LeetCode #966 — MEDIUM

Vowel Spellchecker

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

Solve on LeetCode

The Problem

Problem Statement

Given a wordlist, we want to implement a spellchecker that converts a query word into a correct word.

For a given query word, the spell checker handles two categories of spelling mistakes:

Capitalization: If the query matches a word in the wordlist (case-insensitive), then the query word is returned with the same case as the case in the wordlist.
- Example: wordlist = ["yellow"], query = "YellOw": correct = "yellow"
- Example: wordlist = ["Yellow"], query = "yellow": correct = "Yellow"
- Example: wordlist = ["yellow"], query = "yellow": correct = "yellow"
Vowel Errors: If after replacing the vowels ('a', 'e', 'i', 'o', 'u') of the query word with any vowel individually, it matches a word in the wordlist (case-insensitive), then the query word is returned with the same case as the match in the wordlist.
- Example: wordlist = ["YellOw"], query = "yollow": correct = "YellOw"
- Example: wordlist = ["YellOw"], query = "yeellow": correct = "" (no match)
- Example: wordlist = ["YellOw"], query = "yllw": correct = "" (no match)

In addition, the spell checker operates under the following precedence rules:

When the query exactly matches a word in the wordlist (case-sensitive), you should return the same word back.
When the query matches a word up to capitalization, you should return the first such match in the wordlist.
When the query matches a word up to vowel errors, you should return the first such match in the wordlist.
If the query has no matches in the wordlist, you should return the empty string.

Given some queries, return a list of words answer, where answer[i] is the correct word for query = queries[i].

Example 1:

Input: wordlist = ["KiTe","kite","hare","Hare"], queries = ["kite","Kite","KiTe","Hare","HARE","Hear","hear","keti","keet","keto"]
Output: ["kite","KiTe","KiTe","Hare","hare","","","KiTe","","KiTe"]

Example 2:

Input: wordlist = ["yellow"], queries = ["YellOw"]
Output: ["yellow"]

Constraints:

1 <= wordlist.length, queries.length <= 5000
1 <= wordlist[i].length, queries[i].length <= 7
wordlist[i] and queries[i] consist only of only English letters.

Step 01

Brute Force Baseline

Problem summary: Given a wordlist, we want to implement a spellchecker that converts a query word into a correct word. For a given query word, the spell checker handles two categories of spelling mistakes: Capitalization: If the query matches a word in the wordlist (case-insensitive), then the query word is returned with the same case as the case in the wordlist.

Baseline thinking

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

Pattern signal: Array · Hash Map

Example 1

["KiTe","kite","hare","Hare"]
["kite","Kite","KiTe","Hare","HARE","Hear","hear","keti","keet","keto"]

Example 2

["yellow"]
["YellOw"]

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

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 #966: Vowel Spellchecker
class Solution {
    public String[] spellchecker(String[] wordlist, String[] queries) {
        Set<String> s = new HashSet<>();
        Map<String, String> low = new HashMap<>();
        Map<String, String> pat = new HashMap<>();
        for (String w : wordlist) {
            s.add(w);
            String t = w.toLowerCase();
            low.putIfAbsent(t, w);
            pat.putIfAbsent(f(t), w);
        }
        int m = queries.length;
        String[] ans = new String[m];
        for (int i = 0; i < m; ++i) {
            String q = queries[i];
            if (s.contains(q)) {
                ans[i] = q;
                continue;
            }
            q = q.toLowerCase();
            if (low.containsKey(q)) {
                ans[i] = low.get(q);
                continue;
            }
            q = f(q);
            if (pat.containsKey(q)) {
                ans[i] = pat.get(q);
                continue;
            }
            ans[i] = "";
        }
        return ans;
    }

    private String f(String w) {
        char[] cs = w.toCharArray();
        for (int i = 0; i < cs.length; ++i) {
            char c = cs[i];
            if (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u') {
                cs[i] = '*';
            }
        }
        return String.valueOf(cs);
    }
}

// Accepted solution for LeetCode #966: Vowel Spellchecker
func spellchecker(wordlist []string, queries []string) (ans []string) {
	s := map[string]bool{}
	low := map[string]string{}
	pat := map[string]string{}
	f := func(w string) string {
		res := []byte(w)
		for i := range res {
			if res[i] == 'a' || res[i] == 'e' || res[i] == 'i' || res[i] == 'o' || res[i] == 'u' {
				res[i] = '*'
			}
		}
		return string(res)
	}
	for _, w := range wordlist {
		s[w] = true
		t := strings.ToLower(w)
		if _, ok := low[t]; !ok {
			low[t] = w
		}
		if _, ok := pat[f(t)]; !ok {
			pat[f(t)] = w
		}
	}
	for _, q := range queries {
		if s[q] {
			ans = append(ans, q)
			continue
		}
		q = strings.ToLower(q)
		if s, ok := low[q]; ok {
			ans = append(ans, s)
			continue
		}
		q = f(q)
		if s, ok := pat[q]; ok {
			ans = append(ans, s)
			continue
		}
		ans = append(ans, "")
	}
	return
}

# Accepted solution for LeetCode #966: Vowel Spellchecker
class Solution:
    def spellchecker(self, wordlist: List[str], queries: List[str]) -> List[str]:
        def f(w):
            t = []
            for c in w:
                t.append("*" if c in "aeiou" else c)
            return "".join(t)

        s = set(wordlist)
        low, pat = {}, {}
        for w in wordlist:
            t = w.lower()
            low.setdefault(t, w)
            pat.setdefault(f(t), w)

        ans = []
        for q in queries:
            if q in s:
                ans.append(q)
                continue
            q = q.lower()
            if q in low:
                ans.append(low[q])
                continue
            q = f(q)
            if q in pat:
                ans.append(pat[q])
                continue
            ans.append("")
        return ans

// Accepted solution for LeetCode #966: Vowel Spellchecker
use std::collections::{HashMap, HashSet};

impl Solution {
    pub fn spellchecker(wordlist: Vec<String>, queries: Vec<String>) -> Vec<String> {
        let s: HashSet<String> = wordlist.iter().cloned().collect();
        let mut low: HashMap<String, String> = HashMap::new();
        let mut pat: HashMap<String, String> = HashMap::new();

        let f = |w: &str| -> String {
            w.chars()
                .map(|c| match c {
                    'a' | 'e' | 'i' | 'o' | 'u' => '*',
                    _ => c,
                })
                .collect()
        };

        for w in &wordlist {
            let mut t = w.to_lowercase();
            if !low.contains_key(&t) {
                low.insert(t.clone(), w.clone());
            }
            t = f(&t);
            if !pat.contains_key(&t) {
                pat.insert(t.clone(), w.clone());
            }
        }

        let mut ans: Vec<String> = Vec::new();
        for query in queries {
            if s.contains(&query) {
                ans.push(query);
                continue;
            }
            let mut q = query.to_lowercase();
            if let Some(v) = low.get(&q) {
                ans.push(v.clone());
                continue;
            }
            q = f(&q);
            if let Some(v) = pat.get(&q) {
                ans.push(v.clone());
                continue;
            }
            ans.push("".to_string());
        }
        ans
    }
}

// Accepted solution for LeetCode #966: Vowel Spellchecker
function spellchecker(wordlist: string[], queries: string[]): string[] {
    const s = new Set(wordlist);
    const low = new Map<string, string>();
    const pat = new Map<string, string>();

    const f = (w: string): string => {
        let res = '';
        for (const c of w) {
            if ('aeiou'.includes(c)) {
                res += '*';
            } else {
                res += c;
            }
        }
        return res;
    };

    for (const w of wordlist) {
        let t = w.toLowerCase();
        if (!low.has(t)) {
            low.set(t, w);
        }
        t = f(t);
        if (!pat.has(t)) {
            pat.set(t, w);
        }
    }

    const ans: string[] = [];
    for (let q of queries) {
        if (s.has(q)) {
            ans.push(q);
            continue;
        }
        q = q.toLowerCase();
        if (low.has(q)) {
            ans.push(low.get(q)!);
            continue;
        }
        q = f(q);
        if (pat.has(q)) {
            ans.push(pat.get(q)!);
            continue;
        }
        ans.push('');
    }
    return ans;
}

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(n + m)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.