Successive adding of char to get the longest word in the dictionary [closed]

Answer 1

Assuming you need to do this repeatedly (or you want the answer for every one of the 26 letters), do it backwards:

1. Load a dictionary, and sort it by length, descending
2. Establish a mapping, initially empty, between words and (extension, max_len) tuples.
3. For each word in the sorted list:
   1. If it's already in the mapping, retrieve the max len.
   2. If it's not, set the max len to the word length.
   3. Examine each word produced by deleting a character. If that word is not in the mapping, or our max_len exceeds the max_len of the word already in the mapping, update the mapping with the current word and max_len

Then, to get the chain for a given prefix, simply start with that prefix and repeatedly look it and its extensions up in the dictionary.

Here's the sample Python code:

words = [x.strip().lower() for x in open('/usr/share/dict/words')]
words.sort(key=lambda x:len(x), reverse=True)
word_map = {}  # Maps words to (extension, max_len) tuples

for word in words:
  if word in word_map:
    max_len = word_map[word][1]
  else:
    max_len = len(word)
  for i in range(len(word)):
    new_word = word[:i] + word[i+1:]
    if new_word not in word_map or word_map[new_word][2] < max_len:
      word_map[new_word] = (word, max_len)

# Get a chain for each letter
for term in "abcdefghijklmnopqrstuvwxyz":
  chain = [term]
  while term in word_map:
    term = word_map[term][0]
    chain.append(term)
  print chain

And its output for each letter of the alphabet:

['a', 'ah', 'bah', 'bach', 'brach', 'branch', 'branchi', 'branchia', 'branchiae', 'branchiate', 'abranchiate']
['b', 'ba', 'bac', 'bach', 'brach', 'branch', 'branchi', 'branchia', 'branchiae', 'branchiate', 'abranchiate']
['c', 'ca', 'cap', 'camp', 'campo', 'campho', 'camphor', 'camphory', 'camphoryl', 'camphoroyl']
['d', 'ad', 'cad', 'card', 'carid', 'carida', 'caridea', 'acaridea', 'acaridean']
['e', 'er', 'ser', 'sere', 'secre', 'secret', 'secreto', 'secretor', 'secretory', 'asecretory']
['f', 'fo', 'fot', 'frot', 'front', 'afront', 'affront', 'affronte', 'affronted']
['g', 'og', 'log', 'logy', 'ology', 'oology', 'noology', 'nosology', 'nostology', 'gnostology']
['h', 'ah', 'bah', 'bach', 'brach', 'branch', 'branchi', 'branchia', 'branchiae', 'branchiate', 'abranchiate']
['i', 'ai', 'lai', 'lain', 'latin', 'lation', 'elation', 'delation', 'dealation', 'dealbation']
['j', 'ju', 'jug', 'juga', 'jugal', 'jugale']
['k', 'ak', 'sak', 'sake', 'stake', 'strake', 'straked', 'streaked']
['l', 'la', 'lai', 'lain', 'latin', 'lation', 'elation', 'delation', 'dealation', 'dealbation']
['m', 'am', 'cam', 'camp', 'campo', 'campho', 'camphor', 'camphory', 'camphoryl', 'camphoroyl']
['n', 'an', 'lan', 'lain', 'latin', 'lation', 'elation', 'delation', 'dealation', 'dealbation']
['o', 'lo', 'loy', 'logy', 'ology', 'oology', 'noology', 'nosology', 'nostology', 'gnostology']
['p', 'pi', 'pig', 'prig', 'sprig', 'spring', 'springy', 'springly', 'sparingly', 'sparringly']
['q']
['r', 'ra', 'rah', 'rach', 'brach', 'branch', 'branchi', 'branchia', 'branchiae', 'branchiate', 'abranchiate']
['s', 'si', 'sig', 'spig', 'sprig', 'spring', 'springy', 'springly', 'sparingly', 'sparringly']
['t', 'ut', 'gut', 'gutt', 'gutte', 'guttle', 'guttule', 'guttulae', 'guttulate', 'eguttulate']
['u', 'ut', 'gut', 'gutt', 'gutte', 'guttle', 'guttule', 'guttulae', 'guttulate', 'eguttulate']
['v', 'vu', 'vum', 'ovum']
['w', 'ow', 'low', 'alow', 'allow', 'hallow', 'shallow', 'shallowy', 'shallowly']
['x', 'ox', 'cox', 'coxa', 'coxal', 'coaxal', 'coaxial', 'conaxial']
['y', 'ly', 'loy', 'logy', 'ology', 'oology', 'noology', 'nosology', 'nostology', 'gnostology']
['z', 'za', 'zar', 'izar', 'izard', 'izzard', 'gizzard']

Edit: Given the degree to which branches merge towards the end, I thought it would be interesting to draw a graph to demonstrate this:

An interesting extension of this challenge: It's likely there are several equilength final words for some letters. Which set of chains minimizes the number of final nodes (eg, merges the most letters)?