Get Similar Words | tech::interview


Related: LintCode 623 - K EDIT DISTANCE
Get Similar Words | tech::interview
给定一个word list 和一个target word,要求输出在word list 里跟target word的edit distance 相差不大于k的所有words。
这是Airbnb的电面题。直接用edit distance挨个遍历一遍也能做,但是如果word list很大,那重复计算会非常多,这时候可以用trie来优化。
下面举个例,假设字典为["cs", "ct", "cby"],target word为"cat"k=1
c
/ | \
b s t
/
y
从根节点开始搜索,遍历的单词分别为:c -> cb -> (cby) -> cs -> ct
与普通的Edit distance动态规划算法一样,我们对每一个“路过”的单词记录一个DP table。那么所遍历的单词的DP table应该为(假设当前遍历单词,也就是下面代码中的path为”c”):
c a t
[ 0 1 2 3 ] <- prev_dist
-> c [ 1 0 1 2 ] <- cur_dist
cb [ 2 1 1 2 ]
cs [ 2 1 1 2 ]
ct [ 2 1 1 1 ]
每一行的最后一位数字即为当前单词与target word的edit distance。显然,如果该值小于等于k,则加入到结果中。最终返回的结果只有一个单词,即ct
注意,遍历到单词cb时,edit distance已经为2,且长度已经与cat相等,也就意味着该节点的子树中所包含的单词与target word的edit distance无论如何都不可能小于等于k,因此直接从该子树返回。所以单词cby是没有被遍历到的。这也就是Trie做这题的便利之处,当字典较大时会明显提高效率。
你说遍历到cb 就不继续遍历了 因为与 cat的distance已经有2了, 可是假如下一个字母是t呢,cbt 和 cat 不是distance也是1吗
BFS , level traversal
struct Node {
Node(char val): value{val}{}
~Node() {
for(int i = 0; i < 27; ++i) {
if (children[i]) {
delete children[i];
children[i] = nullptr;
}
}
}
Node* get(char c, bool auto_create = true) {
assert((c >= 'a' && c <= 'z') || c == '\0');
int id = c == '\0' ? 26 : c - 'a';
if (!children[id] && auto_create) {
children[id] = new Node(c);
children[id]->parent = this;
num_children++;
}
return children[id];
}
bool end() {
return get('\0',false);
}
int num_children { 0 };
char value{ 0 };
Node* parent{ nullptr };
Node* children[27] { nullptr };
};
public:
Trie() {
_root = new Node(-1);
}
~Trie() {
if (_root) {
delete _root;
_root = nullptr;
}
}
void add_string(const string& str) {
auto node = _root;
for (auto c : str) {
node = node->get(c);
}
node->get('\0');
}
vector<string> get_similar(const string& s, int threshold) {
vector<string> ret;
function<void(Node*, string&, vector<int>&)> search = [&](Node* node,
string& path, vector<int>& prev_dist) {
if (node->end() && prev_dist.back() <= threshold)
ret.push_back(path);
for(int i = 0; i < 26; ++i) {
if (!node->children[i]) continue;
vector<int> cur_dist = prev_dist;
cur_dist[0]++;
char letter = 'a' + i;
bool go = cur_dist[0] <= threshold;
for (int len = 1; len <= s.size(); ++len) {
int ins_cost = cur_dist[len - 1] + 1;
int del_cost = prev_dist[len] + 1;
int rep_cost = 0;
if (s[len - 1] != letter)
rep_cost = prev_dist[len - 1] + 1;
else
rep_cost = prev_dist[len - 1];
cur_dist[len] = min(min(ins_cost, del_cost), rep_cost);
if (cur_dist[len] <= threshold) go = true;
}
if (go) {
path.push_back(letter);
search(node->children[i], path, cur_dist);
path.pop_back();
}
}
};
if(s.empty()) return ret;
string p;
vector<int> dist(s.size() + 1);
iota(dist.begin(), dist.end(), 0);
search(_root, p, dist);
return ret;
}
private:
Node* _root{ nullptr };
http://www.mitbbs.com/article_t/JobHunting/32692817.html
int editDist(string word1, string word2){
  int len1=word1.length();
  int len2=word2.length();
  int m[len1+1][len2+1];
  //m[0][0]=0;
  //init the matrix
  for(int i=0; i<=len1; i++){
    m[i][0]=i;
  }
  for(int j=0; j<=len2; j++)
     m[0][j]=j;
  //then update the matrix
  for(int i=1; i<=len1; i++){
    for(int j=1; j<=len2; j++){
       m[i][j]=min(m[i-1][j-1]+(int)(word1[i-1]!=word2[j-1]), m[i-1][j]+1);
       m[i][j]=min(m[i][j], m[i][j-1]+1);
      
      
    }
    
  }
  return m[len1][len2];
}
vector<string> similar(string word, vector<string> &word_list, int k){
  vector<string> ret;
  int n= word_list.size();
  if(n<1) return ret;
  for(int i=0; i<n; i++){
    if(editDist(word, word_list[i])<=k){
        cout<<endl<<"dist: "<< editDist(word, word_list[i]);
        ret.push_back(word_list[i]); 
    }
  }
  
  return ret;

}
我看到一篇文章,如果用trie的话,可以避免一些重复的计算,例如如果我们已经计算
了target word和mit之间的距离,当我们再计算target word和mits之间的距离时,就
可以省去一些计算,这是使用trie的意义。

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