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Leetcode Problem 2451: Odd String Difference

Let’s define the “difference” between two strings (s1, s2) of the same length as the number of positions at which the corresponding characters are different. For example, the difference between "abc" and "abx" is 1.

You are given an array of strings words where each string has the same length. Return the string that differs from the rest of the strings in the array by an odd number of positions. It is guaranteed that there is exactly one such string in the array.

Clarifying Questions

  1. Input Constraints: What are the lower and upper limits on the length of the words array and the length of each string within it?
    • Assumption: 2 <= len(words) <= 100 and 1 <= len(words[0]) <= 100.
  2. Character Types: What types of characters do the strings contain?
    • Assumption: The strings contain only lowercase English letters.
  3. Guaranteed Unique Odd Difference: Is it guaranteed that there is exactly one string with an odd difference from all the rest?
    • Clarification: Yes, as specified in the problem statement.

Strategy

  1. Difference Calculation: A helper function to calculate the difference between two strings as the number of positions where their characters differ.
  2. Even vs. Odd Difference: For each string in words, calculate the difference from the first string. Determine if the differences are even or odd.
  3. Identification: Identify the single string with an odd difference from the majority.

Code

def find_odd_string(words):
    def difference(s1, s2):
        """ Calculate the number of differing positions between two strings of the same length """
        return sum(c1 != c2 for c1, c2 in zip(s1, s2))
    
    # Calculate the difference of each string from the first one
    differences = []
    for i in range(1, len(words)):
        diff = difference(words[0], words[i])
        differences.append((diff, i))
    
    # We assume that there is exactly one odd-indexed string
    odd_string_index = -1
    odd_counts = 0
    even_counts = 0
    
    for diff, idx in differences:
        if diff % 2 == 1:
            odd_counts += 1
            odd_string_index = idx
        else:
            even_counts += 1
    
    # Determine if the original string was the odd-one-out
    if odd_counts == 1:
        return words[odd_string_index]
    else:
        return words[0]

# Example Usage
words = ["abc", "abb", "acc"]
print(find_odd_string(words))  # Output: "abb" (as an example)

Time Complexity

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