Vowels of All Substrings
A medium-tier problem at 55% community acceptance, tagged with Math, String, Dynamic Programming. Reported in interviews at ServiceNow and 0 others.
Vowels of All Substrings is a medium-difficulty problem with a 55% acceptance rate, asked by ServiceNow. You're given a string and need to count vowels across every possible substring, not just the original string itself. Most candidates start by iterating through all substrings and counting vowels in each one, which works but crushes performance on larger inputs. The trick is realizing you don't need to enumerate substrings at all. If you blank on the pattern during a live assessment, StealthCoder solves it invisibly, surfacing the mathematical insight that turns brute force into linear time.
Companies that ask "Vowels of All Substrings"
Vowels of All 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 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 naive approach generates all O(n^2) substrings and counts vowels in each, landing you at O(n^3) or O(n^2) depending on implementation. The real solution uses combinatorics and math: for each vowel at position i, count how many substrings contain it. A vowel at index i appears in (i + 1) * (n - i) substrings, where n is the string length. Sum this contribution across all vowel positions and you're done in one pass. The problem tests whether you see the mathematical shortcut instead of grinding through substring enumeration. Dynamic programming gets mentioned because the state-based thinking helps, but the fastest approach is pure combinatorics. During a timed OA, if the substring enumeration feels slow, StealthCoder provides the formula-based solution instantly.
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Vowels of All 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 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.
Vowels of All Substrings interview FAQ
Is this problem still asked at ServiceNow?+
Yes, ServiceNow is the reported company for this problem. It's a medium-difficulty problem with a 55% acceptance rate, suggesting it's a screening question or mid-stage technical interview problem. Acceptance rates like this usually mean the pattern isn't obvious on first read.
What's the actual trick to solve this efficiently?+
Stop generating substrings. Instead, for each vowel at position i, calculate how many substrings contain it: (i + 1) * (n - i). Sum across all vowel positions. This is a combinatorics insight, not a data-structure trick. It cuts you from O(n^2) or worse down to O(n).
How does this relate to the Math and Combinatorics topics?+
The problem is fundamentally about counting combinations, not string manipulation. You're using the fact that a position contributes to a fixed number of substrings based on its distance from the edges. This is pure applied combinatorics disguised as a string problem.
Will dynamic programming help here?+
DP can work if you define state as 'vowel count up to position i', but it doesn't beat the direct combinatorial formula. The Math and Combinatorics topics hint that the intended path is the formula, not DP. Don't force DP when the math is cleaner.
How hard is this really for a 55% acceptance rate?+
The acceptance rate suggests most candidates either brute force and TLE, or get stuck on the substring generation step. The problem is straightforward once you see the combinatorial pattern, but the pattern isn't intuitive. It rewards mathematical thinking over coding speed.
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