README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/2400-2499/2452.Words%20Within%20Two%20Edits%20of%20Dictionary/README_EN.md	1459	Biweekly Contest 90 Q2	Array String

2452. Words Within Two Edits of Dictionary

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Description

You are given two string arrays, queries and dictionary. All words in each array comprise of lowercase English letters and have the same length.

In one edit you can take a word from queries, and change any letter in it to any other letter. Find all words from queries that, after a maximum of two edits, equal some word from dictionary.

Return a list of all words from queries, that match with some word from dictionary after a maximum of two edits. Return the words in the same order they appear in queries.

 

Example 1:

Input: queries = ["word","note","ants","wood"], dictionary = ["wood","joke","moat"]
Output: ["word","note","wood"]
Explanation:
- Changing the 'r' in "word" to 'o' allows it to equal the dictionary word "wood".
- Changing the 'n' to 'j' and the 't' to 'k' in "note" changes it to "joke".
- It would take more than 2 edits for "ants" to equal a dictionary word.
- "wood" can remain unchanged (0 edits) and match the corresponding dictionary word.
Thus, we return ["word","note","wood"].

Example 2:

Input: queries = ["yes"], dictionary = ["not"]
Output: []
Explanation:
Applying any two edits to "yes" cannot make it equal to "not". Thus, we return an empty array.

 

Constraints:

• 1 <= queries.length, dictionary.length <= 100
• n == queries[i].length == dictionary[j].length
• 1 <= n <= 100
• All queries[i] and dictionary[j] are composed of lowercase English letters.

Solutions

Solution 1: Brute Force Enumeration

We directly traverse each word $s$ in the array $\textit{queries}$, and then traverse each word $t$ in the array $\textit{dictionary}$. If there exists a word $t$ whose edit distance from $s$ is less than $3$, we add $s$ to the answer array and then exit the inner loop. If there is no such word $t$, we continue to traverse the next word $s$.

The time complexity is $O(m \times n \times l)$, where $m$ and $n$ are the lengths of the arrays $\textit{queries}$ and $\textit{dictionary}$ respectively, and $l$ is the length of the word.

Python3

class Solution:
    def twoEditWords(self, queries: List[str], dictionary: List[str]) -> List[str]:
        ans = []
        for s in queries:
            for t in dictionary:
                if sum(a != b for a, b in zip(s, t)) < 3:
                    ans.append(s)
                    break
        return ans

Java

class Solution {
    public List<String> twoEditWords(String[] queries, String[] dictionary) {
        List<String> ans = new ArrayList<>();
        int n = queries[0].length();
        for (var s : queries) {
            for (var t : dictionary) {
                int cnt = 0;
                for (int i = 0; i < n; ++i) {
                    if (s.charAt(i) != t.charAt(i)) {
                        ++cnt;
                    }
                }
                if (cnt < 3) {
                    ans.add(s);
                    break;
                }
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    vector<string> twoEditWords(vector<string>& queries, vector<string>& dictionary) {
        vector<string> ans;
        for (auto& s : queries) {
            for (auto& t : dictionary) {
                int cnt = 0;
                for (int i = 0; i < s.size(); ++i) {
                    cnt += s[i] != t[i];
                }
                if (cnt < 3) {
                    ans.emplace_back(s);
                    break;
                }
            }
        }
        return ans;
    }
};

Go

func twoEditWords(queries []string, dictionary []string) (ans []string) {
	for _, s := range queries {
		for _, t := range dictionary {
			cnt := 0
			for i := range s {
				if s[i] != t[i] {
					cnt++
				}
			}
			if cnt < 3 {
				ans = append(ans, s)
				break
			}
		}
	}
	return
}

TypeScript

function twoEditWords(queries: string[], dictionary: string[]): string[] {
    const n = queries[0].length;
    return queries.filter(s => {
        for (const t of dictionary) {
            let diff = 0;
            for (let i = 0; i < n; ++i) {
                if (s[i] !== t[i]) {
                    ++diff;
                }
            }
            if (diff < 3) {
                return true;
            }
        }
        return false;
    });
}

Rust

impl Solution {
    pub fn two_edit_words(queries: Vec<String>, dictionary: Vec<String>) -> Vec<String> {
        queries
            .into_iter()
            .filter(|s| {
                dictionary
                    .iter()
                    .any(|t| s.chars().zip(t.chars()).filter(|&(a, b)| a != b).count() < 3)
            })
            .collect()
    }
}

C#

public class Solution {
    public IList<string> TwoEditWords(string[] queries, string[] dictionary) {
        var ans = new List<string>();
        foreach (var s in queries) {
            foreach (var t in dictionary) {
                int cnt = 0;
                for (int i = 0; i < s.Length; i++) {
                    if (s[i] != t[i]) {
                        cnt++;
                    }
                }
                if (cnt < 3) {
                    ans.Add(s);
                    break;
                }
            }
        }
        return ans;
    }
}