To solve Groups of Special-Equivalent Strings, immediately map each string to a normalized key representing its even and odd characters sorted separately. Use a hash set to store unique keys and return its size. This method directly leverages array scanning plus hash lookup, ensuring correct grouping without redundant comparisons and handles strings of varying arrangements.
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.
How GhostInterview Helps
- Automatically generates normalized keys for even-odd index sorting.
- Tracks unique groups efficiently using hash sets without manual bookkeeping.
- Highlights potential pitfalls like pairwise comparisons or misgrouping.
Topic Pages
Related GhostInterview Pages
- LeetCode Interview Copilot - Use GhostInterview as a live solver when you want direct help with LeetCode-style coding questions.
- Coding Interview Assistant - See how GhostInterview supports array, string, linked list, graph, and tree interview workflows.
- How GhostInterview Works - Review the screenshot, reasoning, and answer flow before using the solver in a live interview.
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.
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Stay in the same pattern family
This problem challenges you to perform multiple string replacements at specific indices using a pattern involving array scanning and hash lookup.
Open problem page#792 Number of Matching SubsequencesGiven a string s and a list of words, count how many words are subsequences of s using efficient array scanning and hash lookup.
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Open problem page