README_EN.md

2451. Odd String Difference

中文文档

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:
        cnt = defaultdict(list)
        for w in words:
            d = [str(ord(b) - ord(a)) for a, b in pairwise(w)]
            cnt[','.join(d)].append(w)
        return next(v[0] for v in cnt.values() if len(v) == 1)

Java

class Solution {
    public String oddString(String[] words) {
        Map<String, List<String>> cnt = new HashMap<>();
        for (var w : words) {
            List<String> d = new ArrayList<>();
            for (int i = 0; i < w.length() - 1; ++i) {
                d.add(String.valueOf(w.charAt(i + 1) - w.charAt(i)));
            }
            cnt.computeIfAbsent(String.join(",", d), k -> new ArrayList<>()).add(w);
        }
        for (var v : cnt.values()) {
            if (v.size() == 1) {
                return v.get(0);
            }
        }
        return "";
    }
}

C++

class Solution {
public:
    string oddString(vector<string>& words) {
        unordered_map<string, vector<string>> cnt;
        for (auto& w : words) {
            string d;
            for (int i = 0; i < w.size() - 1; ++i) {
                d += (char) (w[i + 1] - w[i]);
                d += ',';
            }
            cnt[d].emplace_back(w);
        }
        for (auto& [_, v] : cnt) {
            if (v.size() == 1) {
                return v[0];
            }
        }
        return "";
    }
};

Go

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

TypeScript

function oddString(words: string[]): string {
    const n = words[0].length;
    const map = new Map<string, [boolean, number]>();
    words.forEach((word, i) => {
        const diff: number[] = [];
        for (let j = 1; j < n; j++) {
            diff.push(word[j].charCodeAt(0) - word[j - 1].charCodeAt(0));
        }
        const k = diff.join();
        map.set(k, [!map.has(k), i]);
    });
    for (const [isOnly, i] of map.values()) {
        if (isOnly) {
            return words[i];
        }
    }
    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()
    }
}

...