2451. Odd String Difference

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2451. Odd String Difference

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:

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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:

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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.

Hints/Notes

  • 2025/03/13 Q3
  • string
  • No solution from 0x3F or Leetcode

Solution

Language: C++

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class Solution {
public:
    string oddString(vector<string>& words) {
        vector<int> diff;
        for (int i = 1; i < words[0].size(); i++) {
            diff.push_back(words[0][i] - words[0][i - 1]);
        }
        int notMatchFreq = 0;
        string res;
        for (int i = 1; i < words.size(); i++) {
            string& word = words[i];
            for (int j = 1; j < word.size(); j++) {
                if (word[j] != word[j - 1] + diff[j - 1]) {
                    res = word;
                    notMatchFreq++;
                    break;
                }
            }
        }
        return notMatchFreq == 1 ? res : words[0];
    }
};s