基于词库的最少词汇覆盖给定字母集的求解方案问询

Alright, let's walk through a practical, efficient approach to find the smallest set of words that covers all your target letters—this method balances speed and effectiveness, perfect for large word banks:

Step 1: Prune the Word Bank First #

• Do a quick pass to discard any word that doesn’t contain at least one target letter. There’s zero reason to keep these around; they can’t contribute to covering your desired letters, and trimming them early saves you unnecessary computation later.

Step 2: Calculate Letter Rarity Scores #

• For the remaining words, count how often each target letter appears across the entire filtered word list. Then compute a relative rarity score for each letter: the fewer times a letter shows up, the higher its rarity score.
  • Why? Rare letters are harder to cover with random words—prioritizing them ensures we knock out the "tough ones" first, leaving common letters that are easy to pick up later.

Step 3: Rank Words by "Coverage Value" #

• For each word in the filtered list, calculate a value score:

  Value Score = (Sum of rarity scores for all unique target letters in the word) / Word Length
  

• Sort the words in descending order of this score. The top-ranked words are your best bets—they deliver the most coverage of rare letters per character, meaning you can cross off more hard-to-find letters with fewer words.

Step 4: Iterate Until All Letters Are Covered #

• Pick the top-ranked word, add it to your final set, and remove all the target letters it covers from your "to-do" list.
• Repeat the entire process with the remaining target letters: re-prune the word bank (focusing only on words that contain leftover letters), recalculate rarity scores for the remaining letters, re-rank the words, and pick the next best option.
• Keep looping until every target letter is covered.

Quick Example to Illustrate #

Suppose your target letters are {x, z, a, b} and your filtered word bank has ["axz", "ab", "xyz"]:

1. First round: x and z are rare (appear twice each), a appears twice, b once.
2. "axz" has a score of (x_rarity + z_rarity + a_rarity)/3, which beats "xyz" (since a is more useful than y here) and "ab".
3. Pick "axz", remove x, z, a from targets—only b remains.
4. Next round: only words containing b are left ("ab"), pick it, and you’re done with 2 words total.

内容的提问来源于stack exchange，提问作者kontur