Posts published in “Hashtable”

花花酱 LeetCode 748. Shortest Completing Word

By zxi on December 20, 2017

题目大意: 给你一个由字母和数字组成车牌。另外给你一些单词，让你找一个最短的单词能够覆盖住车牌中的字母（不考虑大小写）。如果有多个解，输出第一个解。

Problem:

Find the minimum length word from a given dictionary words, which has all the letters from the string licensePlate. Such a word is said to complete the given string licensePlate

Here, for letters we ignore case. For example, "P" on the licensePlate still matches "p" on the word.

It is guaranteed an answer exists. If there are multiple answers, return the one that occurs first in the array.

The license plate might have the same letter occurring multiple times. For example, given a licensePlate of "PP", the word "pair" does not complete the licensePlate, but the word "supper" does.

Example 1:

Input: licensePlate = "1s3 PSt", words = ["step", "steps", "stripe", "stepple"]

Output: "steps"

Explanation: The smallest length word that contains the letters "S", "P", "S", and "T".

Note that the answer is not "step", because the letter "s" must occur in the word twice.

Also note that we ignored case for the purposes of comparing whether a letter exists in the word.

Example 2:

Input: licensePlate = "1s3 456", words = ["looks", "pest", "stew", "show"]

Output: "pest"

Explanation: There are 3 smallest length words that contains the letters "s".

We return the one that occurred first.

Note:

licensePlate will be a string with length in range [1, 7].
licensePlate will contain digits, spaces, or letters (uppercase or lowercase).
words will have a length in the range [10, 1000].
Every words[i] will consist of lowercase letters, and have length in range [1, 15].

Idea:

Hashtable

Solution:

C++

Time complexity: 时间复杂度 O(N*26), N is number of words.

Space complexity: 空间复杂度 O(26) = O(1)

// Author: Huahua

// Runtime: 15 ms

class Solution {

public:

string shortestCompletingWord(string licensePlate, vector<string>& words) {

vector<int> l(26, 0);

for (const char ch : licensePlate)

if (isalpha(ch)) ++l[tolower(ch) - 'a'];

string ans;

int min_l = INT_MAX;

for (const string& word : words) {

if (word.length() >= min_l) continue;

if (!matches(l, word)) continue;

min_l = word.length();

ans = word;

}

return ans;

}

private:

bool matches(const vector<int>& l, const string& word) {

vector<int> c(26, 0);

for (const char ch : word)

++c[tolower(ch) - 'a'];

for (int i = 0; i < 26; ++i)

if (c[i] < l[i]) return false;

return true;

}

};

花花酱 LeetCode 745. Prefix and Suffix Search

By zxi on December 11, 2017

Link: https://leetcode.com/problems/prefix-and-suffix-search/description/

Problem:

Given many words, words[i] has weight i.

Design a class WordFilter that supports one function, WordFilter.f(String prefix, String suffix). It will return the word with given prefix and suffix with maximum weight. If no word exists, return -1.

Examples:

Input:

WordFilter(["apple"])

WordFilter.f("a", "e") // returns 0

WordFilter.f("b", "") // returns -1

Note:

words has length in range [1, 15000].
For each test case, up to words.length queries WordFilter.f may be made.
words[i] has length in range [1, 10].
prefix, suffix have lengths in range [0, 10].
words[i] and prefix, suffix queries consist of lowercase letters only.

Idea:

Construct all possible filters

Solution1:

C++

Time complexity: O(NL^3 + QL) where N is the number of words, L is the max length of the word, Q is the number of queries.

Space complexity: O(NL^3)

// Author: Huahua

// Runtime: 482 ms

class WordFilter {

public:

WordFilter(const vector<string>& words) {

int index = 0;

for (const string& word : words) {

int n = word.length();

string prefix;

for (int i = 0; i <= n; ++i) {

if (i > 0) prefix += word[i - 1];

string suffix;

for (int j = n; j >= 0; --j) {

if (j != n) suffix = word[j] + suffix;

const string key = prefix + "_" + suffix;

filters_[key] = index;

}

++index;

}

int f(string prefix, string suffix) {

const string key = prefix + "_" + suffix;

auto it = filters_.find(key);

if (it != filters_.end()) return it->second;

return -1;

}

private:

unordered_map<string, int> filters_;

};

Version #2

// Author: Huahua

// Runtime: 499 ms

class WordFilter {

public:

WordFilter(const vector<string>& words) {

int index = 0;

for (const string& word : words) {

int n = word.length();

vector<string> prefixes(n + 1, "");

vector<string> suffixes(n + 1, "");

for (int i = 0; i < n; ++i) {

prefixes[i + 1] = prefixes[i] + word[i];

suffixes[i + 1] = word[n - i - 1] + suffixes[i];

}

for (const string& prefix : prefixes)

for (const string& suffix : suffixes)

filters_[prefix + "_" + suffix] = index;

++index;

}

int f(string prefix, string suffix) {

const string key = prefix + "_" + suffix;

auto it = filters_.find(key);

if (it != filters_.end()) return it->second;

return -1;

}

private:

unordered_map<string, int> filters_;

};

Solution 2:

C++ / Trie

Time complexity: O(NL^2 + QL) where N is the number of words, L is the max length of the word, Q is the number of queries.

Space complexity: O(NL^2)

// Author: Huahua

// Runtime: 572 ms

class Trie {

public:

/** Initialize your data structure here. */

Trie(): root_(new TrieNode()) {}

/** Inserts a word into the trie. */

void insert(const string& word, int index) {

TrieNode* p = root_.get();

for (const char c : word) {

if (!p->children[c - 'a'])

p->children[c - 'a'] = new TrieNode();

p = p->children[c - 'a'];

// Update index

p->index = index;

}

p->is_word = true;

}

/** Returns the index of word that has the prefix. */

int startsWith(const string& prefix) const {

auto node = find(prefix);

if (!node) return -1;

return node->index;

}

private:

struct TrieNode {

TrieNode():index(-1), is_word(false), children(27, nullptr){}

~TrieNode() {

for (TrieNode* child : children)

if (child) delete child;

}

int index;

int is_word;

vector<TrieNode*> children;

};

const TrieNode* find(const string& prefix) const {

const TrieNode* p = root_.get();

for (const char c : prefix) {

p = p->children[c - 'a'];

if (p == nullptr) break;

}

return p;

}

std::unique_ptr<TrieNode> root_;

};

class WordFilter {

public:

WordFilter(vector<string> words) {

for (int i = 0; i < words.size(); ++i) {

const string& w = words[i];

string key = "{" + w;

trie_.insert(key, i);

for (int j = 0; j < w.size(); ++j) {

key = w[w.size() - j - 1] + key;

trie_.insert(key, i);

}

int f(string prefix, string suffix) {

return trie_.startsWith(suffix + "{" + prefix);

}

private:

Trie trie_;

};

Related Problems:

花花酱 LeetCode 737. Sentence Similarity II

By zxi on November 28, 2017

Problem:

Given two sentences words1, words2 (each represented as an array of strings), and a list of similar word pairs pairs, determine if two sentences are similar.

For example, words1 = ["great", "acting", "skills"] and words2 = ["fine", "drama", "talent"] are similar, if the similar word pairs are pairs = [["great", "good"], ["fine", "good"], ["acting","drama"], ["skills","talent"]].

Note that the similarity relation is transitive. For example, if “great” and “good” are similar, and “fine” and “good” are similar, then “great” and “fine” are similar.

Similarity is also symmetric. For example, “great” and “fine” being similar is the same as “fine” and “great” being similar.

Also, a word is always similar with itself. For example, the sentences words1 = ["great"], words2 = ["great"], pairs = [] are similar, even though there are no specified similar word pairs.

Finally, sentences can only be similar if they have the same number of words. So a sentence like words1 = ["great"] can never be similar to words2 = ["doubleplus","good"].

Note:

The length of words1 and words2 will not exceed 1000.
The length of pairs will not exceed 2000.
The length of each pairs[i] will be 2.
The length of each words[i] and pairs[i][j] will be in the range [1, 20].

题目大意：

给你两个句子（由单词数组表示）和一些近义词对，问你这两个句子是否相似，即每组相对应的单词都要相似。

注意相似性可以传递，比如只给你”great”和”fine”相似、”fine”和”good”相似，能推断”great”和”good”也相似。

Idea:

DFS / Union Find

Solution1:

Time complexity: O(|Pairs| * |words1|)

Space complexity: O(|Pairs|)

C++ / DFS

// Author: Huahua

// Runtime: 346 ms

class Solution {

public:

bool areSentencesSimilarTwo(vector<string>& words1, vector<string>& words2, vector<pair<string, string>>& pairs) {

if (words1.size() != words2.size()) return false;

g_.clear();

for (const auto& p : pairs) {

g_[p.first].insert(p.second);

g_[p.second].insert(p.first);

}

unordered_set<string> visited;

for (int i = 0; i < words1.size(); ++i) {

visited.clear();

if (!dfs(words1[i], words2[i], visited)) return false;

}

return true;

}

private:

bool dfs(const string& src, const string& dst, unordered_set<string>& visited) {

if (src == dst) return true;

visited.insert(src);

for (const auto& next : g_[src]) {

if (visited.count(next)) continue;

if (dfs(next, dst, visited)) return true;

}

return false;

}

unordered_map<string, unordered_set<string>> g_;

};

Time complexity: O(|Pairs| + |words1|)

Space complexity: O(|Pairs|)

C++ / DFS Optimized

// Author: Huahua

// Runtime 229 ms

class Solution {

public:

bool areSentencesSimilarTwo(vector<string>& words1, vector<string>& words2, vector<pair<string, string>>& pairs) {

if (words1.size() != words2.size()) return false;

g_.clear();

ids_.clear();

for (const auto& p : pairs) {

g_[p.first].insert(p.second);

g_[p.second].insert(p.first);

}

int id = 0;

for (const auto& p : pairs) {

if(!ids_.count(p.first)) dfs(p.first, ++id);

if(!ids_.count(p.second)) dfs(p.second, ++id);

}

for (int i = 0; i < words1.size(); ++i) {

if (words1[i] == words2[i]) continue;

auto it1 = ids_.find(words1[i]);

auto it2 = ids_.find(words2[i]);

if (it1 == ids_.end() || it2 == ids_.end()) return false;

if (it1->second != it2->second) return false;

}

return true;

}

private:

bool dfs(const string& curr, int id) {

ids_[curr] = id;

for (const auto& next : g_[curr]) {

if (ids_.count(next)) continue;

if (dfs(next, id)) return true;

}

return false;

}

unordered_map<string, int> ids_;

unordered_map<string, unordered_set<string>> g_;

};

Solution2:

Time complexity: O(|Pairs| + |words1|)

Space complexity: O(|Pairs|)

C++ / Union Find

// Author: Huahua

// Runtime: 219 ms

class UnionFindSet {

public:

bool Union(const string& word1, const string& word2) {

const string& p1 = Find(word1, true);

const string& p2 = Find(word2, true);

if (p1 == p2) return false;

parents_[p1] = p2;

return true;

}

const string& Find(const string& word, bool create = false) {

if (!parents_.count(word)) {

if (!create) return word;

return parents_[word] = word;

}

string w = word;

while (w != parents_[w]) {

parents_[w] = parents_[parents_[w]];

w = parents_[w];

}

return parents_[w];

}

private:

unordered_map<string, string> parents_;

};

class Solution {

public:

bool areSentencesSimilarTwo(vector<string>& words1, vector<string>& words2, vector<pair<string, string>>& pairs) {

if (words1.size() != words2.size()) return false;

UnionFindSet s;

for (const auto& pair : pairs)

s.Union(pair.first, pair.second);

for (int i = 0; i < words1.size(); ++i)

if (s.Find(words1[i]) != s.Find(words2[i])) return false;

return true;

}

};

C++ / Union Find, Optimized

// Author: Huahua

// Runtime: 175 ms (< 98.45%)

class UnionFindSet {

public:

UnionFindSet(int n) {

parents_ = vector<int>(n + 1, 0);

ranks_ = vector<int>(n + 1, 0);

for (int i = 0; i < parents_.size(); ++i)

parents_[i] = i;

}

bool Union(int u, int v) {

int pu = Find(u);

int pv = Find(v);

if (pu == pv) return false;

if (ranks_[pu] > ranks_[pv]) {

parents_[pv] = pu;

} else if (ranks_[pv] > ranks_[pu]) {

parents_[pu] = pv;

} else {

parents_[pu] = pv;

++ranks_[pv];

}

return true;

}

int Find(int id) {

if (id != parents_[id])

parents_[id] = Find(parents_[id]);

return parents_[id];

}

private:

vector<int> parents_;

vector<int> ranks_;

};

class Solution {

public:

bool areSentencesSimilarTwo(vector<string>& words1, vector<string>& words2, vector<pair<string, string>>& pairs) {

if (words1.size() != words2.size()) return false;

UnionFindSet s(pairs.size() * 2);

unordered_map<string, int> indies; // word to index

for (const auto& pair : pairs) {

int u = getIndex(pair.first, indies, true);

int v = getIndex(pair.second, indies, true);

s.Union(u, v);

}

for (int i = 0; i < words1.size(); ++i) {

if (words1[i] == words2[i]) continue;

int u = getIndex(words1[i], indies);

int v = getIndex(words2[i], indies);

if (u < 0 || v < 0) return false;

if (s.Find(u) != s.Find(v)) return false;

}

return true;

}

private:

int getIndex(const string& word, unordered_map<string, int>& indies, bool create = false) {

auto it = indies.find(word);

if (it == indies.end()) {

if (!create) return INT_MIN;

int index = indies.size();

indies.emplace(word, index);

return index;

}

return it->second;

}

};

Related Problems:

花花酱 LeetCode 734. Sentence Similarity

By zxi on November 28, 2017

Problem:

Given two sentences words1, words2 (each represented as an array of strings), and a list of similar word pairs pairs, determine if two sentences are similar.

For example, “great acting skills” and “fine drama talent” are similar, if the similar word pairs are pairs = [["great", "fine"], ["acting","drama"], ["skills","talent"]].

Note that the similarity relation is not transitive. For example, if “great” and “fine” are similar, and “fine” and “good” are similar, “great” and “good” are not necessarily similar.

However, similarity is symmetric. For example, “great” and “fine” being similar is the same as “fine” and “great” being similar.

Also, a word is always similar with itself. For example, the sentences words1 = ["great"], words2 = ["great"], pairs = [] are similar, even though there are no specified similar word pairs.

Finally, sentences can only be similar if they have the same number of words. So a sentence like words1 = ["great"] can never be similar to words2 = ["doubleplus","good"].

Note:

The length of words1 and words2 will not exceed 1000.
The length of pairs will not exceed 2000.
The length of each pairs[i] will be 2.
The length of each words[i] and pairs[i][j] will be in the range [1, 20].

题目大意：

给你两个句子（由单词数组表示）和一些近义词对，问你这两个句子是否相似，即每组相对应的单词都要相似。

注意相似性不能传递，比如给只你”great”和”fine”相似、”fine”和”good”相似，这种情况下”great”和”good”不相似。

Idea:

Use hashtable to store mapping from word to its similar words.

Solution: HashTable

Time complexity: O(|pairs| + |words1|)

Space complexity: O(|pairs|)

C++

// Author: Huahua

// Runtime: 9 ms

class Solution {

public:

bool areSentencesSimilar(vector<string>& words1, vector<string>& words2, vector<pair<string, string>> pairs) {

if (words1.size() != words2.size()) return false;

unordered_map<string, unordered_set<string>> similar_words;

for (const auto& pair : pairs) {

similar_words[pair.first].insert(pair.second);

similar_words[pair.second].insert(pair.first);

}

for (int i = 0; i < words1.size(); ++i) {

if (words1[i] == words2[i]) continue;

if (!similar_words[words1[i]].count(words2[i])) return false;

}

return true;

}

};

Java

// Author: Huahua

// Runtime: 10 ms

class Solution {

public boolean areSentencesSimilar(String[] words1, String[] words2, String[][] pairs) {

if (words1.length != words2.length) return false;

Map<String, Set<String>> similar_words = new HashMap<>();

for (String[] pair : pairs) {

if (!similar_words.containsKey(pair[0]))

similar_words.put(pair[0], new HashSet<>());

if (!similar_words.containsKey(pair[1]))

similar_words.put(pair[1], new HashSet<>());

similar_words.get(pair[0]).add(pair[1]);

similar_words.get(pair[1]).add(pair[0]);

}

for (int i = 0; i < words1.length; ++i) {

if (words1[i].equals(words2[i])) continue;

if (!similar_words.containsKey(words1[i])) return false;

if (!similar_words.get(words1[i]).contains(words2[i])) return false;

}

return true;

}

Python

"""

Author: Huahua

Runtime: 62 ms

"""

class Solution:

def areSentencesSimilar(self, words1, words2, pairs):

if len(words1) != len(words2): return False

similar_words = {}

for w1, w2 in pairs:

if not w1 in similar_words: similar_words[w1] = set()

if not w2 in similar_words: similar_words[w2] = set()

similar_words[w1].add(w2)

similar_words[w2].add(w1)

for w1, w2 in zip(words1, words2):

if w1 == w2: continue

if w1 not in similar_words: return False

if w2 not in similar_words[w1]: return False

return True

Related Problems:

[解题报告] LeetCode 737. Sentence Similarity II

花花酱 LeetCode 1. Two Sum

By zxi on November 23, 2017

Problem:

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

Idea:

Brute force / Hashtable

Solution1:

Brute force / C++

Time complexity: O(n^2)

Space complexity: O(1)

// Author: Huahua

// Runtime: 166 ms

class Solution {

public:

vector<int> twoSum(vector<int>& nums, int target) {

for (int i = 0; i < nums.size(); ++i) {

for (int j = i + 1;j < nums.size(); ++j) {

int sum = nums[i] + nums[j];

if (sum == target) {

return {i, j};

}

return {};

}

};

Solution 2:

Hashtable / C++

Time complexity: O(n)

Space complexity: O(n)

// Author: Huahua

// Runtime: 6 ms

class Solution {

public:

vector<int> twoSum(vector<int>& nums, int target) {

unordered_map<int, int> indies;

for (int i = 0; i < nums.size(); ++i)

indies[nums[i]] = i;

for (int i = 0; i < nums.size(); ++i) {

int left = target - nums[i];

if (indies.count(left) && indies[left] != i) {

return {i, indies[left]};

}

return {};

}

};