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2451. Odd String Difference

Difficulty Topics

Description

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.

Solution

odd-string-difference.py
class Solution:
    def oddString(self, words: List[str]) -> str:
        diff = defaultdict(list)

        for word in words:
            d = []

            for i in range(1, len(word)):
                d.append(ord(word[i]) - ord(word[i - 1]))

            diff[tuple(d)].append(word)

        for k, v in diff.items():
            if len(v) == 1:
                return v[0]