Autocorrect Prototype
Reported by candidates from Goldman Sachs's online assessment. Pattern, common pitfall, and the honest play if you blank under the timer.
You've got an autocorrect problem from Goldman Sachs, reported January 2024. The setup looks straightforward: match query strings against a word list and return anagrams, sorted alphabetically. The trap is efficiency. Brute-force anagram checking on every query against every word will TLE. You need a canonical form (sorted letters or character count) as your lookup key, built once upfront. StealthCoder can be your safety net if the logic fuzzes during the live OA.
The problem
\ Complete the implementation of an autocorrect function. Given\ a search query string, the function should return all words which are anagrams.\ \ Given 2 arrays, words[n], and queries[q], for each\ query, return an array of the strings that are anagrams, sorted alphabetically ascending.\ \ Note: An anagram is any string that can be formed be\ rearranging the letters of a string.\ \ Function Description\ \ Complete the function getSearchResults in the editor.\
Reported by candidates. Source: FastPrep
Pattern and pitfall
The pattern is hash-table with a preprocessing step. Sort each word's letters to create a canonical anagram key. Store all words under their key in a map. For each query, compute its canonical key and retrieve the bucket instantly. Sort results alphabetically before returning. The common mistake is recalculating anagram groups per query or doing naive character-by-character comparisons. A second pitfall: forgetting to sort the output alphabetically, even though the spec says so. Use a hash map (dictionary in Python, map in JavaScript) keyed by sorted letters. This scales from O(n*m*log m + q*log q) with sorting to O(n*m*log m) preprocessing and O(q*log q) per query. StealthCoder handles the canonical-key insight if you blank on optimization during the assessment.
If you see this problem in your OA tomorrow, the play is to recognize the pattern in 30 seconds. StealthCoder buys you that recognition.
You can drill Autocorrect Prototype cold, or you can hedge it. StealthCoder runs invisibly during screen share and surfaces a working solution in under 2 seconds. The proctor sees the IDE. They don't see what's behind it. Built by an Amazon engineer who passed his OA cold and still thinks the filter is broken.
Get StealthCoderRelated leaked OAs
This OA pattern shows up on LeetCode as group anagrams. If you have time before the OA, drill that.
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Make sure you actually pass Goldman Sachs's OA.
Goldman Sachs reuses patterns across OAs. Built by an Amazon engineer who passed his OA cold and still thinks the filter is broken. Works on HackerRank, CodeSignal, CoderPad, and Karat.
Autocorrect Prototype FAQ
What makes this an anagram problem and not just string matching?+
Anagrams have the same letters in any order. 'listen' and 'silent' are anagrams. The trick is that you can't compare letter-by-letter; you need a fingerprint. Sorting the letters or counting character frequencies gives you that fingerprint instantly.
How do I group anagrams without checking every pair?+
Preprocess: for each word, sort its letters alphabetically to get a canonical key. Store the word in a hash map under that key. Now all anagrams of a query share the same key. Lookup is O(1) average case. No nested loops needed.
Do I have to sort the output for each query?+
Yes. The spec says 'sorted alphabetically ascending.' Forgetting this is a common miss. After you retrieve the anagram bucket from your map, sort it before returning. It's one line of code but mandatory.
Should I use character count instead of sorting letters?+
Both work. Sorted letters is simpler to code and debug. A frequency map or counter is also valid but requires more lines. For a live OA, go with sorted letters unless you're very comfortable with counters.
What if a query isn't in the word list? Should I return an empty array?+
The spec doesn't explicitly say, but the safe assumption is yes: return an empty array. Your hash map will have no entry for a novel anagram key, so you return an empty result. This is idiomatic behavior.