Groups of Special-Equivalent Strings in Python

In Python, finding groups of special-equivalent strings involves understanding that two strings are special-equivalent if we can transform one into the other by swapping characters at even positions with each other, and characters at odd positions with each other.

A special-equivalent group is a collection of strings where every pair in the group is special-equivalent, and the group cannot be made larger by adding another string from the array.

Understanding Special-Equivalent Strings

Two strings are special-equivalent if:

• Characters at even indices (0, 2, 4, ...) can be rearranged among themselves
• Characters at odd indices (1, 3, 5, ...) can be rearranged among themselves

For example, "zzxy" and "xyzz" are special-equivalent because we can swap positions 0?2 and 1?3.

Algorithm Approach

The key insight is that strings belong to the same group if they have the same sorted characters at even positions and the same sorted characters at odd positions. We create a unique code for each string by concatenating its sorted even-indexed and odd-indexed characters.

Implementation

def numSpecialEquivGroups(strings):
    codes = set()
    
    for word in strings:
        # Extract characters at even and odd positions
        even_chars = word[::2]   # Characters at indices 0, 2, 4, ...
        odd_chars = word[1::2]   # Characters at indices 1, 3, 5, ...
        
        # Create unique code by sorting both parts
        code = ''.join(sorted(even_chars)) + ''.join(sorted(odd_chars))
        codes.add(code)
    
    return len(codes)

# Test with example
strings = ["abcd", "cdab", "cbad", "xyzz", "zzxy", "zzyx"]
result = numSpecialEquivGroups(strings)
print(f"Number of special-equivalent groups: {result}")

Number of special-equivalent groups: 3

Step-by-Step Breakdown

Let's trace through how the algorithm works with our example:

def show_grouping_process(strings):
    codes = {}
    
    for word in strings:
        even_chars = word[::2]
        odd_chars = word[1::2]
        code = ''.join(sorted(even_chars)) + ''.join(sorted(odd_chars))
        
        if code not in codes:
            codes[code] = []
        codes[code].append(word)
        
        print(f"Word: {word}")
        print(f"  Even positions: {even_chars} ? sorted: {''.join(sorted(even_chars))}")
        print(f"  Odd positions: {odd_chars} ? sorted: {''.join(sorted(odd_chars))}")
        print(f"  Code: {code}")
        print()
    
    print("Groups formed:")
    for i, (code, group) in enumerate(codes.items(), 1):
        print(f"Group {i} (code: {code}): {group}")
    
    return len(codes)

strings = ["abcd", "cdab", "cbad", "xyzz", "zzxy", "zzyx"]
result = show_grouping_process(strings)

Word: abcd
  Even positions: ac ? sorted: ac
  Odd positions: bd ? sorted: bd
  Code: acbd

Word: cdab
  Even positions: ca ? sorted: ac
  Odd positions: db ? sorted: bd
  Code: acbd

Word: cbad
  Even positions: ca ? sorted: ac
  Odd positions: bd ? sorted: bd
  Code: acbd

Word: xyzz
  Even positions: xz ? sorted: xz
  Odd positions: yz ? sorted: yz
  Code: xzyz

Word: zzxy
  Even positions: zx ? sorted: xz
  Odd positions: zy ? sorted: yz
  Code: xzyz

Word: zzyx
  Even positions: zy ? sorted: yz
  Odd positions: zx ? sorted: xz
  Code: yzxz

Groups formed:
Group 1 (code: acbd): ['abcd', 'cdab', 'cbad']
Group 2 (code: xzyz): ['xyzz', 'zzxy']
Group 3 (code: yzxz): ['zzyx']

Optimized Solution

Here's a more concise version of the solution:

def numSpecialEquivGroups(strings):
    return len({
        ''.join(sorted(word[::2])) + ''.join(sorted(word[1::2]))
        for word in strings
    })

# Test the optimized solution
strings = ["abcd", "cdab", "cbad", "xyzz", "zzxy", "zzyx"]
print(numSpecialEquivGroups(strings))

3

Time and Space Complexity

• Time Complexity: O(N × M log M), where N is the number of strings and M is the average length of strings (due to sorting)
• Space Complexity: O(N × M) for storing the unique codes in the set

Conclusion

The solution efficiently groups special-equivalent strings by creating unique codes based on sorted even and odd positioned characters. This approach uses a set to automatically handle duplicate codes and count distinct groups.