LeetCode #2452 — MEDIUM

Words Within Two Edits of Dictionary

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Trie

Example 1

["word","note","ants","wood"]
["wood","joke","moat"]

Example 2

["yes"]
["not"]

Core Insight

What unlocks the optimal approach

Try brute-forcing the problem.
For each word in queries, try comparing to each word in dictionary.
If there is a maximum of two edit differences, the word should be present in answer.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2452: Words Within Two Edits of Dictionary
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;
    }
}

// Accepted solution for LeetCode #2452: Words Within Two Edits of Dictionary
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
}

# Accepted solution for LeetCode #2452: Words Within Two Edits of Dictionary
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

// Accepted solution for LeetCode #2452: Words Within Two Edits of Dictionary
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()
    }
}

// Accepted solution for LeetCode #2452: Words Within Two Edits of Dictionary
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;
    });
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(L)

Space

O(N × L)

Approach Breakdown

HASH SET

O(N × L) time

O(N × L) space

Store all N words in a hash set. Each insert/lookup hashes the entire word of length L, giving O(L) per operation. Prefix queries require checking every stored word against the prefix — O(N × L) per prefix search. Space is O(N × L) for storing all characters.

TRIE

O(L) time

O(N × L) space

Each operation (insert, search, prefix) takes O(L) time where L is the word length — one node visited per character. Total space is bounded by the sum of all stored word lengths. Tries win over hash sets when you need prefix matching: O(L) prefix search vs. checking every stored word.

Shortcut: One node per character → O(L) per operation. Prefix queries are what make tries worthwhile.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.