2451. Odd String Difference

Problem

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 (Java)

class Solution {
    public String oddString(String[] words) {
        Map<List<Integer>, List<String>> map = new HashMap<>();

        for (int i = 0; i < words.length; i++) {
            char[] ch = words[i].toCharArray();
            List<Integer> list = new ArrayList<>();

            for (int j = 0; j < ch.length - 1; j++) {
                list.add(ch[j + 1] - ch[j]);
            }

            if (map.containsKey(list)) {
                map.get(list).add(words[i]);
            }
            else {
                List<String> l = new ArrayList<>();
                l.add(words[i]);
                map.put(list, l);
            }
        }

        for (List<Integer> key : map.keySet()) {
            if (map.get(key).size() == 1) {
                return map.get(key).get(0);
            }
        }

        return "";
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).