Sunday, May 20, 2018

LeetCode 737 - Sentence Similarity II


http://zxi.mytechroad.com/blog/hashtable/leetcode-737-sentence-similarity-ii/
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].
Time complexity: O(|Pairs| + |words1|)
Space complexity: O(|Pairs|)
(1) 用union find的关键就是要给每个pair里的word分配一个unique id,然后用map存string到id的关系
(2) 在遍历words1和words2,如果map里面没有words1或words2当前的词或者通过find我们发现当前word1和word2不在一个集合里,return false
Union Find: 时间复杂度O(k + n), 空间复杂度O(k)
class Solution {
    public boolean areSentencesSimilarTwo(String[] words1, String[] words2, String[][] pairs) {
        if (words1.length != words2.length) {
            return false;
        }

        UnionFind uf = new UnionFind(2 * pairs.length);

        Map<String, Integer> map = new HashMap<>();
        int id = 0;

        for (String[] pair : pairs) {
            for (String word : pair) {
                if (!map.containsKey(word)) {
                    map.put(word, id);
                    ++id;
                }
            }

            uf.union(map.get(pair[0]), map.get(pair[1]));
        }


        for (int i = 0; i < words1.length; i++) {
            String word1 = words1[i];
            String word2 = words2[i];

            if (word1.equals(word2)) {
                continue;
            }

            if (!map.containsKey(word1) || !map.containsKey(word2) || uf.find(map.get(word1)) != uf.find(map.get(word2))) {
                return false;
            }
        }
        return true;
    }

    class UnionFind {
        int[] sets;
        int[] size;
        int count;

        public UnionFind(int n) {
            sets = new int[n];
            size = new int[n];
            count = n;

            for (int i = 0; i < n; i++) {
                sets[i] = i;
                size[i] = 1;
            }
        }

        public int find(int node) {
            while (node != sets[node]) {
                node = sets[node];
            }
            return node;
        }

        public void union(int i, int j) {
            int node1 = find(i);
            int node2 = find(j);

            if (node1 == node2) {
                return;
            }

            if (size[node1] < size[node2]) {
                sets[node1] = node2;
                size[node2] += size[node1];
            }
            else {
                sets[node2] = node1;
                size[node1] += size[node2];
            }
            --count;
        }
    }


C++ / Union Find
    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;
    }
https://www.jianshu.com/p/43fe7cc6797b
题解:用map<String, String>来实现union and find
    public boolean areSentencesSimilarTwo(String[] words1, String[] words2, String[][] pairs) {
        if(words1.length!=words2.length) return false;
        Map<String, String> map = new HashMap<>();
        for(String[] pair : pairs){
            String parent0 = find(pair[0], map);
            String parent1 = find(pair[1], map);
            if(!parent0.equals(parent1)) map.put(parent0, parent1);
        }
        int n = words1.length;
        for(int i=0; i<n; i++){
            if (!words1[i].equals(words2[i]) && !find(words1[i], map).equals(find(words2[i], map))) return false;
        }
        return true;
    }
    
    private String find(String word, Map<String, String> map){
        if(!map.containsKey(word)) return word;
        String str = word;
        while(map.containsKey(str)){
            str = map.get(str);
        }
        map.put(word, str);
        return str;
    }

X. 
DFS: 时间复杂度O(nk), 其中n为words的长度, k为pairs的长度,空间复杂度O(k)
    public boolean areSentencesSimilarTwo(String[] words1, String[] words2, String[][] pairs) {
        if (words1.length != words2.length) {
            return false;
        }

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

        for (String[] pair : pairs) {
            graph.putIfAbsent(pair[0], new HashSet<>());
            graph.putIfAbsent(pair[1], new HashSet<>());

            graph.get(pair[0]).add(pair[1]);
            graph.get(pair[1]).add(pair[0]);
        }

        for (int i = 0; i < words1.length; i++) {
            if (words1[i].equals(words2[i])) {
                continue;
            }

            if (!graph.containsKey(words1[i]) || !graph.containsKey(words2[i])) {
                return false;
            }
            Set<String> visited = new HashSet<>();

            if (!dfs(words1[i], words2[i], graph, visited)) {
                return false;
            }
        }
        return true;
    }

    public boolean dfs(String startWord, String endWord, Map<String, Set<String>> graph, Set<String> visited) {
        if (graph.get(startWord).contains(endWord)) {
            return true;
        }

        if (visited.contains(startWord)) {
            return false;
        }

        visited.add(startWord);
        for (String nextWord : graph.get(startWord)) {
            if (dfs(nextWord, endWord, graph, visited)) {
                return true;
            }
        }
        return false;
    }
http://zxi.mytechroad.com/blog/hashtable/leetcode-737-sentence-similarity-ii/
Time complexity: O(|Pairs| + |words1|)
Space complexity: O(|Pairs|)
C++ / DFS Optimized
    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_;



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