Check if All Characters Have Equal Number of Occurrences
A easy-tier problem at 79% community acceptance, tagged with Hash Table, String, Counting. Reported in interviews at Bolt and 0 others.
You're staring at a string, asked to verify that every character appears the same number of times. Sounds trivial. It is, until you overthink it or miss the counting step. Bolt has asked this one. With a 78% acceptance rate, most people get it right, but the ones who don't usually skip the verification phase entirely. The trick isn't algorithmic complexity. It's making sure your count map actually validates what you claim. If this problem shows up live and you blank on structure, StealthCoder surfaces a working solution in seconds, invisible to the proctor.
Companies that ask "Check if All Characters Have Equal Number of Occurrences"
Check if All Characters Have Equal Number of Occurrences 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 because the OA filter rejects engineers who'd pass the on-site. That's a broken filter. This is the workaround.
Get StealthCoderBuild a frequency map (hash table) of every character in the string, then check if all values in that map are identical. The pitfall is assuming a single pass is enough. You need two logical steps: count, then validate. Some candidates count and immediately return true on the first duplicate they see, missing characters that appear a different number of times. String manipulation and Counting go hand in hand here. Hash Table is your data structure because lookup and insertion are both O(1). The obvious approach works. There's no secret greedy algorithm or dynamic programming trap. Where candidates fail is in incomplete validation logic or off-by-one errors in the counting loop. StealthCoder is the hedge if you freeze on the validation logic during the live OA.
Pattern tags
You know the problem.
Make sure you actually pass it.
Check if All Characters Have Equal Number of Occurrences recycles across companies for a reason. It's easy-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 because the OA filter rejects engineers who'd pass the on-site. That's a broken filter. This is the workaround. Works on HackerRank, CodeSignal, CoderPad, and Karat.
Check if All Characters Have Equal Number of Occurrences interview FAQ
Is this actually an easy problem or is there a trick?+
It's genuinely easy. No trick. The 78% acceptance rate reflects that most people correctly implement count and validate. Failures come from incomplete validation or sloppy iteration, not hidden complexity. Two passes: count all characters, then verify all counts are equal.
Do I need to sort the character counts after building the hash table?+
No. Sorting is unnecessary. Use a set to check if the count map has exactly one unique value. Or iterate the map once and compare each count to the first one. Sorting adds O(n log n) overhead for zero benefit.
What if the string is empty or has one character?+
Empty string: vacuously true, all zero characters have equal occurrences. Single character: that character appears once, so yes, all characters (one) have equal occurrences. Edge cases pass trivially once your validation logic is correct.
How does this problem relate to Counting and Hash Table topics?+
Counting is the core skill. You must tally each character's frequency accurately. Hash Table is the vehicle, giving you O(1) lookups and insertions. String work is incidental. The problem tests whether you pick the right data structure and implement counting without errors.
Will Bolt or other companies ask harder follow-ups after this?+
Likely. This is often a warm-up. Expect follow-ups: count characters with specific constraints, validate anagrams across multiple strings, or optimize for space. Master the base problem first so you don't waste time on easy logic during the live OA.
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