#893

Medium

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Groups of Special-Equivalent Strings

Determine the number of special-equivalent string groups using character swaps at even or odd indices efficiently.

Problem Statement

Given an array of strings where each string has the same length, you can perform a move by swapping any two characters at even indices or any two characters at odd indices within a string. Determine how many groups of special-equivalent strings exist.

Two strings are special-equivalent if you can make them identical by performing any number of such moves. Your task is to compute the total count of unique groups formed under this rule for the given array.

Examples

Example 1

Input: words = ["abcd","cdab","cbad","xyzz","zzxy","zzyx"]

Output: 3

One group is ["abcd", "cdab", "cbad"], since they are all pairwise special equivalent, and none of the other strings is all pairwise special equivalent to these. The other two groups are ["xyzz", "zzxy"] and ["zzyx"]. Note that in particular, "zzxy" is not special equivalent to "zzyx".

Example 2

Input: words = ["abc","acb","bac","bca","cab","cba"]

Output: 3

Example details omitted.

Constraints

1 <= words.length <= 1000
1 <= words[i].length <= 20
words[i] consist of lowercase English letters.
All the strings are of the same length.

Solution Approach

Normalize Strings Using Even-Odd Character Sorting

For each string, separate characters at even and odd indices, sort each group independently, and combine them to form a normalized key. This ensures that all special-equivalent strings map to the same key.

Use a Hash Set to Track Unique Groups

Store each normalized key in a hash set while scanning the array. The number of unique keys in the set after processing all strings equals the number of special-equivalent groups.

Iterate Efficiently and Avoid Redundant Comparisons

Scan each string once and avoid pairwise comparisons. Sorting even and odd characters separately reduces complexity and guarantees correct grouping regardless of string arrangement.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity is O(N * L log L), where N is the number of strings and L is the string length due to sorting characters at even and odd indices. Space complexity is O(N * L) to store normalized keys in the hash set.

What Interviewers Usually Probe

Focus on array scanning rather than naive pairwise string comparison.
Look for efficient grouping using hash-based normalization.
Be prepared to explain why sorting even and odd characters separately guarantees correctness.

Common Pitfalls or Variants

Common pitfalls

Attempting pairwise comparisons between all strings leading to O(N^2) time.
Failing to separate even and odd indices when generating keys.
Assuming identical sorted full strings is sufficient for equivalence.

Follow-up variants

Consider strings of varying lengths and how normalization would change.
Compute group sizes instead of just the number of groups.
Count groups under additional constraints, such as limiting moves.

FAQ

What defines a special-equivalent string in this problem?

Two strings are special-equivalent if you can swap any even-indexed characters or any odd-indexed characters to make them identical.

Why do we sort even and odd indices separately?

Sorting ensures all strings that can transform into each other through allowed swaps produce the same normalized key for hashing.

Can this approach handle arrays with 1000 strings of length 20?

Yes, because it scans each string once and uses hash sets, keeping operations within acceptable time and space limits.

Is pairwise comparison necessary to determine groups?

No, using normalized keys with hash sets eliminates the need for O(N^2) pairwise checks.

How does this solution tie to the 'Array scanning plus hash lookup' pattern?

It scans each array element once to compute a key and uses a hash lookup to track unique groups efficiently, matching the pattern.

terminal

Solution

Solution 1

#### Python3

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class Solution:
    def numSpecialEquivGroups(self, words: List[str]) -> int:
        s = {''.join(sorted(word[::2]) + sorted(word[1::2])) for word in words}
        return len(s)

Continue Practicing

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