Monday, August 8, 2016

Google – Longest String Pair that Forms a Cycle


Google – Longest String Pair that Forms a Cycle
给一个 sorted dictionary, [about, apple, google, leg, lemma, time]
要求返回最长的一对pair,使得Pair中的两个单词满足:
第一个单词的最后两个字母等于第二个单词的头两个
第二个单词的最后一个字母等于第一个单词的头一个
就相当于形成一个cycle。
GFG – Check if Strings can be Chained to Form a Cycle 可以算这题的一个follow up。这题用hash map就可以搞定了,但是GFG这题还是比较难的。
[Solution]
最直接的办法就是用一个hash table来hash头两个字母以及单词。然后遍历一遍dict找最长的pair。这种做法时间复杂度为O(n),空间也为O(n)。
不过注意题目写了是sorted dictionary. 如果被问到有没有其他方法?自然会想到binary search.
如果再仔细想想这道题的binary search并不能优化时间复杂度,反而降低了。只有在空间上有所提升。
[Solution #1] – binary search
对于每个字典里的string,根据它的最后两个字母,在整个字典里做binary search。
但是当mid的头两个字母和curr的后两个字母一样时,不能直接丢掉左边或右边,得从mid开始往两边扫。所以这样binary search的时间复杂度为O(n), 即使用递归binary search, 递归式为T(n) = 2T(n / 2),画个tree就知道也是O(n).
这样每个string做一遍binary search就是O(n^2).
不过递归binary search的代码不大写,值得好好看看。
// O(n^2) time, O(logn) space
class Solution {
  int result = 0;
 
  public int longestPair(String[] dict) {
    if (dict == null || dict.length == 0) {
      return 0;
    }
 
    for (String word : dict) {
      String end = word.substring(word.length() - 2, word.length());
      char first = word.charAt(0);
      binarySearch(word, end, first, 0, dict.length - 1, dict);
    }
    return result;
  }
 
  private void binarySearch(String word, String start, char last, int l, int r, String[] dict) {
    if (l > r) {
      return;
    }
 
    while (l <= r) {
      int mid = (l + r) / 2;
      String curr = dict[mid];
 
      if (curr.startsWith(start)) {
        if (curr.charAt(curr.length() - 1) == last) {
          result = Math.max(result, word.length() + curr.length());
        }
        binarySearch(word, start, last, l, mid - 1, dict);
        binarySearch(word, start, last, mid + 1, r, dict);
        return;
      }
      else if (curr.substring(0, 2).compareTo(start) < 0) {
        l = mid + 1;
      }
      else {
        r = mid - 1;
      }
    }
  }
}
 
 
/*
  If the dict is not sorted, then HashMap is required
  O(n) time, O(n) space
*/
class Solution2 {
  public int longestPair(String[] dict) {
    if (dict == null || dict.length == 0) {
      return 0;
    }
 
    Map<String, List<Integer>> map = new HashMap<>();
    for (int i = 0; i < dict.length; i++) {
      String word = dict[i];
      int len = word.length();
      if (len <= 2) continue;
      String key = word.substring(len - 2, len)  + word.charAt(0);
      map.putIfAbsent(key, new ArrayList<>());
      map.get(key).add(i);
    }
 
    int result = 0;
    for (int i = 0; i < dict.length; i++) {
      String word = dict[i];
      int len = word.length();
      String _key = word.substring(0, 2) + word.charAt(word.length() - 1);
      if (map.containsKey(_key)) {
        for (int j : map.get(_key)) {
          result = Math.max(result, len + dict[j].length());
        }
      }
    }
    return result;
  }
}
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