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Leetcode #893: Groups of Special-Equivalent Strings

In this guide, we solve Leetcode #893 Groups of Special-Equivalent Strings in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

Problem Statement

You are given an array of strings of the same length words. In one move, you can swap any two even indexed characters or any two odd indexed characters of a string words[i].

Quick Facts

  • Difficulty: Medium
  • Premium: No
  • Tags: Array, Hash Table, String, Sorting

Intuition

Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.

By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.

Approach

Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.

This keeps the solution linear while remaining easy to explain in an interview setting.

Steps:

  • Initialize a hash map for seen items or counts.
  • Iterate through the input, querying/updating the map.
  • Return the first valid result or the final computed value.

Example

Input: words = ["abcd","cdab","cbad","xyzz","zzxy","zzyx"] Output: 3 Explanation: 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".

Python Solution

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)

Complexity

The time complexity is O(n). The space complexity is O(n).

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.


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