2451

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.

Companies: Visa

Related Topics:
Array, Hash Table, String

Similar Questions:

• Minimum Rounds to Complete All Tasks (Medium)

Solution 1.

// OJ: https://leetcode.com/problems/odd-string-difference
// Author: github.com/lzl124631x
// Time: O(NW)
// Space: O(W)
class Solution {
public:
    string oddString(vector<string>& A) {
        unordered_map<string, vector<int>> m;
        for (int i = 0; i < A.size(); ++i) {
            auto s = A[i];
            int d = s[0] - 'a';
            for (char &c : s) c = 'a' + (c - 'a' - d + 26) % 26;
            if (m[s].size() < 2) m[s].push_back(i);
        }
        for (auto &[s, v] : m) {
            if (v.size() == 1) return A[v[0]];
        }
        return "";
    }
};