Word Frequency in a Large Text
Reported by candidates from Robinhood's online assessment. Pattern, common pitfall, and the honest play if you blank under the timer.
Robinhood reported this one in July 2026, and the detail that matters is the output format. You don't return a map or a list of pairs. You return strings like "hello 2", sorted by frequency descending, then alphabetically. It's a hash table problem wearing a string parsing costume. If the OA lands in your inbox this week, expect to spend more time on separators and sort order than on the counting. StealthCoder sits invisibly on your screen during the live OA as a safety net if you blank, but this one is very doable with a clean plan.
The problem
Given a large string text, split it into words using punctuation characters and whitespace as separators. Result Rules Matching is case-insensitive. Punctuation is not part of a word. Return each result as "word frequency" using the lowercase word. Sort by descending frequency. When frequencies are equal, sort the words lexicographically. Function wordFrequency(text: String) → String[] Examples Example 1 text = "hello world, hello computer" return = ["hello 2","computer 1","world 1"] hello appears twice. computer and world appear once, so their tie is resolved lexicographically.
Reported by candidates. Source: FastPrep
Pattern and pitfall
The trick is three steps. First, tokenize: walk the text and treat any non-letter, non-digit character as a separator, lowercasing as you go. Second, count with a hash map from word to frequency. Third, sort the entries with a comparator: higher count first, then word ascending. Then format each as word + space + count. The common pitfall is splitting only on spaces, so "world," becomes a different word than "world". Another is forgetting empty tokens when two separators sit side by side, or sorting by count alone and losing the lexicographic tiebreak. On a large text, avoid regex-heavy splits that build huge intermediate arrays if you can scan once. Complexity is O(n) for the scan plus O(k log k) for sorting k distinct words. If you freeze mid-OA, StealthCoder can hand you the comparator and tokenizer so you just verify edge cases.
The honest play: practice the pattern, and have StealthCoder ready for the one you didn't see coming.
You can drill Word Frequency in a Large Text 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 for the candidate who saw this exact problem leak two days before his OA and wondered if anyone had a play.
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Word Frequency in a Large Text FAQ
How hard is the Robinhood word frequency question really?+
Easy to medium. The counting is trivial. The points are lost on tokenizing punctuation correctly and getting the two-level sort right. If you've written a custom comparator before, you can finish this quickly and spend the rest of your time on edge cases.
What's the trick to this problem?+
Normalize first, count second, sort third. Lowercase everything, treat any non-alphanumeric character as a separator, count in a hash map, then sort by count descending and word ascending. Most failures come from skipping the tiebreak or leaving punctuation attached to words.
How should I handle punctuation and empty tokens?+
Scan character by character and build a word buffer. When you hit a separator, push the buffer if it's non-empty and reset it. Flush once more at the end. This avoids empty strings from consecutive separators and trailing punctuation without needing a regex.
Is this hash table pattern still asked in 2026?+
Yes. Frequency counting with a custom sort shows up constantly because it tests parsing, maps, and comparators together. Robinhood reported this in July 2026, so it's current. Expect variants like top K words or different tie rules.
How do I prepare in 48 hours?+
Write this once from scratch in your language. Practice a comparator that sorts by two keys, and test inputs with repeated punctuation, mixed case, and a single word. Then do two or three top-K frequency problems. That covers nearly every variation of this question.