README.md

2451. Odd String Difference

You are given an array of equal-length strings words. Assume that the length of each string is n.

Each string words[i] can be converted into a difference integer array difference[i] of length n - 1 where difference[i][j] = words[i][j+1] - words[i][j] where 0 <= j <= n - 2. Note that the difference between two letters is the difference between their positions in the alphabet i.e. the position of 'a' is 0, 'b' is 1, and 'z' is 25.

• For example, for the string "acb", the difference integer array is [2 - 0, 1 - 2] = [2, -1].

All the strings in words have the same difference integer array, except one. You should find that string.

Return the string in words that has different difference integer array.

Example 1:

Input: words = ["adc","wzy","abc"]
Output: "abc"
Explanation:
- The difference integer array of "adc" is [3 - 0, 2 - 3] = [3, -1].
- The difference integer array of "wzy" is [25 - 22, 24 - 25]= [3, -1].
- The difference integer array of "abc" is [1 - 0, 2 - 1] = [1, 1].
The odd array out is [1, 1], so we return the corresponding string, "abc".

Example 2:

Input: words = ["aaa","bob","ccc","ddd"]
Output: "bob"
Explanation: All the integer arrays are [0, 0] except for "bob", which corresponds to [13, -13].

Constraints:

• 3 <= words.length <= 100
• n == words[i].length
• 2 <= n <= 20
• words[i] consists of lowercase English letters.

Solutions (Python)

1. Solution

class Solution:
    def oddString(self, words: List[str]) -> str:
        n = len(words[0])
        difference = []

        for i in range(len(words)):
            difference.append([])
            for j in range(n - 1):
                difference[i].append(ord(words[i][j + 1]) - ord(words[i][j]))

            if i >= 2:
                prev = difference[i - 1] == difference[i - 2]
                curr = difference[i] == difference[i - 1]
                if prev and not curr:
                    return words[i]
                elif not prev and curr:
                    return words[i - 2]
                elif not prev and not curr:
                    return words[i - 1]