425. Word Squares

Given a set of words (without duplicates), find all word squares you can build from them.
A sequence of words forms a valid word square if the kth row and column read the exact same string, where 0 ≤ k < max(numRows, numColumns).
For example, the word sequence ["ball","area","lead","lady"] forms a word square because each word reads the same both horizontally and vertically.
b a l l
a r e a
l e a d
l a d y
Note:

1. There are at least 1 and at most 1000 words.
2. All words will have the exact same length.
3. Word length is at least 1 and at most 5.
4. Each word contains only lowercase English alphabet a-z.

Example 1:
Input:
["area","lead","wall","lady","ball"]

Output:
[
  [ "wall",
    "area",
    "lead",
    "lady"
  ],
  [ "ball",
    "area",
    "lead",
    "lady"
  ]
]

Explanation:
The output consists of two word squares. The order of output does not matter (just the order of words in each word square matters).

Example 2:
Input:
["abat","baba","atan","atal"]

Output:
[
  [ "baba",
    "abat",
    "baba",
    "atan"
  ],
  [ "baba",
    "abat",
    "baba",
    "atal"
  ]
]

Explanation:
The output consists of two word squares. The order of output does not matter (just the order of words in each word square matters).











小班课件

https://www.youtube.com/watch?v=TdX3EgW4_eM



https://leetcode.com/problems/word-squares/discuss/91333/Explained.-My-Java-solution-using-Trie-126ms-1616




这个trie , trienode class 的结构
做这个trie 的结构的时候， 注意  List<String> startWith 把 所有以这个prefix 的 word 都加了进来
我觉得我的 ood 还不扎实。。

class Solution {
    class TrieNode{
        List<String> startWith;
        TrieNode[] children;
        
        TrieNode(){
            startWith = new ArrayList<>();
            children = new TrieNode[26];
        }
    }
    
    class Trie{
        TrieNode root;
        
        Trie(String[] words){
            root = new TrieNode();
            for(String w : words){
                TrieNode cur = root;
                for(char ch : w.toCharArray()){
                    int index = ch - 'a';
                    if(cur.children[index] == null){
                        cur.children[index] = new TrieNode();
                    }
                    cur.children[index].startWith.add(w);
                    cur = cur.children[index];
                }
                
            }
        }
        List<String> getCandidates(String prefix){
            List<String> ans = new ArrayList<>();
            TrieNode cur = root;
            for(char ch : prefix.toCharArray()){
                int index = ch - 'a';
                if(cur.children[index] == null){
                    return ans;
                }
                cur = cur.children[index];
            }
            ans.addAll(cur.startWith);
            return ans;
        }
    }
    public List<List<String>> wordSquares(String[] words) {
        List<List<String>> ans = new ArrayList<>();
        if(words == null || words.length == 0) return ans;
        int len = words[0].length();
        Trie trie = new Trie(words);
        List<String> solution = new ArrayList<>();
        for(String w : words){
            solution.add(w);
            search(solution, ans, trie, len);
            solution.remove(solution.size() - 1);
        }
        return ans;
        
    }
    
    private void search(List<String> solution, List<List<String>> ans, Trie trie, int len){
        // base case 
        if(solution.size() == len){
            ans.add(new ArrayList<>(solution));
            return;
        }
        
        
        // dfs
        int index = solution.size();
        StringBuilder prefix = new StringBuilder();
        for(String word : solution){
            prefix.append(word.charAt(index));
        }
        
        List<String> candidates = trie.getCandidates(prefix.toString());
        for(String candidate : candidates){
            solution.add(candidate);
            search(solution, ans, trie, len);
            solution.remove(solution.size() - 1);
        }  
    }
}