Word Abbreviation
A hard-tier problem at 62% community acceptance, tagged with Array, String, Greedy. Reported in interviews at Applied Intuition and 1 others.
Word Abbreviation is a hard-rated problem that asks you to generate unique abbreviations for a list of words, where each abbreviation is shorter than the original word. Applied Intuition and Snap have both asked it in interviews. At first glance, it looks like a straightforward string compression task. In reality, the challenge is ensuring uniqueness without collisions, and your greedy approach will fail on overlapping abbreviation patterns. The 62% acceptance rate reflects the gap between a naive solution and one that actually handles all edge cases. If you haven't seen the collision trap before, StealthCoder will surface the working pattern in seconds during your live assessment.
Companies that ask "Word Abbreviation"
Word Abbreviation is the kind of problem that decides whether you pass. StealthCoder reads the problem on screen and surfaces a working solution in under 2 seconds. Invisible to screen share. The proctor sees nothing. Built by an engineer at a top-10 tech company who can solve these problems cold but didn't want to trust himself in a 90-minute screen share.
Get StealthCoderThe core trick is understanding that abbreviating a word by its first letter, length, and last letter (like 'internationalization' to 'i18n') sounds clever until two words abbreviate identically. You can't just greedily pick the shortest abbreviation for each word in isolation. You need to generate candidate abbreviations in increasing length (one letter, two letter prefix plus length plus last, three letter prefix, and so on) and filter them against all other words' candidates. This is where a Trie or sorted comparison becomes essential. Many candidates try to optimize too early or rely on a single abbreviation per word without checking conflicts. The pattern is greedy, but the greedy step is at the candidate-generation level, not the word level. StealthCoder hedges the live interview if you freeze on how to structure the conflict-resolution loop.
Pattern tags
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Word Abbreviation recycles across companies for a reason. It's hard-tier, and most candidates blank under the timer. StealthCoder is the hedge: an AI overlay invisible during screen share. It reads the problem and surfaces a working solution in under 2 seconds. Built by an engineer at a top-10 tech company who can solve these problems cold but didn't want to trust himself in a 90-minute screen share. Works on HackerRank, CodeSignal, CoderPad, and Karat.
Word Abbreviation interview FAQ
Is Word Abbreviation still asked at FAANG?+
It's been reported by Applied Intuition and Snap. It's not in the top-tier rotation like TwoSum or MergeIntervals, but it does cycle through mid-tier companies, especially those focused on string processing and systems design. Expect it less frequently than other hard problems, but don't skip it.
What's the collision trap everyone walks into?+
Abbreviating each word independently, like 'integer' to 'i6' and 'ineligible' to 'i9', then realizing both could also abbreviate as 'in5' or 'ine4'. You must generate all valid abbreviations per word, then find the shortest one unique across the entire list. Single-pass greedy fails here.
Do I need a Trie or sorting?+
Both are viable. Trie excels if you're building a common-prefix structure to prune candidates early. Sorting the word list and comparing candidates lexicographically also works and is often simpler to code. Either approach handles the collision detection correctly.
How does this relate to array and string topics?+
You're iterating over an array of strings, generating substring abbreviations, and comparing them for uniqueness. The array part is trivial; the real work is the string generation and conflict resolution. Trie and sorting help you scale the comparison phase.
What's a realistic time complexity?+
If n is word count and m is average word length, generating all abbreviations is O(n * m^2) or O(n * m) depending on how you prune. Comparing candidates and resolving conflicts across all words adds another pass. Expect O(n^2 * m) or better with careful implementation. Hard problems rarely optimize to linear.
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