Solution: Vowel Spellchecker

Leetcode Solutions (161 Part Series)

1 Solution: Next Permutation
2 Solution: Trim a Binary Search Tree
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3 Leetcode Solutions Index
4 Solution: Minimize Deviation in Array
5 Solution: Vertical Order Traversal of a Binary Tree
6 Solution: Count Ways to Make Array With Product
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10 Solution: Concatenation of Consecutive Binary Numbers
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13 Solution: Simplify Path
14 Solution: Building Boxes
15 Solution: Decode XORed Permutation
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22 Solution: Copy List with Random Pointer
23 Solution: Valid Anagram
24 Solution: Number of Steps to Reduce a Number to Zero
25 Solution: Shortest Path in Binary Matrix
26 Solution: Is Graph Bipartite?
27 Solution: Maximum Score From Removing Substrings (ver. 1)
28 Solution: Maximum Score From Removing Substrings (ver. 2)
29 Solution: Sort the Matrix Diagonally
30 Solution: The K Weakest Rows in a Matrix (ver. 1)
31 Solution: The K Weakest Rows in a Matrix (ver. 2)
32 Solution: Letter Case Permutation
33 Solution: Container With Most Water
34 Solution: Arithmetic Slices
35 Solution: Minimum Remove to Make Valid Parentheses
36 Solution: Roman to Integer
37 Solution: Broken Calculator
38 Solution: Find the Most Competitive Subsequence
39 Solution: Longest Word in Dictionary through Deleting
40 Solution: Search a 2D Matrix II
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43 Solution: Validate Stack Sequences
44 Solution: Divide Two Integers (ver. 1)
45 Solution: Divide Two Integers (ver. 2)
46 Solution: Maximum Frequency Stack
47 Solution: Distribute Candies
48 Solution: Set Mismatch (ver. 1)
49 Solution: Set Mismatch (ver. 2)
50 Solution: Missing Number
51 Solution: Intersection of Two Linked Lists
52 Solution: Average of Levels in Binary Tree
53 Solution: Short Encoding of Words (ver. 1)
54 Solution: Design HashMap (ver. 1)
55 Solution: Short Encoding of Words (ver. 2)
56 Solution: Design HashMap (ver. 2)
57 Solution: Remove Palindromic Subsequences
58 Solution: Add One Row to Tree
59 Solution: Integer to Roman
60 Solution: Coin Change
61 Solution: Check If a String Contains All Binary Codes of Size K
62 Solution: Binary Trees With Factors
63 Solution: Swapping Nodes in a Linked List
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66 Solution: Generate Random Point in a Circle
67 Solution: Wiggle Subsequence
68 Solution: Keys and Rooms
69 Solution: Design Underground System
70 Solution: Reordered Power of 2
71 Solution: Vowel Spellchecker
72 Solution: 3Sum With Multiplicity
73 Solution: Advantage Shuffle
74 Solution: Pacific Atlantic Water Flow
75 Solution: Word Subsets
76 Solution: Palindromic Substrings
77 Solution: Reconstruct Original Digits from English
78 Solution: Flip Binary Tree To Match Preorder Traversal
79 Solution: Russian Doll Envelopes
80 Solution: Stamping The Sequence
81 Solution: Palindrome Linked List
82 Solution: Ones and Zeroes
83 Solution: Longest Valid Parentheses
84 Solution: Design Circular Queue
85 Solution: Global and Local Inversions
86 Solution: Minimum Operations to Make Array Equal
87 Solution: Determine if String Halves Are Alike
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90 Solution: Longest Increasing Path in a Matrix
91 Solution: Deepest Leaves Sum
92 Solution: Beautiful Arrangement II
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95 Solution: Fibonacci Number
96 Solution: Remove All Adjacent Duplicates in String II
97 Solution: Number of Submatrices That Sum to Target
98 Solution: Remove Nth Node From End of List
99 Solution: Combination Sum IV
100 Solution: N-ary Tree Preorder Traversal
101 Solution: Triangle
102 Solution: Brick Wall
103 Solution: Count Binary Substrings
104 Solution: Critical Connections in a Network
105 Solution: Rotate Image
106 Solution: Furthest Building You Can Reach
107 Solution: Power of Three
108 Solution: Unique Paths II
109 Solution: Find First and Last Position of Element in Sorted Array
110 Solution: Powerful Integers
111 Solution: Prefix and Suffix Search
112 Solution: Course Schedule III
113 Solution: Running Sum of 1d Array
114 Solution: Non-decreasing Array
115 Solution: Jump Game II
116 Solution: Convert Sorted List to Binary Search Tree
117 Solution: Delete Operation for Two Strings
118 Solution: Super Palindromes
119 Solution: Construct Target Array With Multiple Sums
120 Solution: Count Primes
121 Solution: Maximum Points You Can Obtain from Cards
122 Solution: Range Sum Query 2D – Immutable
123 Solution: Ambiguous Coordinates
124 Solution: Flatten Binary Tree to Linked List
125 Solution: Valid Number
126 Solution: Binary Tree Cameras
127 Solution: Longest String Chain
128 Solution: Find Duplicate File in System
129 Solution: Minimum Moves to Equal Array Elements II
130 Solution: Binary Tree Level Order Traversal
131 Solution: Find and Replace Pattern
132 Solution: N-Queens
133 Solution: To Lower Case
134 Solution: Evaluate Reverse Polish Notation
135 Solution: Partitioning Into Minimum Number Of Deci-Binary Numbers
136 Solution: Maximum Product of Word Lengths
137 Solution: Maximum Erasure Value
138 Solution: N-Queens II
139 Solution: Maximum Gap
140 Solution: Search Suggestions System
141 Solution: Max Area of Island
142 Solution: Interleaving String
143 Solution: Maximum Area of a Piece of Cake After Horizontal and Vertical Cuts
144 Solution: Open the Lock
145 Solution: Maximum Performance of a Team
146 Solution: Longest Consecutive Sequence
147 Solution: Min Cost Climbing Stairs
148 Solution: Construct Binary Tree from Preorder and Inorder Traversal
149 Solution: Jump Game VI
150 Solution: My Calendar I
151 Solution: Stone Game VII
152 Solution: Minimum Number of Refueling Stops
153 Solution: Palindrome Pairs
154 Solution: Maximum Units on a Truck
155 Solution: Matchsticks to Square
156 Solution: Generate Parentheses
157 Solution: Number of Subarrays with Bounded Maximum
158 Solution: Swim in Rising Water
159 Solution: Pascal’s Triangle
160 Solution: Out of Boundary Paths
161 Solution: Redundant Connection

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Leetcode Problem #966 (Medium): Vowel Spellchecker

Description:

(Jump to: Solution Idea || Code: JavaScript | Python | Java | C++)

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 capitlization, 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].

Examples:

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 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”]

Constraints:

1 <= wordlist.length <= 5000

1 <= queries.length <= 5000

1 <= wordlist[i].length <= 7

1 <= queries[i].length <= 7

All strings in wordlist and queries consist only of english letters.

Idea:

(Jump to: Problem Description || Code: JavaScript | Python | Java | C++)

This problem can be broken up into a couple steps of increasing difficulty. The first step is to check whether or not the words in the query list (Q) exists in the word list (W). For that, we can use the simplest form of value-lookup data structure, which is a Set.

Next, we need to check if each query has a case-insensitive match in W. For case-insensitive matching, the easiest thing to do is to lowercase (or uppercase) both terms before comparing. In this case, since we want to match one term, but return another, we should use a Map data structure, where the key is the lowercased term and the value is the matching word.

But here we encounter an issue, as it is possible for two words to have the same lowercase form. Per the rules we want to favor the one that appears first in W, so we can either iterate through W forwards and repeatedly check to make sure we’re not overwriting an existing entry, or we can simply iterate through W backwards and just automatically overwrite entries. This will force the first occurance to be the one that “sticks”.

For the third check, we need to match the word except for the vowels. Whenever you need to selectively match strings by only a portion, the easiest way to do it is with a mask. In this case, we can use regex to replace all vowel occurrances with a character mask, such as “#”. For example, we can check if “tail” and “tool” would match by applying the character masks to both terms and seeing that “t##l” == “t##l”.

This calls for another map structure. We could technically reuse the earlier one, as there will be no overlaps, but navigating two separate, smaller maps is generally more efficient than one large one. Since we’ll also want to iterate backwards through W for this map, we migtht as well do it at the same time as the other one.

Then we can just iterate through Q and check for matches in the correct order. As is generally the case with query lists, we can replace the queries in Q with their result in order to save on space complexity.

Then, when we’re done, we just return Q.

Implementation:

Javascript can use logical OR chaining to shorten the assignment of the proper result in Q.

Regex is much slower in Java and C++, so we can use a helper function to do the same thing for us.

C++ will also need a helper to lowercase the words.

Javascript Code:

(Jump to: Problem Description || Solution Idea)

const regex = /[aeiou]/g
var spellchecker = function(W, Q) {
    let orig = new Set(W), lower = new Map(), mask = new Map()
    for (let i = W.length - 1; ~i; i--) {
        let word = W[i], wlow = word.toLowerCase()
        lower.set(wlow, word)
        mask.set(wlow.replace(regex, "*"), word)
    }
    for (let i in Q) {
        let query = Q[i], qlow = query.toLowerCase(),
            qmask = qlow.replace(regex, "*")
        if (orig.has(query)) continue
        else Q[i] = lower.get(qlow) || mask.get(qmask) || ""
    }
    return Q
};

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Python Code:

(Jump to: Problem Description || Solution Idea)

class Solution:
    def spellchecker(self, W: List[str], Q: List[str]) -> List[str]:
        orig, lcase, mask = set(W), defaultdict(), defaultdict()
        regex = r'[aeiou]'
        for i in range(len(W)-1,-1,-1):
            word = W[i]
            wlow = word.lower()
            lcase[wlow] = word
            mask[re.sub(regex, '*', wlow)] = word
        for i in range(len(Q)):
            query = Q[i]
            qlow = query.lower()
            qmask = re.sub(regex, '*', qlow)
            if query in orig: continue
            elif qlow in lcase: Q[i] = lcase[qlow]
            elif qmask in mask: Q[i] = mask[qmask]
            else: Q[i] = ""
        return Q

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Java Code:

(Jump to: Problem Description || Solution Idea)

class Solution {
    public String[] spellchecker(String[] W, String[] Q) {
        Set<String> orig = new HashSet<>(Arrays.asList(W));
        Map<String, String> lower = new HashMap<>(), mask = new HashMap<>();
        for (int i = W.length - 1; i >= 0; i--) {
            String word = W[i], wlow = word.toLowerCase();
            lower.put(wlow, word);
            mask.put(vmask(wlow), word);
        }
        for (int i = 0; i < Q.length; i++) {
            String query = Q[i], qlow = query.toLowerCase(),
                qmask = vmask(qlow);
            if (orig.contains(query)) continue;
            else if (lower.containsKey(qlow)) Q[i] = lower.get(qlow);
            else if (mask.containsKey(qmask)) Q[i] = mask.get(qmask);
            else Q[i] = "";
        }
        return Q;
    }
    public String vmask(String str) {
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < str.length(); i++) {
            char c = str.charAt(i);
            if (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u') c = '*';
            sb.append(c);
        }
        return sb.toString();
    }
}

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C++ Code:

(Jump to: Problem Description || Solution Idea)

class Solution {
public:
    vector<string> spellchecker(vector<string>& W, vector<string>& Q) {
        set<string> orig (W.begin(), W.end());
        unordered_map<string, string> lower, mask;
        for (int i = W.size() - 1; ~i; i--) {
            string word = W[i], wlow = lcase(word);
            lower[wlow] = word, mask[vmask(wlow)] = word;
        }
        for (string &query : Q) {
            string qlow = lcase(query), qmask = vmask(qlow);
            if (orig.count(query)) continue;
            else if (lower.count(qlow)) query = lower[qlow];
            else if (mask.count(qmask)) query = mask[qmask];
            else query = "";
        }
        return Q;
    }
    static string vmask(string str) {
        for (char &c : str)
            if (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u')
                c = '*';
        return str;
    }
    static string lcase(string str) {
        for (char &c : str) c = tolower(c);
        return str;
    }
};

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Leetcode Solutions (161 Part Series)

原文链接：Solution: Vowel Spellchecker

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