LeetCode 411: Minimum Unique Word Abbreviation

Note: This solution is combined with Trie + Generialized Abbreviation:

1. Since it looks for smallest one, we need persist a string to find minimum length.

2. Lots of mistakes:

index not moving (index++)

index is messed up. (s.charAt(index) - 'a')

data is not intialized.

class Solution {
class TrieNode {
TrieNode[] children = new TrieNode[26];
boolean isWord = false;
}

private TrieNode root;
private List<String> result;
private String minL = "";
public String minAbbreviation(String target, String[] dictionary) {
if (dictionary.length == 0) return String.valueOf(target.length());
root = new TrieNode();
result = new ArrayList<>();
for (String s : dictionary) addWord(s);
generateComb(target, "", 0, 0);
for (String s : result) {
if (!search(s, root, 0, 0) && (minL.length() == 0 || s.length() < minL.length())) {
minL = s;
}
}
return minL;
}

private void addWord(String s) {
TrieNode current = root;
for (int i = 0; i < s.length(); i++) {
if (current.children[s.charAt(i) - 'a'] == null) {
current.children[s.charAt(i) - 'a'] = new TrieNode();
}
current = current.children[s.charAt(i) - 'a'];
}
current.isWord = true;
}

private boolean search(String s, TrieNode root, int index, int num) {
if (root == null) return false;
if (num > 0) {
for (int i = 0; i < 26; i++) {
if (search(s, root.children[i], index, num - 1)) return true;
}
return false;
}
if (index == s.length()) return root.isWord;
if (Character.isDigit(s.charAt(index))) {
int current = 0;
while (index < s.length() && Character.isDigit(s.charAt(index))) {
current = current * 10 + (s.charAt(index++) - '0');
}
return search(s, root, index, current);
}
return search(s, root.children[s.charAt(index) - 'a'], index + 1, num);
}

private void generateComb(String word, String current, int count, int index) {
if (index == word.length()) {
if (count > 0) {
current += String.valueOf(count);
}
result.add(current);
return;
}

generateComb(word, current, count + 1, index + 1);
generateComb(word, current + (count > 0 ? count : "") + word.charAt(index), 0, index + 1);
}
}