Challenge 288 solutions by Packy Anderson

  * Raku that maybe looks like Raku, but mostly like Perl
  * Perl
  * Python that definitely looks like Perl
  * Elixir solution for Task 1 (for now)

1 Blog post

2 files changed, 138 insertions, 0 deletions

diff --git a/challenge-288/packy-anderson/python/ch-1.py b/challenge-288/packy-anderson/python/ch-1.py
new file mode 100755
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+++ b/challenge-288/packy-anderson/python/ch-1.py
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+#!/usr/bin/env python
+
+def isPalindrome(num):
+  # convert numerics to Strings, then reverse one of them
+  return str(num) == str(num)[::-1]
+
+def closestPalindrome(string):
+  num = int(string)
+  distance = 1
+  while True:
+    # is the smaller number at this distance a palindrome?
+    if isPalindrome(num - distance):
+      return str(num - distance)
+
+    # is the larger number at this distance a palindrome?
+    if isPalindrome(num + distance):
+      return str(num + distance)
+
+    # step 1 number futher away
+    distance += 1
+
+def solution(string):
+    print(f'Input: $str = "{string}"')
+    print(f'Output: "{closestPalindrome(string)}"')
+
+print('Example 1:')
+solution("123")
+
+print('\nExample 2:')
+solution("2")
+
+print('\nExample 3:')
+solution("1400")
+
+print('\nExample 4:')
+solution("1001")
+
+print('\nExample 5: (it doesn\'t say the input is a POSITIVE int)')
+solution("-1")
diff --git a/challenge-288/packy-anderson/python/ch-2.py b/challenge-288/packy-anderson/python/ch-2.py
new file mode 100755
index 0000000000..fd60c27664
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+++ b/challenge-288/packy-anderson/python/ch-2.py
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+#!/usr/bin/env python
+
+# helper functions to determine if the adjacent cells
+# BEFORE this one have the same value
+def prevXSame(matrix, x, y):
+  return x > 0 and matrix[y][x] == matrix[y][x-1]
+
+def prevYSame(matrix, x, y):
+  return y > 0 and matrix[y][x] == matrix[y-1][x]
+
+def contiguousBlock(matrix):
+  # first, find out the size of the matrix
+  height = len(matrix)
+  width  = len(matrix[0])
+  # start a counter for the number of blocks
+  next_block = 0
+  # a matrix of blocks
+  blocks = [[None for x in range(width)] for y in range(height)]
+
+  for y in range(height):
+    for x in range(width):
+      if prevXSame(matrix, x, y):
+        # make this cell's block number match
+        # the one above it
+        blocks[y][x] = blocks[y][x-1]
+
+      if prevYSame(matrix, x, y):
+        # if we've already assigned a block number
+        # based on the prev X being the same, and
+        # it's a DIFFERENT block than the prev Y
+        if (blocks[y][x] is not None and
+            blocks[y][x] != blocks[y-1][x]):
+          # renumber the block for the prev X to
+          # match the block for the prev Y
+          new = blocks[y-1][x]
+          old = blocks[y][x-1]
+          for y2 in range(y+1):
+            for x2 in range(width):
+              if blocks[y2][x2] == old:
+                blocks[y2][x2] = new
+
+        # make this cell's block number match
+        # the one before it
+        blocks[y][x] = blocks[y-1][x]
+
+      if blocks[y][x] is None:
+        # neither previous adjacent cell matches,
+        # assign a new block number to this cell
+        blocks[y][x] = next_block
+        next_block += 1
+
+  # now let's count the elements in each block
+  counts = [0 for x in range(next_block)]
+  for y in range(height):
+    for x in range(width):
+      counts[blocks[y][x]] += 1
+
+  return max(counts)
+
+def formatMatrix(matrix, indent=17):
+  output = []
+  for row in matrix:
+    output_row = ' ' * indent + '  ['
+    output_row += ', '.join(map(lambda i: f"'{i}'", row))
+    output_row += ']'
+    output.append(output_row)
+
+  return(
+    "[\n" + ",\n".join(output) + "\n" +
+    ' ' * indent + ']'
+  )
+
+def solution(matrix):
+    print(f'Input: $matrix = {formatMatrix(matrix)}')
+    print(f'Output: { contiguousBlock(matrix) }')
+
+print('Example 1:')
+solution([
+           ['x', 'x', 'x', 'x', 'o'],
+           ['x', 'o', 'o', 'o', 'o'],
+           ['x', 'o', 'o', 'o', 'o'],
+           ['x', 'x', 'x', 'o', 'o'],
+         ])
+
+print('\nExample 2:')
+solution([
+           ['x', 'x', 'x', 'x', 'x'],
+           ['x', 'o', 'o', 'o', 'o'],
+           ['x', 'x', 'x', 'x', 'o'],
+           ['x', 'o', 'o', 'o', 'o'],
+         ])
+
+print('\nExample 3:')
+solution([
+           ['x', 'x', 'x', 'o', 'o'],
+           ['o', 'o', 'o', 'x', 'x'],
+           ['o', 'x', 'x', 'o', 'o'],
+           ['o', 'o', 'o', 'x', 'x'],
+         ])