Question

1258. Synonymous Sentences

中文文档

Description

You are given a list of equivalent string pairs synonyms where synonyms[i] = [si, ti] indicates that si and ti are equivalent strings. You are also given a sentence text.

Return _all possible synonymous sentences sorted lexicographically_.

 

Example 1:

Input: synonyms = [["happy","joy"],["sad","sorrow"],["joy","cheerful"]], text = "I am happy today but was sad yesterday"
Output: ["I am cheerful today but was sad yesterday","I am cheerful today but was sorrow yesterday","I am happy today but was sad yesterday","I am happy today but was sorrow yesterday","I am joy today but was sad yesterday","I am joy today but was sorrow yesterday"]

Example 2:

Input: synonyms = [["happy","joy"],["cheerful","glad"]], text = "I am happy today but was sad yesterday"
Output: ["I am happy today but was sad yesterday","I am joy today but was sad yesterday"]

 

Constraints:

• 0 <= synonyms.length <= 10
• synonyms[i].length == 2
• 1 <= si.length, ti.length <= 10
• si != ti
• text consists of at most 10 words.
• All the pairs of synonyms are unique.
• The words of text are separated by single spaces.

Solutions

Solution 1: Union-Find + DFS

We can notice that the synonyms in the problem are transitive, i.e., if a and b are synonyms, and b and c are synonyms, then a and c are also synonyms. Therefore, we can use a union-find set to find the connected components of synonyms, where all the words in each connected component are synonyms and are sorted in lexicographical order.

Next, we split the string text into a word array sentence by spaces. For each word sentence[i], if it is a synonym, we replace it with all the words in the connected component, otherwise, we do not replace it. In this way, we can get all the sentences. This can be implemented by DFS search.

We design a function $dfs(i)$, which represents starting from the $i$th word of sentence, replacing it with all the words in the connected component, and then recursively processing the following words.

If $i$ is greater than or equal to the length of sentence, it means that we have processed all the words, and at this time, we add the current sentence to the answer array. Otherwise, if sentence[i] is not a synonym, we do not replace it, directly add it to the current sentence, and then recursively process the following words. Otherwise, we replace sentence[i] with all the words in the connected component, and also recursively process the following words.

Finally, return the answer array.

The time complexity is $O(n^2)$, and the space complexity is $O(n)$. Where $n$ is the number of words.

class UnionFind:
  def __init__(self, n):
    self.p = list(range(n))
    self.size = [1] * n

  def find(self, x):
    if self.p[x] != x:
      self.p[x] = self.find(self.p[x])
    return self.p[x]

  def union(self, a, b):
    pa, pb = self.find(a), self.find(b)
    if pa != pb:
      if self.size[pa] > self.size[pb]:
        self.p[pb] = pa
        self.size[pa] += self.size[pb]
      else:
        self.p[pa] = pb
        self.size[pb] += self.size[pa]


class Solution:
  def generateSentences(self, synonyms: List[List[str]], text: str) -> List[str]:
    def dfs(i):
      if i >= len(sentence):
        ans.append(' '.join(t))
        return
      if sentence[i] not in d:
        t.append(sentence[i])
        dfs(i + 1)
        t.pop()
      else:
        root = uf.find(d[sentence[i]])
        for j in g[root]:
          t.append(words[j])
          dfs(i + 1)
          t.pop()

    words = list(set(chain.from_iterable(synonyms)))
    d = {w: i for i, w in enumerate(words)}
    uf = UnionFind(len(d))
    for a, b in synonyms:
      uf.union(d[a], d[b])
    g = defaultdict(list)
    for i in range(len(words)):
      g[uf.find(i)].append(i)
    for k in g.keys():
      g[k].sort(key=lambda i: words[i])
    sentence = text.split()
    ans = []
    t = []
    dfs(0)
    return ans

class UnionFind {
  private int[] p;
  private int[] size;

  public UnionFind(int n) {
    p = new int[n];
    size = new int[n];
    for (int i = 0; i < n; ++i) {
      p[i] = i;
      size[i] = 1;
    }
  }

  public int find(int x) {
    if (p[x] != x) {
      p[x] = find(p[x]);
    }
    return p[x];
  }

  public void union(int a, int b) {
    int pa = find(a), pb = find(b);
    if (pa != pb) {
      if (size[pa] > size[pb]) {
        p[pb] = pa;
        size[pa] += size[pb];
      } else {
        p[pa] = pb;
        size[pb] += size[pa];
      }
    }
  }
}

class Solution {
  private List<String> ans = new ArrayList<>();
  private List<String> t = new ArrayList<>();
  private List<String> words;
  private Map<String, Integer> d;
  private UnionFind uf;
  private List<Integer>[] g;
  private String[] sentence;

  public List<String> generateSentences(List<List<String>> synonyms, String text) {
    Set<String> ss = new HashSet<>();
    for (List<String> pairs : synonyms) {
      ss.addAll(pairs);
    }
    words = new ArrayList<>(ss);
    int n = words.size();
    d = new HashMap<>(n);
    for (int i = 0; i < words.size(); ++i) {
      d.put(words.get(i), i);
    }
    uf = new UnionFind(n);
    for (List<String> pairs : synonyms) {
      uf.union(d.get(pairs.get(0)), d.get(pairs.get(1)));
    }
    g = new List[n];
    Arrays.setAll(g, k -> new ArrayList<>());
    for (int i = 0; i < n; ++i) {
      g[uf.find(i)].add(i);
    }
    for (List<Integer> e : g) {
      e.sort((i, j) -> words.get(i).compareTo(words.get(j)));
    }
    sentence = text.split(" ");
    dfs(0);
    return ans;
  }

  private void dfs(int i) {
    if (i >= sentence.length) {
      ans.add(String.join(" ", t));
      return;
    }
    if (!d.containsKey(sentence[i])) {
      t.add(sentence[i]);
      dfs(i + 1);
      t.remove(t.size() - 1);
    } else {
      for (int j : g[uf.find(d.get(sentence[i]))]) {
        t.add(words.get(j));
        dfs(i + 1);
        t.remove(t.size() - 1);
      }
    }
  }
}

class UnionFind {
public:
  UnionFind(int n) {
    p = vector<int>(n);
    size = vector<int>(n, 1);
    iota(p.begin(), p.end(), 0);
  }

  void unite(int a, int b) {
    int pa = find(a), pb = find(b);
    if (pa != pb) {
      if (size[pa] > size[pb]) {
        p[pb] = pa;
        size[pa] += size[pb];
      } else {
        p[pa] = pb;
        size[pb] += size[pa];
      }
    }
  }

  int find(int x) {
    if (p[x] != x) {
      p[x] = find(p[x]);
    }
    return p[x];
  }

private:
  vector<int> p, size;
};

class Solution {
public:
  vector<string> generateSentences(vector<vector<string>>& synonyms, string text) {
    unordered_set<string> ss;
    for (auto& pairs : synonyms) {
      ss.insert(pairs[0]);
      ss.insert(pairs[1]);
    }
    vector<string> words{ss.begin(), ss.end()};
    unordered_map<string, int> d;
    int n = words.size();
    for (int i = 0; i < n; ++i) {
      d[words[i]] = i;
    }
    UnionFind uf(n);
    for (auto& pairs : synonyms) {
      uf.unite(d[pairs[0]], d[pairs[1]]);
    }
    vector<vector<int>> g(n);
    for (int i = 0; i < n; ++i) {
      g[uf.find(i)].push_back(i);
    }
    for (int i = 0; i < n; ++i) {
      sort(g[i].begin(), g[i].end(), [&](int a, int b) {
        return words[a] < words[b];
      });
    }
    vector<string> sentence;
    string s;
    istringstream iss(text);
    while (iss >> s) {
      sentence.emplace_back(s);
    }
    vector<string> ans;
    vector<string> t;
    function<void(int)> dfs = [&](int i) {
      if (i >= sentence.size()) {
        string s;
        for (int j = 0; j < t.size(); ++j) {
          if (j) {
            s += " ";
          }
          s += t[j];
        }
        ans.emplace_back(s);
        return;
      }
      if (!d.count(sentence[i])) {
        t.emplace_back(sentence[i]);
        dfs(i + 1);
        t.pop_back();
      } else {
        for (int j : g[uf.find(d[sentence[i]])]) {
          t.emplace_back(words[j]);
          dfs(i + 1);
          t.pop_back();
        }
      }
    };
    dfs(0);
    return ans;
  }
};

type unionFind struct {
  p, size []int
}

func newUnionFind(n int) *unionFind {
  p := make([]int, n)
  size := make([]int, n)
  for i := range p {
    p[i] = i
    size[i] = 1
  }
  return &unionFind{p, size}
}

func (uf *unionFind) find(x int) int {
  if uf.p[x] != x {
    uf.p[x] = uf.find(uf.p[x])
  }
  return uf.p[x]
}

func (uf *unionFind) union(a, b int) {
  pa, pb := uf.find(a), uf.find(b)
  if pa != pb {
    if uf.size[pa] > uf.size[pb] {
      uf.p[pb] = pa
      uf.size[pa] += uf.size[pb]
    } else {
      uf.p[pa] = pb
      uf.size[pb] += uf.size[pa]
    }
  }
}

func generateSentences(synonyms [][]string, text string) (ans []string) {
  ss := map[string]bool{}
  for _, pairs := range synonyms {
    ss[pairs[0]] = true
    ss[pairs[1]] = true
  }
  words := []string{}
  for w := range ss {
    words = append(words, w)
  }
  d := map[string]int{}
  for i, w := range words {
    d[w] = i
  }
  uf := newUnionFind(len(words))
  for _, pairs := range synonyms {
    uf.union(d[pairs[0]], d[pairs[1]])
  }
  g := make([][]int, len(words))
  for i := range g {
    g[uf.find(i)] = append(g[uf.find(i)], i)
  }
  for i := range g {
    sort.Slice(g[i], func(a, b int) bool { return words[g[i][a]] < words[g[i][b]] })
  }
  t := []string{}
  sentences := strings.Split(text, " ")
  var dfs func(int)
  dfs = func(i int) {
    if i >= len(sentences) {
      ans = append(ans, strings.Join(t, " "))
      return
    }
    if _, ok := ss[sentences[i]]; !ok {
      t = append(t, sentences[i])
      dfs(i + 1)
      t = t[:len(t)-1]
      return
    } else {
      for _, j := range g[uf.find(d[sentences[i]])] {
        t = append(t, words[j])
        dfs(i + 1)
        t = t[:len(t)-1]
      }
    }
  }
  dfs(0)
  return
}