Saturday, November 28, 2015

Airbnb: K Edit Distance


Related: LeetCode 161 - One Edit DistanceLeetCode 72 - Edit Distance
Buttercola: Airbnb: K Edit Distance
Given a list of word and a target word, output all the words for each the edit distance with the target no greater than k.

e.g. [abc, abd, abcd, adc], target "ac", k = 1,

output = [abc, adc]

Naive Solution:
A naive solution would be, for each word in the list, calculate the edit distance with the target word. If it is equal or less than k, output to the result list. 

If we assume that the average length of the words is m, and the total number of words in the list is n, the total time complexity is O(n * m^2). 

A better solution with a trie:
The problem with the previous solution is if the given list of the words is like ab, abc, abcd, each time we need to repeatedly calculate the edit distance with the target word. If we can combine the prefix of all words together, we can save lots of time. 
  public List<String> getKEditDistance(String[] words, String target, int k) {
    List<String> result = new ArrayList<>();
     
    if (words == null || words.length == 0 ||
        target == null || target.length() == 0 ||
        k < 0) {
      return result;
    }
     
    Trie trie = new Trie();
    for (String word : words) {
      trie.add(word);
    }
     
    TrieNode root = trie.root;
     
    int[] prev = new int[target.length() + 1];
    for (int i = 0; i < prev.length; i++) {
      prev[i] = i;
    }
     
    getKEditDistanceHelper("", target, k, root, prev, result);
     
    return result;
  }
   
  private void getKEditDistanceHelper(String curr, String target, int k, TrieNode root,
                                      int[] prevDist, List<String> result) {
    if (root.isLeaf) {
      if (prevDist[target.length()] <= k) {
        result.add(curr);
      } else {
        return;
      }
    }
     
    for (int i = 0; i < 26; i++) {
      if (root.children[i] == null) {
        continue;
      }
       
      int[] currDist = new int[target.length() + 1];
      currDist[0] = curr.length() + 1;
      for (int j = 1; j < prevDist.length; j++) {
        if (target.charAt(j - 1) == (char) (i + 'a')) {
          currDist[j] = prevDist[j - 1];
        } else {
          currDist[j] = Math.min(Math.min(prevDist[j - 1], prevDist[j]),
                                 currDist[j - 1]) + 1;
        }
      }
       
      getKEditDistanceHelper(curr + (char) (i + 'a'), target, k,
         root.children[i], currDist, result);
    }
  }
   
  public static void main(String[] args) {
    Solution solution = new Solution();
    String[] input = new String[]{"abcd", "abc", "abd", "ad"};
    String target = "ac";
    int k = 1;
     
    List<String> result = solution.getKEditDistance(input, target, k);
     
    for (String s : result) {
      System.out.println(s);
    }
  }
   
  class TrieNode {
    TrieNode[] children;
    boolean isLeaf;
     
    public TrieNode() {
      children = new TrieNode[26];
       
    }
  }
   
  class Trie {
    TrieNode root;
     
    public Trie() {
      root = new TrieNode();
    }
     
    // Add a word into trie
    public void add(String s) {
      if (s == null || s.length() == 0) {
        return;
      }
       
      TrieNode p = root;
      for (int i = 0; i < s.length(); i++) {
        char c = s.charAt(i);
        if (p.children[c - 'a'] == null) {
          p.children[c - 'a'] = new TrieNode();
        }
         
        if (i == s.length() - 1) {
          p.children[c - 'a'].isLeaf = true;
        }
         
        p = p.children[c - 'a'];
      }
    }
  }
https://robinliu.gitbooks.io/leetcode/content/trie.html
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。 首先建Trie:
     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。
http://ninefu.github.io/blog/KEditDistance/
 public static void main(String[] args) {
  KEditDistance kEditDistance = new KEditDistance();
  TrieNode root = kEditDistance.new TrieNode();
  String[] lists = new String[] {"abcd", "abc", "abd","ad"};
  String word = "ac";
  int k = 1;
  List<String> result = kEditDistance.getKEditDistance(lists, word, k, root);
  for (String s : result) {
   System.out.println(s);
  }
 }
 
 public List<String> getKEditDistance(String[] lists, String target, int k, TrieNode root) {
  List<String> result = new LinkedList<>();
  if (lists == null || lists.length == 0 || target == null || target.length() == 0 || k < 0) {
   return result;
  }
  for (String s : lists) {
   root.add(s);
  }
  int[] prevDist = new int[target.length() + 1];
  for (int i = 0; i < prevDist.length; i++) {
   prevDist[i] = i;
  }
  
  computeDistance("", prevDist, target, k, root, result);
  return result;
 }
 
 public void computeDistance(String cur, int[] prevDist, String target, int k, TrieNode node, List<String> result) {
  if (node.isWord) {
   if (prevDist[target.length()] <= k) {
    result.add(cur);
   } else {
    return;
   }
  }

  for (int i = 0; i < 26; i++) {
   if (node.children[i] == null) {
    continue;
   }
//   print(node);

   int[] curDist = new int[target.length() + 1];
   curDist[0] = cur.length() + 1;
   for (int j = 1; j < prevDist.length; j++) {
    if (target.charAt(j - 1) == (char) (i + 'a')) {
     curDist[j] = prevDist[j - 1];
    } else {
     curDist[j] = Math.min(Math.min(prevDist[j - 1], prevDist[j]), curDist[j - 1]) + 1;
    }
   }
   
   computeDistance(cur + (char)(i + 'a'), curDist, target, k, node.children[i], result);
  }
 }
 
 public  void print(TrieNode root) {
  TrieNode cur = root;
  int level = 0;
  Queue<TrieNode> queue = new LinkedList<>();
  queue.add(cur);
  while (!queue.isEmpty()) {
   int size = queue.size();
   while (size > 0) {
    cur = queue.poll();
    System.out.println("level " + level + cur.character);
    for (int i = 0; i < cur.children.length; i++) {
     if (cur.children[i] != null) {
      queue.add(cur.children[i]);
     }
    }
    size--;
   }
   level++;
  }
  
 }
 
 class TrieNode {
  boolean isWord;
  TrieNode[] children;
  char character;
  
  public TrieNode() {
   children = new TrieNode[26];
  }
  
  public void add(String s) {
   if (s == null || s.length() == 0) {
    return;
   }
   TrieNode cur = this;
   for (int i = 0; i < s.length(); i++) {
    char c = s.charAt(i);
    if (cur.children[c - 'a'] == null) {
     cur.children[c - 'a'] = new TrieNode();
    }
    cur = cur.children[c - 'a'];
    cur.character = c;
   }
   cur.isWord = true;
  }
 }
http://elise.co.nf/?p=955
和LC的 edit distance一样,只不过加入了trie和dfs,与dp结合,难度更上一层
 public List<String> findKEditDistance(String[] strs, String target, int k) {
  List<String> res = new ArrayList<>();
  if(strs.length==0) return res;
  Trie trie = new Trie();
  TrieNode root = trie.buildTrie(strs);
  int[] distance = new int[target.length()+1];
  for(int i = 0;i<distance.length;i++) distance[i] = i;
  helper(target,res,root,distance,k,"");
  return res;
 }
 private void helper(String target, List<String> res, TrieNode root, 
   int[] distance, int k, String tmp) { 
  if(root.isWord) {
   if(distance[target.length()]<=k) res.add(tmp);
   return;
  }
  
  for(int i = 0;i<26;i++) {
   if(root.children[i]==null) continue;
   int[] curdistance = new int[target.length()+1];
   //this +1 is to add the processing character, which is (char)(i+'a')
   curdistance[0] = tmp.length()+1;
   //use curdistance and distance to represent edit distance for tmp and target
   //distance[j] means edit distance between tmp and target[0,j)
   for(int j = 1;j<distance.length;j++) {
    if(target.charAt(j-1)==(char)(i+'a')) curdistance[j] = distance[j-1];
    // if character matches, no insert, delete or replace. so no increasing edit distance
    //current edit distance for tmp and target[0,j) is distance[j-1]
    else {
     //not match, either delete, insert or replace character on tmp
     curdistance[j] = Math.min(Math.min(distance[j-1], distance[j]), curdistance[j-1])+1; //+1?
     //distance[j-1] means replace tmp.charAt(j-1) with target.charAt(j-1)
     //distance[j] means delete tmp.charAt(j-1)
     //curdistance[j-1] means add target.charAt(j-1) after tmp.charAt(j-1)
    }
   }
   helper(target,res,root.children[i],curdistance,k,tmp+(char)(i+'a'));
  }
 }
}
class TrieNode {
 public TrieNode[] children = new TrieNode[26];
 public boolean isWord = false;
}
class Trie {
 public void insert(TrieNode root, String str) {
  TrieNode cur = root;
  for(int i = 0;i<str.length();i++) {
   char c = str.charAt(i);
   if(cur.children[c-'a']==null) cur.children[c-'a'] = new TrieNode();
   cur = cur.children[c-'a'];
  }
  cur.isWord = true;
 }
 public TrieNode buildTrie(String[] strs) {
  TrieNode root = new TrieNode();
  for(String str:strs) insert(root,str);
  return root;
 }
}
http://lhearen.top/2016/10/16/K-Edit-Distance/

http://stevehanov.ca/blog/index.php?id=114
Read full article from Buttercola: Airbnb: K Edit Distance

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