LeetCode 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.

hashmap可以做list跟string的映射，之前都没有试过，第一次尝试，用起来还可以，如果这样的话，

可以判断重复的list，

Runtime3 ms

Beats

60.75%

Memory39.9 MB

Beats

93.18%