Groups of Special-Equivalent Strings

Groups of Special-Equivalent Strings

What is this problem about?

The Groups of Special-Equivalent Strings interview question defines two strings as "special-equivalent" if you can swap any two even-indexed characters, or any two odd-indexed characters, any number of times to make the strings equal. Given an array of strings, you need to find how many groups of special-equivalent strings exist.

Why is this asked in interviews?

Meta uses this String and Hash Table coding problem to test your ability to define a "canonical form" or "signature" for complex equivalence relations. It evaluates if you realize that infinite swapping within a specific subset (even/odd indices) simply means the sorted version of that subset uniquely identifies it.

Algorithmic pattern used

This problem relies on Canonical Signature Hashing.

1. For every string, extract all characters at even indices and sort them.
2. Extract all characters at odd indices and sort them.
3. Concatenate the sorted even characters and sorted odd characters to create a unique signature string.
4. Add this signature to a Hash Set.
5. Since special-equivalent strings will always generate the exact same signature, the number of groups is simply the size of the Hash Set.

Example explanation

Strings: ["abcd", "cdab", "cbad", "xyzz", "zzxy", "zzyx"]

1. "abcd":
   • Evens: 'a', 'c' -> sorted: "ac"
   • Odds: 'b', 'd' -> sorted: "bd"
   • Signature: "acbd"
2. "cdab":
   • Evens: 'c', 'a' -> sorted: "ac"
   • Odds: 'd', 'b' -> sorted: "bd"
   • Signature: "acbd" (Matches "abcd"!) By generating these signatures, we find that the first three strings form one group, and the last three form another. Result: 2 groups.

Common mistakes candidates make

• Simulating Swaps: Trying to write a backtracking or BFS algorithm to generate all possible swaps to see if one string can reach another. This is $O(N!)$ and will instantly Time Limit Exceed.
• Sorting the Whole String: Sorting the entire string "abcd" to "abcd" destroys the even/odd index constraint. "abcd" and "badc" are NOT special-equivalent, but sorting the whole string would make them look the same.
• Inefficient String Building: Creating many substrings and concatenations in a loop. Using a size-52 array (26 for even counts, 26 for odd counts) as the hash key is $O(L)$ per string and avoids sorting entirely.

Interview preparation tip

Any time a problem allows "any number of swaps" between elements of a specific group, it means the order doesn't matter, only the frequency of elements in that group matters. Use frequency counts or sorting to create a hashable signature.