https://www.geeksforgeeks.org/find-all-shortest-unique-prefixes-to-represent-each-word-in-a-given-list/
Random r = new Random();
StringBuilder ret = new StringBuilder(len);
while (len-- > 0) {
ret.append((char) (r.nextInt(26) + 'a'));
}
return ret.toString();
}
public static String findShortestPrefix(String s, Set<String> D) {
// Builds a trie according to given dictionary D.
Trie T = new Trie();
for (String word : D) {
T.insert(word);
}
return T.getShortestUniquePrefix(s);
}
public static class Trie {
private static class TrieNode {
private boolean isString = false;
private Map<Character, TrieNode> leaves = new HashMap<>();
public boolean getIsString() {
return isString;
}
public void setIsString(boolean string) {
isString = string;
}
public Map<Character, TrieNode> getLeaves() {
return leaves;
}
}
private TrieNode root = new TrieNode();
public boolean insert(String s) {
TrieNode p = root;
for (char c : s.toCharArray()) {
if (!p.getLeaves().containsKey(c)) {
p.getLeaves().put(c, new TrieNode());
}
p = p.getLeaves().get(c);
}
// s already existed in this trie.
if (p.getIsString()) {
return false;
} else { // p.getIsString() == false
p.setIsString(true); // Inserts s into this trie.
return true;
}
}
public String getShortestUniquePrefix(String s) {
TrieNode p = root;
StringBuilder prefix = new StringBuilder();
for (char c : s.toCharArray()) {
prefix.append(c);
if (!p.getLeaves().containsKey(c)) {
return prefix.toString();
}
p = p.getLeaves().get(c);
}
return "";
}
}
Given an array of words, find all shortest unique prefixes to represent each word in the given array. Assume that no word is prefix of another.
Examples:
Input: arr[] = {"zebra", "dog", "duck", "dove"}
Output: dog, dov, du, z
Explanation: dog => dog
dove = dov
duck = du
z => zebra
A Simple Solution is to consider every prefix of every word (starting from the shortest to largest), and if a prefix is not prefix of any other string, then print it.
看到prefix肯定想到的是trie咯，在trie node里面多加一个field count，表示
这个字符出现多少次
1，insert word into the trie
2, search the word, 找到第一个count为1的node，返回
因为说明这个node下面没有分支了，他就应该是唯一的
这个字符出现多少次
1，insert word into the trie
2, search the word, 找到第一个count为1的node，返回
因为说明这个node下面没有分支了，他就应该是唯一的
public static final int R = 256; private Node root; private class Node{ private int count; private boolean isEnd; private Node next[] = new Node[R]; public Node(){ count = 0; isEnd = false; } public Node(int count, boolean isEnd){ this.count = count; this.isEnd = isEnd; } } public void insert(String str){ if(root == null) root = new Node(); Node curr = root; for(int i = 0; i < str.length(); i++){ char c = str.charAt(i); if(curr.next[c] == null){ curr.next[c] = new Node(1, false); }else{ curr.next[c].count++; } curr = curr.next[c]; } curr.isEnd = true; } public String shortestPrefix(String str){ Node curr = root; int len = 0; for(int i = 0; i < str.length(); i++){ char c = str.charAt(i); curr = curr.next[c]; len++; if(curr.count == 1) break; } return str.substring(0, len); } public static void main(String[] args){ TriePrefix trie = new TriePrefix(); String[] words = {"by", "sea", "sells", "she", "shells", "shore", "the"}; String[] words1 = {"zebra", "dog", "duck", "dove"}; for(String word : words1){ trie.insert(word); } String[] res = new String[words1.length]; for(int i = 0; i < words1.length; i++){ res[i] = trie.shortestPrefix(words1[i]); System.out.println(res[i]); } }
https://www.geeksforgeeks.org/find-shortest-unique-prefix-every-word-given-list-set-2-using-sorting/
In this post a sorting based approach is discussed. On comparing the string with 2 other most similar strings in the array, we can find its shortest unique prefix. For example, if we sort the array {“zebra”, “dog”, “duck”, “dove”}, we get {“dog”, “dove”, “duck”, “zebra”}. The shortest unique prefix for the string “dove” can be found as:
Compare “dove” to “dog” –> unique prefix for dove is “dov”
Compare “dove” to “duck” –> unique prefix for dove is “do”
Now, the shortest unique prefix for “dove” is the one with greater length between “dov” and “do”. So, it is “dov”.
The shortest unique prefix for first and last string can be found by comparing them with only 1 most similar neighbor on right and left, respectively.
We can sort the array of strings and keep on doing this for every string of the array.
Compare “dove” to “dog” –> unique prefix for dove is “dov”
Compare “dove” to “duck” –> unique prefix for dove is “do”
Now, the shortest unique prefix for “dove” is the one with greater length between “dov” and “do”. So, it is “dov”.
The shortest unique prefix for first and last string can be found by comparing them with only 1 most similar neighbor on right and left, respectively.
We can sort the array of strings and keep on doing this for every string of the array.
If we want to output the prefixes as the order of strings in the input array, we can store the string and its corresponding index in the hashmap. While adding the prefix to result array, we can get the index of the corresponding string from hashmap and add the prefix to that index.
http://www.codebytes.in/2014/10/finding-shortest-prefixes-for-strings.html
[We are using two arrays, the one given and the one we'll be using for the prefixes]
1. Make another array for prefixes. Store first characters of original strings in this array.
2. For all the prefixes to the left of it (in the prefix array), check whether this prefix has been used somewhere.
3. If it has been used, check whether you can add on character to the previous duplicate prefix, if not, add one character to the one that is being checked for duplicates.
4. Follow the same procedure for this new updated prefix (whether it was at some previous location in the array or the current one that was being checked, which is now updated)[Recursion]
ShortestUniquePrefix.java
private static String randString(int len) {[We are using two arrays, the one given and the one we'll be using for the prefixes]
1. Make another array for prefixes. Store first characters of original strings in this array.
2. For all the prefixes to the left of it (in the prefix array), check whether this prefix has been used somewhere.
3. If it has been used, check whether you can add on character to the previous duplicate prefix, if not, add one character to the one that is being checked for duplicates.
4. Follow the same procedure for this new updated prefix (whether it was at some previous location in the array or the current one that was being checked, which is now updated)[Recursion]
public static void checkPrefix(String[] strings, String[] pre, String s, int index){ //System.out.println(Arrays.toString(pre)); for(int i=index-1;i>=0;--i){ if(s.matches(pre[i])){ if(s.length()==strings[i].length()){ //System.out.println("Can't update the previous one, need to update this one"); pre[index] = strings[index].substring(0, s.length()+1); checkPrefix(strings, pre, pre[index], index); return; } else if(s.length() < strings[i].length()){ //System.out.println("Can update the previous one"); pre[i] = strings[i].substring(0, s.length()+1); checkPrefix(strings, pre, pre[i], i); return; } } } } public static void findPrefixes(String[] strings){ System.out.println(); String[] pre = new String[strings.length]; for(int i=0;i<pre.length;++i){ if(i>0){ if(strings[i].matches(strings[i-1])){ System.out.println("Duplicate string - Error!"); continue; } } pre[i]=Character.toString(strings[i].charAt(0)); checkPrefix(strings, pre, pre[i], i); } System.out.println(Arrays.toString(strings)); System.out.println(Arrays.toString(pre)); }
Random r = new Random();
StringBuilder ret = new StringBuilder(len);
while (len-- > 0) {
ret.append((char) (r.nextInt(26) + 'a'));
}
return ret.toString();
}
public static String findShortestPrefix(String s, Set<String> D) {
// Builds a trie according to given dictionary D.
Trie T = new Trie();
for (String word : D) {
T.insert(word);
}
return T.getShortestUniquePrefix(s);
}
public static class Trie {
private static class TrieNode {
private boolean isString = false;
private Map<Character, TrieNode> leaves = new HashMap<>();
public boolean getIsString() {
return isString;
}
public void setIsString(boolean string) {
isString = string;
}
public Map<Character, TrieNode> getLeaves() {
return leaves;
}
}
private TrieNode root = new TrieNode();
public boolean insert(String s) {
TrieNode p = root;
for (char c : s.toCharArray()) {
if (!p.getLeaves().containsKey(c)) {
p.getLeaves().put(c, new TrieNode());
}
p = p.getLeaves().get(c);
}
// s already existed in this trie.
if (p.getIsString()) {
return false;
} else { // p.getIsString() == false
p.setIsString(true); // Inserts s into this trie.
return true;
}
}
public String getShortestUniquePrefix(String s) {
TrieNode p = root;
StringBuilder prefix = new StringBuilder();
for (char c : s.toCharArray()) {
prefix.append(c);
if (!p.getLeaves().containsKey(c)) {
return prefix.toString();
}
p = p.getLeaves().get(c);
}
return "";
}
}
