Count Pairs Of Similar Strings
A easy-tier problem at 72% community acceptance, tagged with Array, Hash Table, String. Reported in interviews at IBM and 0 others.
Count Pairs of Similar Strings is an easy problem that shows up in IBM interviews and similar screening rounds. You're given an array of strings and need to count pairs where the two strings contain the exact same set of characters, ignoring frequency and order. The problem looks trivial at first glance, just compare character sets, but the real skill tested is choosing the right representation and avoiding the trap of overthinking the comparison logic. With a 72% acceptance rate, most people solve it, but a decent number trip on the encoding step or inefficient pair checking. If this problem hits your live assessment and the character-set trick doesn't jump out, StealthCoder solves it in seconds, invisible to the proctor.
Companies that ask "Count Pairs Of Similar Strings"
Count Pairs Of Similar Strings 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. Made by a working FAANG engineer who treats the OA the way companies treat hiring: as a game with rules you should know.
Get StealthCoderThe pattern is simple: normalize each string to a canonical form (usually sorted characters or a bitmask) so two strings with the same character set have the same representation. Then count how many strings share each representation using a hash table. The trick candidates miss is realizing that frequency doesn't matter, 'abc' and 'aabbcc' are similar. Some people waste time building full character-count maps when they only need to know which characters are present. Others iterate through all pairs and compare, which is slower than grouping by canonical form first. The bitmask approach is clever but not required; a sorted string works fine. The bigger gotcha is off-by-one errors in pair counting if you're manually iterating. Hash table counting sidesteps that entirely. This is a 5-to-10 minute problem if you've internalized the 'normalize and group' pattern, but a 20-plus minute slog if you're building pair comparisons from scratch.
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Count Pairs Of Similar Strings 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. Made by a working FAANG engineer who treats the OA the way companies treat hiring: as a game with rules you should know. Works on HackerRank, CodeSignal, CoderPad, and Karat.
Count Pairs Of Similar Strings interview FAQ
Is this problem still asked at IBM and similar companies?+
Yes. IBM has reportedly asked it. It's a typical easy-tier screening question, often used to filter candidates who can't translate a simple requirement into code. It's not trendy or algorithmic, just competence under light time pressure.
What's the trick I'm missing if the obvious approach feels slow?+
You're probably comparing every pair directly instead of grouping by character set first. Normalize each string (sort it or use a bitmask), put each into a hash table, then count pairs. O(n) instead of O(n^2). The character set, not frequency, is what makes strings similar.
Does frequency matter when checking if two strings are similar?+
No. Two strings are similar if they contain the same characters, regardless of how many times each appears. 'a' and 'aaa' are similar. This is the core insight; miss it and you'll build overly complex comparison logic.
Should I use a bitmask or a hash table?+
Either works. Bitmask is slightly faster and more elegant if all characters are lowercase letters. Hash table is more general. Sorting the string works too and is often clearer. Pick what you can code fastest without bugs under pressure.
Why is this classified as easy if it requires bit manipulation as a topic?+
Bit manipulation is listed as a topic, but you don't need it to solve the problem. You can normalize strings using sort or character sets instead. It's easy because the logic is straightforward once you understand that only unique characters matter, not frequency.
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