LC244. Shortest Word Distance II
This is a follow up of Shortest Word Distance. The only difference is now you are given the list of words and your method will be called repeatedly many times with different parameters. How would you optimize it?
Design a class which receives a list of words in the constructor, and implements a method that takes two words word1 and word2 and return the shortest distance between these two words in the list.
For example,
Assume that words = ["practice", "makes", "perfect", "coding", "makes"].
Given word1 = “coding”, word2 = “practice”, return 3.
Given word1 = "makes", word2 = "coding", return 1.
Note: You may assume that word1 does not equal to word2, and word1 and word2 are both in the list.
求一个array中两个元素的最短距离是多少，那就用hashmap存就行。
TreeSet solution：
TreeSet的add应该是O(lgn)的，但是具有higher和lower函数，应该也是O(lgn)的，会比遍历两个ArrayList更快一些。这样shorest的复杂度应该是O(mlgn) (m < n).
public class WordDistance {
String[] word;
Map<String, TreeSet<Integer>> map;
public WordDistance(String[] words) {
word = words;
map = new HashMap<>();
for(int i = 0; i < words.length; i++){
if(!map.containsKey(words[i])){
map.put(words[i], new TreeSet<Integer>());
}
map.get(words[i]).add(i);
} // for
}
public int shortest(String word1, String word2) {
TreeSet<Integer> set1 = map.get(word1);
TreeSet<Integer> set2 = map.get(word2);
if(set1.size() > set2.size()){
return shortest(word2, word1);
}
int res = word.length;
for(Integer in: set1){
Integer hi = set2.higher(in);
if(hi != null){
res = Math.min(hi.intValue() - in.intValue(), res);
}
Integer lo = set2.lower(in);
if(lo != null){
res = Math.min(in.intValue() - lo.intValue(), res);
}
} // for
return res;
}
}
// Your WordDistance object will be instantiated and called as such:
// WordDistance wordDistance = new WordDistance(words);
// wordDistance.shortest("word1", "word2");
// wordDistance.shortest("anotherWord1", "anotherWord2");
Another better solution:
public class WordDistance {
private Map<String, List<Integer>> map;
public WordDistance(String[] words) {
map = new HashMap<String, List<Integer>>();
for(int i = 0; i < words.length; i++) {
String w = words[i];
if(map.containsKey(w)) {
map.get(w).add(i);
} else {
List<Integer> list = new ArrayList<Integer>();
list.add(i);
map.put(w, list);
}
}
}
public int shortest(String word1, String word2) {
List<Integer> list1 = map.get(word1);
List<Integer> list2 = map.get(word2);
int ret = Integer.MAX_VALUE;
for(int i = 0, j = 0; i < list1.size() && j < list2.size(); ) {
int index1 = list1.get(i), index2 = list2.get(j);
if(index1 < index2) {
ret = Math.min(ret, index2 - index1);
i++;
} else {
ret = Math.min(ret, index1 - index2);
j++;
}
}
return ret;
}
}
Discuss的这个答案，遍历的时候list1, list2是排序的array，只需要O(m + n)就可以找到最小的差值，而不需要O(mn).
🌰
word1Indexes = 10,50,70 word2Indexes = 22,30,60,71
The left over portion of the larger length array need to be again checked. In the above cases, the min diff between 70 and 71 will be left out.
Also the code can be optimized further by binary search; Consider the following arrays; 1,2,3,4,7,9,10 6,8,11,12
after the first iteration, diff between word1 and word2 = 6 - 1 = 5
Hence in array1, no point in looking at values 2,3,4 one by one
Instead use binary search to find the value in array1 that is larger than the difference of 5 (which is 7).