README_EN.md

2451. Odd String Difference

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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:

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.

Solutions

Python3

class Solution:
    def oddString(self, words: List[str]) -> str:
        d = defaultdict(list)
        for s in words:
            t = tuple(ord(b) - ord(a) for a, b in pairwise(s))
            d[t].append(s)
        return next(ss[0] for ss in d.values() if len(ss) == 1)

Java

class Solution {
    public String oddString(String[] words) {
        var d = new HashMap<String, List<String>>();
        for (var s : words) {
            int m = s.length();
            var cs = new char[m - 1];
            for (int i = 0; i < m - 1; ++i) {
                cs[i] = (char) (s.charAt(i + 1) - s.charAt(i));
            }
            var t = String.valueOf(cs);
            d.putIfAbsent(t, new ArrayList<>());
            d.get(t).add(s);
        }
        for (var ss : d.values()) {
            if (ss.size() == 1) {
                return ss.get(0);
            }
        }
        return "";
    }
}

C++

class Solution {
public:
    string oddString(vector<string>& words) {
        unordered_map<string, vector<string>> d;
        for (auto& s : words) {
            int m = s.size();
            string t(m - 1, 0);
            for (int i = 0; i < m - 1; ++i) {
                t[i] = s[i + 1] - s[i];
            }
            d[t].push_back(s);
        }
        for (auto& [_, ss] : d) {
            if (ss.size() == 1) {
                return ss[0];
            }
        }
        return "";
    }
};

Go

func oddString(words []string) string {
	d := map[string][]string{}
	for _, s := range words {
		m := len(s)
		cs := make([]byte, m-1)
		for i := 0; i < m-1; i++ {
			cs[i] = s[i+1] - s[i]
		}
		t := string(cs)
		d[t] = append(d[t], s)
	}
	for _, ss := range d {
		if len(ss) == 1 {
			return ss[0]
		}
	}
	return ""
}

TypeScript

function oddString(words: string[]): string {
    const d: Map<string, string[]> = new Map();
    for (const s of words) {
        const cs: number[] = [];
        for (let i = 0; i < s.length - 1; ++i) {
            cs.push(s[i + 1].charCodeAt(0) - s[i].charCodeAt(0));
        }
        const t = cs.join(',');
        if (!d.has(t)) {
            d.set(t, []);
        }
        d.get(t)!.push(s);
    }
    for (const [_, ss] of d) {
        if (ss.length === 1) {
            return ss[0];
        }
    }
    return '';
}

Rust

use std::collections::HashMap;
impl Solution {
    pub fn odd_string(words: Vec<String>) -> String {
        let n = words[0].len();
        let mut map: HashMap<String, (bool, usize)> = HashMap::new();
        for (i, word) in words.iter().enumerate() {
            let mut k = String::new();
            for j in 1..n {
                k.push_str(&(word.as_bytes()[j] - word.as_bytes()[j - 1]).to_string());
                k.push(',');
            }
            let new_is_only = !map.contains_key(&k);
            map.insert(k, (new_is_only, i));
        }
        for (is_only, i) in map.values() {
            if *is_only {
                return words[*i].clone();
            }
        }
        String::new()
    }
}

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