Problem Description

Given an array of strings A, two strings S and T are considered special-equivalent if you can transform S into T by swapping any number of pairs of characters at even indices, and/or any number of pairs of characters at odd indices.

Your task is to determine how many groups of special-equivalent strings exist in A. Each string belongs to exactly one group, and no string is in more than one group.

Constraints:

• Each string in A has the same length.
• All characters are lowercase English letters.
• Each string appears at most once in A.

Thought Process

At first glance, the problem seems to require checking all possible swaps for each string to compare with others, which would be computationally expensive. However, upon closer inspection, we realize that swapping characters at even indices among themselves and odd indices among themselves means that the order within even and odd positions does not matter—only their multisets do.

Therefore, instead of brute-forcing all swaps, we can try to find a canonical form or signature for each string. If two strings have the same canonical form (i.e., the same characters at even indices, and the same at odd indices, up to reordering), they belong to the same special-equivalent group.

This insight allows us to reduce the problem to grouping strings by their canonical forms.

Solution Approach

Let's break down the optimized solution step by step:

1. Extract Even and Odd Indexed Characters:
   • For each string, collect all characters at even indices (0, 2, 4, ...).
   • Collect all characters at odd indices (1, 3, 5, ...).
2. Sort Each Collection:
   • Sort the even-indexed characters and the odd-indexed characters separately. This is because within even (or odd) indices, any permutation is allowed, so their sorted order uniquely represents the group.
3. Form a Signature:
   • Combine the sorted even and sorted odd sequences into a tuple or string. This tuple acts as a unique signature for the special-equivalent group.
4. Group by Signature:
   • Use a set (or hash set) to collect all unique signatures.
5. Count Unique Groups:
   • The number of unique signatures is the number of special-equivalent groups.

This approach leverages the mathematical property that swapping within a group is equivalent to treating the group as a multiset, and sorting is a simple way to create a unique identifier for a multiset.

Example Walkthrough

Let's consider the input A = ["abc","acb","bac","bca","cab","cba"].

1. For each string, split into even and odd indexed characters:
   • "abc": even = ['a', 'c'], odd = ['b']
   • "acb": even = ['a', 'b'], odd = ['c']
   • "bac": even = ['b', 'c'], odd = ['a']
   • "bca": even = ['b', 'a'], odd = ['c']
   • "cab": even = ['c', 'b'], odd = ['a']
   • "cba": even = ['c', 'a'], odd = ['b']
2. Sort the even and odd lists:
   • "abc": ('a','c'), ('b')
   • "acb": ('a','b'), ('c')
   • "bac": ('b','c'), ('a')
   • "bca": ('a','b'), ('c')
   • "cab": ('b','c'), ('a')
   • "cba": ('a','c'), ('b')
3. Combine as signatures:
   • "abc": ('ac','b')
   • "acb": ('ab','c')
   • "bac": ('bc','a')
   • "bca": ('ab','c')
   • "cab": ('bc','a')
   • "cba": ('ac','b')
4. Unique signatures: {('ac','b'), ('ab','c'), ('bc','a')}
5. Result: There are 3 groups of special-equivalent strings.

Time and Space Complexity

Brute-force approach:

• Would require comparing every pair of strings and potentially generating all permutations of even and odd indices, which is factorial time and infeasible for large inputs.

• For each string of length n, splitting and sorting takes O(n log n) time.
• If there are m strings, total time is O(m * n log n).
• Space complexity is O(m * n) due to storing signatures for each string.
• Using a set for uniqueness has O(m) additional space.

This is efficient and works well within typical Leetcode constraints.

Summary

The key insight is that swapping within even or odd indices is equivalent to treating those characters as unordered groups. By sorting the characters at even and odd indices separately, we can generate a unique signature for each special-equivalent group. Grouping strings by these signatures gives us the answer efficiently, avoiding brute-force permutations and comparisons.

This solution is elegant because it reduces a seemingly complex equivalence relation to a simple signature-based grouping problem.

Code Implementation

class Solution:
    def numSpecialEquivGroups(self, A):
        signatures = set()
        for word in A:
            even = sorted(word[::2])
            odd = sorted(word[1::2])
            signature = (''.join(even), ''.join(odd))
            signatures.add(signature)
        return len(signatures)
      

class Solution {
public:
    int numSpecialEquivGroups(vector<string>& A) {
        unordered_set<string> signatures;
        for (const string& word : A) {
            string even, odd;
            for (int i = 0; i < word.size(); ++i) {
                if (i % 2 == 0) even += word[i];
                else odd += word[i];
            }
            sort(even.begin(), even.end());
            sort(odd.begin(), odd.end());
            signatures.insert(even + "|" + odd);
        }
        return signatures.size();
    }
};
      

class Solution {
    public int numSpecialEquivGroups(String[] A) {
        Set<String> signatures = new HashSet<>();
        for (String word : A) {
            char[] even = new char[(word.length() + 1) / 2];
            char[] odd = new char[word.length() / 2];
            for (int i = 0; i < word.length(); ++i) {
                if (i % 2 == 0) even[i / 2] = word.charAt(i);
                else odd[i / 2] = word.charAt(i);
            }
            Arrays.sort(even);
            Arrays.sort(odd);
            String signature = new String(even) + "|" + new String(odd);
            signatures.add(signature);
        }
        return signatures.size();
    }
}
      

var numSpecialEquivGroups = function(A) {
    const signatures = new Set();
    for (let word of A) {
        let even = [];
        let odd = [];
        for (let i = 0; i < word.length; ++i) {
            if (i % 2 === 0) even.push(word[i]);
            else odd.push(word[i]);
        }
        even.sort();
        odd.sort();
        const signature = even.join('') + '|' + odd.join('');
        signatures.add(signature);
    }
    return signatures.size;
};