LeetCode-in-Java

2451. Odd String Difference

Easy

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

public class Solution {
    public String oddString(String[] w) {
        int n = w[0].length() - 1;
        int[] x = new int[n];
        int s = 1;
        int y = 0;
        int index = 1;
        for (int i = 0; i < n; i++) {
            x[i] = w[0].charAt(i + 1) - w[0].charAt(i);
        }
        for (int i = 1; y * s == 0 || s + y < 3; i++) {
            boolean b = true;
            for (int j = 0; j < n; j++) {
                if (x[j] != w[i].charAt(j + 1) - w[i].charAt(j)) {
                    b = false;
                    break;
                }
            }
            if (b) {
                s++;
            } else {
                y++;
                index = i;
            }
        }
        return s == 1 ? w[0] : w[index];
    }
}

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