Count Number of Homogenous Substrings
A medium-tier problem at 57% community acceptance, tagged with Math, String. Reported in interviews at Virtu Financial and 0 others.
Count Number of Homogenous Substrings shows up in your OA and you've got a few minutes to recognize the pattern. This problem asks you to count all contiguous substrings where every character is identical. It's a Math and String problem that Virtu Financial has asked. The trap is overthinking it: most candidates try to generate substrings explicitly, which works but wastes time. The actual trick is mathematical. If you have a run of n identical characters, the number of homogenous substrings in that run is n*(n+1)/2. Once you spot that formula, the solution collapses from O(n^2) to O(n). If this problem hits your live assessment and the brute force approach feels slow, StealthCoder surfaces the optimized logic in seconds, invisible to the proctor.
Companies that ask "Count Number of Homogenous Substrings"
Count Number of Homogenous Substrings 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 Amazon engineer who realized the OA tests how well you memorized 200 problems, not how well you code.
Get StealthCoderThe pattern here is grouping consecutive identical characters into runs, then applying the summation formula to each run. A run of length 3 (like 'aaa') contains 6 substrings: 'a', 'a', 'a', 'aa', 'aa', 'aaa'. That's 1+2+3, or 3*4/2. Most candidates either iterate through the string building actual substrings (slow and error-prone) or jump straight to a formula without justifying it. The missed middle ground is: iterate once, track the current run length, and whenever the character changes, apply the formula and reset. Edge cases trip people up: single characters (length 1 yields 1 substring), empty input, and off-by-one errors when terminating the last run. With 57% acceptance, plenty of candidates solve it but many lose points to inefficiency or logic bugs. StealthCoder's hedge here is the formula itself. If you blank on it during screen share, you get a correct O(n) solution that looks like you've done this before.
Pattern tags
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Make sure you actually pass it.
Count Number of Homogenous Substrings recycles across companies for a reason. It's medium-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 Amazon engineer who realized the OA tests how well you memorized 200 problems, not how well you code. Works on HackerRank, CodeSignal, CoderPad, and Karat.
Count Number of Homogenous Substrings interview FAQ
Is this really a 'medium' or is it harder than typical mediums?+
It's genuinely medium once you spot the n*(n+1)/2 formula. The acceptance rate of 57% suggests the formula isn't obvious on first read. Without it, you'll waste time on brute force. With it, it's a one-pass iteration and arithmetic.
Does Virtu Financial ask this often in their interviews?+
Data shows only one company in the reports for this problem, so frequency is likely lower. But if it appears in Virtu's pipeline, you should recognize the pattern fast. It's the kind of 'trick once, always easy' problem.
What's the real pitfall most people hit?+
Not recognizing it's a counting formula problem, not a substring generation problem. Candidates build or iterate substrings explicitly, which passes but is slow. The formula is the insight.
How does this relate to other String problems at tech companies?+
It teaches pattern recognition in continuous runs of data. That skill shows up in compression, deduplication, and signal-processing problems. Once you see one, you spot similar patterns faster.
Can I solve this without math if I'm more confident with code?+
Yes, but it'll be slower and riskier in a timed assessment. Explicit iteration works for small inputs. For large strings, O(n) with the formula is the safe path. Either way, knowing the formula saves you from a re-interview.
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