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Problem Statement

Given two strings s1 and s2, find the minimum number of single-character edits (insertions, deletions, or replacements) required to change s1 into any permutation of s2.

Clarifying Questions

1. Length of Strings: Are the lengths of s1 and s2 the same?
   • If yes, this simplifies the problem considerably.
   • If not, do we need to consider insertions and deletions?
2. Character Set: Are the strings composed of ASCII characters or Unicode?
3. Empty Strings: How should we handle empty string cases for s1 or s2?

For this problem, let’s assume:

• Both strings have the same length.
• The strings are composed of lowercase English letters.
• Neither string is empty.

Strategy

Since we are converting s1 into any permutation of s2, the order of characters in s2 doesn’t matter. We just need to ensure that s1 contains the same characters in the same frequencies as s2.

To achieve this, we can use the concept of frequency counts:

1. Count the frequency of each character in s1 and s2.
2. Calculate the difference in frequencies for each character.
3. The number of edits needed is the sum of absolute differences divided by 2 (since each edit can correct a discrepancy in both s1 and s2).

Steps

1. Calculate the frequency of each character in s1 and s2.
2. Compute the difference in these frequencies.
3. Sum the absolute differences and divide by 2 to get the minimum edits required.

Code

Here’s the implementation in Python:

from collections import Counter

def min_edits_to_permutation(s1: str, s2: str) -> int:
    if len(s1) != len(s2):
        raise ValueError("Both strings must have the same length.")
    
    # Count frequency of each character in both strings
    counter1 = Counter(s1)
    counter2 = Counter(s2)
    
    # Calculate the total differences
    total_difference = 0
    all_characters = set(counter1.keys()).union(set(counter2.keys()))
    
    for char in all_characters:
        total_difference += abs(counter1.get(char, 0) - counter2.get(char, 0))
    
    # Each edit can fix discrepancies in both strings, thus divide by 2
    return total_difference // 2

# Example usage
s1 = "abcde"
s2 = "ebcda"
print(min_edits_to_permutation(s1, s2))  # Output: 0, as both are already permutations

Time Complexity

• Counting Frequencies: O(n), where n is the length of the strings.
• Calculating Differences: O(1), since the set of characters (English lowercase letters) is fixed and independent of input size.
• Overall Complexity: O(n), which is efficient for this problem.

This approach ensures that we can efficiently calculate the number of edits needed to transform s1 into any permutation of s2.

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