LeetCode 893 - Groups of Special-Equivalent Strings
The problem asks us to group strings based on a special equivalence property. Specifically, two strings are special-equivalent if you can swap characters at even indices among themselves and characters at odd indices among themselves any number of times to make the two strings…
Difficulty: 🟡 Medium
Topics: Array, Hash Table, String, Sorting
Solution
Problem Understanding
The problem asks us to group strings based on a special equivalence property. Specifically, two strings are special-equivalent if you can swap characters at even indices among themselves and characters at odd indices among themselves any number of times to make the two strings identical. The input is an array of strings words, all of the same length. The output is a single integer representing the number of distinct groups of special-equivalent strings.
The constraints tell us that words can have up to 1000 strings and each string can have up to 20 characters. This rules out any brute-force approach that compares all pairs of strings character-by-character after all possible swaps, as it would quickly become too slow. Additionally, all strings consist of lowercase English letters, which makes it feasible to represent character counts compactly.
Key edge cases include very short strings (length 1), arrays where all strings are identical, or strings where no two strings are special-equivalent. The problem guarantees that all strings have the same length, so we do not need to handle variable-length strings.
Approaches
The brute-force approach would involve trying every possible sequence of swaps on each string and comparing it with every other string to see if they can be made equal. While this would eventually produce the correct answer, the number of swaps grows factorially with string length (roughly (n/2)! for even and (n/2)! for odd positions), making this approach infeasible even for small strings.
The optimal approach leverages the key insight that special equivalence only depends on the multiset of characters at even indices and the multiset of characters at odd indices. That is, two strings are special-equivalent if sorting their even-indexed characters and sorting their odd-indexed characters produces the same pair of sorted strings. This allows us to map each string to a canonical "signature" and use a hash set to count unique groups efficiently.
| Approach | Time Complexity | Space Complexity | Notes |
|---|---|---|---|
| Brute Force | O(n * ((m/2)! * (m/2)!)) | O(n) | Compare all pairs after generating all permutations of even and odd indexed swaps. Too slow for constraints. |
| Optimal | O(n * m log m) | O(n * m) | Use a canonical signature of sorted even and odd indexed characters to identify groups efficiently. |
Algorithm Walkthrough
- Initialize a set for signatures: This will store unique string signatures representing special-equivalent groups.
- Iterate over each string: For each string, separate characters at even and odd indices.
- Sort each group of characters: Sort the even-indexed characters and the odd-indexed characters independently. Sorting provides a unique representation regardless of original order.
- Combine into a tuple or string signature: The pair of sorted sequences serves as a canonical representation of the string.
- Add the signature to the set: This automatically ensures uniqueness since sets cannot contain duplicates.
- Return the size of the set: The number of unique signatures is the number of special-equivalent groups.
Why it works: Sorting even and odd characters produces a canonical form that is invariant under allowed swaps. Two strings with the same sorted even and odd characters can always be rearranged to match each other. This guarantees correctness and avoids unnecessary permutations.
Python Solution
from typing import List
class Solution:
def numSpecialEquivGroups(self, words: List[str]) -> int:
signatures = set()
for word in words:
even_chars = sorted(word[i] for i in range(0, len(word), 2))
odd_chars = sorted(word[i] for i in range(1, len(word), 2))
signature = (tuple(even_chars), tuple(odd_chars))
signatures.add(signature)
return len(signatures)
The Python implementation separates the even and odd characters using slicing, sorts them to create a canonical signature, and uses a set to track unique groups. The use of tuple ensures hashability for storing in a set.
Go Solution
import (
"sort"
"strings"
)
func numSpecialEquivGroups(words []string) int {
signatures := make(map[string]struct{})
for _, word := range words {
var evenChars, oddChars []string
for i, c := range word {
if i%2 == 0 {
evenChars = append(evenChars, string(c))
} else {
oddChars = append(oddChars, string(c))
}
}
sort.Strings(evenChars)
sort.Strings(oddChars)
signature := strings.Join(evenChars, "") + "|" + strings.Join(oddChars, "")
signatures[signature] = struct{}{}
}
return len(signatures)
}
In Go, slices are used for even and odd characters, sorted using sort.Strings, and joined into a single string with a delimiter. A map with empty structs efficiently tracks unique signatures.
Worked Examples
Example 1:
Input: ["abcd","cdab","cbad","xyzz","zzxy","zzyx"]
| Word | Even Chars | Odd Chars | Signature | Added to Set |
|---|---|---|---|---|
| "abcd" | a, c | b, d | ("ac","bd") | yes |
| "cdab" | c, a | d, b | ("ac","bd") | no |
| "cbad" | c, a | b, d | ("ac","bd") | no |
| "xyzz" | x, z | y, z | ("xz","yz") | yes |
| "zzxy" | z, y | z, x | ("yz","xz") | yes |
| "zzyx" | z, y | z, x | ("yz","xz") | no |
Number of unique signatures: 3
Example 2:
Input: ["abc","acb","bac","bca","cab","cba"]
| Word | Even Chars | Odd Chars | Signature |
|---|---|---|---|
| "abc" | a, c | b | ("ac","b") |
| "acb" | a, b | c | ("ab","c") |
| "bac" | b, c | a | ("bc","a") |
| "bca" | b, a | c | ("ab","c") |
| "cab" | c, b | a | ("bc","a") |
| "cba" | c, a | b | ("ac","b") |
Unique signatures: 3
Complexity Analysis
| Measure | Complexity | Explanation |
|---|---|---|
| Time | O(n * m log m) | Sorting the even and odd characters for each string of length m, for n strings |
| Space | O(n * m) | Storing unique signatures for each string; each signature is O(m) |
Sorting dominates the runtime, but it is acceptable given the constraints of m ≤ 20 and n ≤ 1000.
Test Cases
# Basic examples
assert Solution().numSpecialEquivGroups(["abcd","cdab","cbad","xyzz","zzxy","zzyx"]) == 3 # Example 1
assert Solution().numSpecialEquivGroups(["abc","acb","bac","bca","cab","cba"]) == 3 # Example 2
# Edge cases
assert Solution().numSpecialEquivGroups(["a"]) == 1 # Single word
assert Solution().numSpecialEquivGroups(["aa","aa","aa"]) == 1 # All identical
assert Solution().numSpecialEquivGroups(["ab","ba"]) == 1 # Swap makes them equal
assert Solution().numSpecialEquivGroups(["ab","cd","ef"]) == 3 # No matches
assert Solution().numSpecialEquivGroups(["abc","def","ghi","adg","beh","cfi"]) == 6 # Distinct groups
# Max size test
words = ["".join(chr((i+j)%26 + 97) for j in range(20)) for i in range(1000)]
assert Solution().numSpecialEquivGroups(words) == 1000 # All unique
| Test | Why |
|---|---|
| Example 1 | Standard case with multiple groups |
| Example 2 | Standard case with multiple groups |
| ["a"] | Single string edge case |
| ["aa","aa","aa"] | Identical strings edge case |
| ["ab","ba"] | Minimal swap equivalence |
| ["ab","cd","ef"] | Completely distinct strings |
| Large 1000x20 | Stress test for constraints |
Edge Cases
Single-character strings: A string of length 1 has no odd or even index swaps. All single-character strings are only equivalent to identical strings. The solution handles this naturally because sorting a single character returns the same character.
Identical strings: Multiple identical strings should be counted as one group. The set data structure ensures duplicates are automatically merged into one group.
No special equivalences: If no two strings share the same even and odd character multisets, each string forms its own group