Massive Algorithms: LIKE CODING: LeetCode [245] Shortest Word Distance III

LIKE CODING: LeetCode [245] Shortest Word Distance III

http://sbzhouhao.net/LeetCode/LeetCode-Shortest-Word-Distance-III.html
Related: LeetCode [244] Shortest Word Distance II

Given a list of words and two words word1 and word2, return the shortest distance between these two words in the list.

word1 and word2 may be the same and they represent two individual words in the list.

For example,

Assume that words = ["practice", "makes", "perfect", "coding", "makes"].

Given word1 = "makes", word2 = "coding", return 1. Given word1 = "makes", word2 = "makes", return 3.

Note:

You may assume word1 and word2 are both in the list.

https://leetcode.com/discuss/50715/12-16-lines-java-c
public int shortestWordDistance(String[] words, String word1, String word2) { long dist = Integer.MAX_VALUE, i1 = dist, i2 = -dist; boolean same = word1.equals(word2); for (int i=0; i<words.length; i++) { if (words[i].equals(word1)) { if (same) { i1 = i2; i2 = i; } else { i1 = i; } } else if (words[i].equals(word2)) { i2 = i; } dist = Math.min(dist, Math.abs(i1 - i2)); } return (int) dist; }
https://leetcode.com/discuss/70327/short-solution-lines-modified-from-shortest-word-distance

http://buttercola.blogspot.com/2015/08/leetcode-shortest-word-distance-iii.html
The problem is an extension of the previous one. The only diff is the word1 and word2 can be the same. It can still be easily solved by using a hash map. The question is, can we solve it by using the one-pass of the array?

The key is we cannot update the two pointers simultaneously, if they are the same. We could update one, compare the distance, and then update the other.

    public int shortestWordDistance(String[] words, String word1, String word2) {

        int posA = -1;

        int posB = -1;

        int minDistance = Integer.MAX_VALUE;

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

            String word = words[i];

            if (word.equals(word1)) {

                posA = i;

            } else if (word.equals(word2)) {

                posB = i;

}

            if (posA != -1 && posB != -1 && posA != posB) {

                minDistance = Math.min(minDistance, Math.abs(posA - posB));

}

            if (word1.equals(word2)) {

                posB = posA;

}

}

        return minDistance;

}

https://wxx5433.gitbooks.io/interview-preparation/content/part_ii_leetcode_lintcode/array/shortest_word_distance_iii.html

  public int shortestWordDistance(String[] words, String word1, String word2) {
    int index1 = -1;
    int index2 = -1;
    int minDistance = Integer.MAX_VALUE;
    for (int i = 0; i < words.length; i++) {
      if (words[i].equals(word1)) {
        if (index1 != -1 && word1.equals(word2)) {
          minDistance = Math.min(minDistance, Math.abs(i - index1));
        }
        index1 = i;
      } else if (words[i].equals(word2)) {
        index2 = i;
      }
      if (index1 != -1 && index2 != -1) {
        minDistance = Math.min(minDistance, Math.abs(index1 - index2));
      }
    }
    return minDistance;
  }

https://segmentfault.com/a/1190000003906667

这题和I是一样的，唯一不同的是对于word1和word2相同的时候，我们要区分第一次遇到和第二次遇到这个词。这里加入了一个turns，如果是相同单词的话，每次遇到一个单词turn加1，这样可以根据turn来判断是否要switch。

http://likesky3.iteye.com/blog/2236128

public int shortestWordDistance1(String[] words, String word1, String word2) {
if (words == null) return -1;
if (word1.equals(word2)) return shortestWordDistanceCaseSame(words, word1);
int idx1 = -1, idx2 = -1;
int diff = words.length;
for (int i = 0; i < words.length; i++) {
if (words[i].equals(word1)) {
idx1 = i;
if (idx2 != -1) {
diff = Math.min(diff, idx1 - idx2);
}
} else if (words[i].equals(word2)) {
idx2 = i;
if (idx1 != -1) {
diff = Math.min(diff, idx2 - idx1);
}
}
}
return diff;
}
public int shortestWordDistanceCaseSame(String[] words, String word) {
int prev = -1, curr = -1;
int diff = words.length;
for (int i = 0; i < words.length; i++) {
if (words[i].equals(word)) {
curr = i;
if (prev != -1)
diff = Math.min(curr - prev, diff);
prev = curr;
}
}
return diff;
}
// Method 2: all in one method
public int shortestWordDistance(String[] words, String word1, String word2) {
if (words == null) return -1;
int idx1 = -1, idx2 = -1;
int diff = words.length;
for (int i = 0; i < words.length; i++) {
if (words[i].equals(word1)) {
if (word1.equals(word2)) {
if (idx1 != -1)
diff = Math.min(diff, i - idx1);
idx1 = i;
} else {
idx1 = i;
if (idx2 != -1) {
diff = Math.min(diff, idx1 - idx2);
}
}
} else if (words[i].equals(word2)) {
idx2 = i;
if (idx1 != -1) {
diff = Math.min(diff, idx2 - idx1);
}
}
}
return diff;
}

LIKE CODING: LeetCode [245] Shortest Word Distance III

    int shortestWordDistance(vector<string>& words, string word1, string word2) {

        int n = words.size();

        int p1 = -1, p2 = -1, dist = INT_MAX;

        for(int i=0; i<n; ++i){

            if(word1==word2){

                if(words[i]==word1){

                    if(p1>p2) p2 = i;

                    else p1 = i;

}

            }else{

                if(words[i]==word1) p1 = i;

                if(words[i]==word2) p2 = i;

}

            if(p1>=0 && p2>=0)

                dist = min(dist, abs(p1-p2));

}

        return dist;

}

https://github.com/algorhythms/LeetCode/blob/master/245%20Shortest%20Word%20Distance%20III.py
X. Using HashMap
https://hzhou.me/LeetCode/LeetCode-Shortest-Word-Distance-III.html

    public int shortestWordDistance(String[] words, String word1, String word2) {
        Map<String, List<Integer>> map = new HashMap<>();

        for (int i = 0; i < words.length; i++) {
            String s = words[i];
            List<Integer> list;
            if (map.containsKey(s)) {
                list = map.get(s);
            } else {
                list = new ArrayList<>();
            }
            list.add(i);
            map.put(s, list);
        }
        List<Integer> l1 = map.get(word1);
        List<Integer> l2 = map.get(word2);

        int min = Integer.MAX_VALUE;

        for (int a : l1) {
            for (int b : l2) {
               int dist = Math.abs(b - a);
                if (dist != 0) {
                    min = Math.min(dist, min);
                }
            }
        }

        return min;
    }

http://www.1point3acres.com/bbs/thread-109587-1-1.html
只有一道题，就是minimum word index distance，一看题就乐了，重复率太高。
/* This class will be given a list of words (such as might be tokenized
* from a paragraph of text), and will provide a method that takes two
* words and returns the shortest distance (in words) between those two
* words in the provided text.
* Example:
* WordDistanceFinder finder = new WordDistanceFinder(Arrays.asList("the", "quick", "brown", "fox", "quick"));
* assert(finder.distance("fox","the") == 3);
* assert(finder.distance("quick", "fox") == 1);
*/
1. 先来了一发On的
2. 如果distance方法被call很多次怎么改进
3. 然后说如果没有重复单词怎么改进。。
4. 如果想preprocessing这个list怎么做，给两种方法，（第一种就是来一发hashmap，对所有的word pair，call distance记录下来，第二种就是来一个map，key是单词，value是array of indexes，然后问这样distance怎么算）
5. 如果cache太大memory装不下了怎么办（LRU）
6. 来实现一下LRU
Read full article from LIKE CODING: LeetCode [245] Shortest Word Distance III

LIKE CODING: LeetCode [245] Shortest Word Distance III

For example,

Note:

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