
+    print(f'Output: {len(solutions)}')
+    if solutions:
+        print("")
+        for triplet in solutions:
+            i, j, k = triplet
+            print(
+                f"({i}, {j}, {k}) because " +
+                f"{ints[i]} != {ints[j]} != {ints[k]}"
+            )
+
+print("Example 1:")
+solution([4, 4, 2, 4, 3])
+
+print("\nExample 2:")
+solution([1, 1, 1, 1, 1])
+
+print("\nExample 3:")
+solution([4, 7, 1, 10, 7, 4, 1, 1])
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