Leetcode #966: Vowel Spellchecker

In this guide, we solve Leetcode #966 Vowel Spellchecker in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

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.

Quick Facts

Difficulty: Medium
Premium: No
Tags: Array, Hash Table, String

Intuition

Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.

By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.

Approach

Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.

This keeps the solution linear while remaining easy to explain in an interview setting.

Steps:

Initialize a hash map for seen items or counts.
Iterate through the input, querying/updating the map.
Return the first valid result or the final computed value.

Example

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"]

Python Solution

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

Complexity

The time complexity is $O(n + m)$ , and the space complexity is $O(n)$ , where $n$ and $m$ are the lengths of $\textit{wordlist}$ and $\textit{queries}$ , respectively. The space complexity is $O(n)$ , where $n$ and $m$ are the lengths of $\textit{wordlist}$ and $\textit{queries}$ , respectively.

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.

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.

Approach

Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.

This keeps the solution linear while remaining easy to explain in an interview setting.

Steps:

Initialize a hash map for seen items or counts.

Iterate through the input, querying/updating the map.

Return the first valid result or the final computed value.

Python Solution

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

Complexity

The time complexity is

O(n + m)

, and the space complexity is

O(n)

, where

n

and

m

are the lengths of

\textit{wordlist}

and

\textit{queries}

, respectively. The space complexity is

O(n)

, where

n

and

m

are the lengths of

\textit{wordlist}

and

\textit{queries}

, respectively.

Leetcode #966: Vowel Spellchecker

Problem Statement

Quick Facts

Intuition

Approach

Example

Python Solution

Complexity

Edge Cases and Pitfalls

Summary

Ace your next coding interview

Leetcode #966: Vowel Spellchecker

Problem Statement

Quick Facts

Intuition

Approach

Example

Python Solution

Complexity

Edge Cases and Pitfalls

Summary

Ace your next coding interview