algoadvance

You are given an array words of strings. A move consists of choosing any two even indexed characters of a string and swapping them, or choosing any two odd indexed characters of a string and swapping them.

Two strings are special-equivalent if after any number of moves, they are equal. For example, “abcde” and “adcbe” are special-equivalent because we can swap the ‘b’ and ‘d’ which are both at odd indices to make the strings equal.

Return the number of groups of special-equivalent strings from the input array words.

Example:

Input: ["abcd","cdab","cbad","xyzz","zzxy","zzyx"]
Output: 3
Explanation: 
One group is ["abcd", "cdab", "cbad"], since they are all special-equivalent.
The other two groups are ["xyzz", "zzxy"] and ["zzyx"].

Constraints:

1. 1 <= words.length <= 1000
2. 1 <= words[i].length <= 20
3. All words[i] have the same length.
4. All words[i] consist of only lowercase English letters.

Clarifying Questions

1. Are all given words have the same length in the input array?
   • Yes, as specified in the constraints.
2. Is there any specific performance constraint or maximum length of strings we should consider?
   • Given the constraints, we can expect up to 1000 words each of up to length 20 which should be manageable within the given constraints.

Strategy

1. Character Classification:
   • Separate each string into characters at even and odd indices.
   • Sort the characters at even indices and odd indices separately.
2. Normalization:
   • Combine the sorted even indices and sorted odd indices into a tuple to create a normalized representation of each string.
3. Set Utilization:
   • Use a set to keep track of all unique normalized tuples. The size of this set at the end will be the number of unique groups.

Code

def numSpecialEquivGroups(words):
    def normalize(word):
        even_chars = sorted(word[0::2])
        odd_chars = sorted(word[1::2])
        return (tuple(even_chars), tuple(odd_chars))
    
    normalized_set = set(normalize(word) for word in words)
    return len(normalized_set)

# Example Usage
words = ["abcd", "cdab", "cbad", "xyzz", "zzxy", "zzyx"]
print(numSpecialEquivGroups(words))  # Output: 3

Time Complexity

Let’s analyze the time complexity:

• Normalization step:
  • Separating even and odd index characters takes O(n) time for each word where n is the length of the word.
  • Sorting the characters takes O(n log n).
• Overall for each word:
  • The overall time complexity for normalizing one word is O(n log n).
• Considering all words:
  • Processing all words is O(m * n log n) where m is the number of words and n is the length of each word.
• Storage:
  • The space complexity arises from storing up to m tuples in the set, each containing up to n characters, which is O(m * n).

Thus, the final time complexity is O(m * n log n) and the space complexity is O(m * n).

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