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Problem solution · Java

Word Squares

Word Squares: a Java solution using depth-first search. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Depth-first search
Source
walkccc LeetCode Solutions
Length
82 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Depth-first search

For Word Squares, the implementation follows one branch at a time, making it suitable for components, trees, backtracking, or dependency exploration.

  1. Define the state carried into one recursive or stack frame.
  2. Mark or choose the current state before exploring children.
  3. Combine child results or undo the choice when the branch finishes.

Code notes

  • 82 lines of Java from the credited upstream file 425.java.
  • The implementation visibly relies on sequence storage.
  • 6 loop blocks detected, together with recursive traversal.

Complexity

Count unique states for graph traversal; for backtracking, count the branching factor and maximum depth.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeWord Squares · JavaJava
Use this to learn the idea, then write your own version.
class TrieNode {  public TrieNode[] children = new TrieNode[26];  public List<String> startsWith = new ArrayList<>();} class Trie {  public Trie(final String[] words) {    for (final String word : words)      insert(word);  }   public List<String> findBy(final String prefix) {    TrieNode node = root;    for (final char c : prefix.toCharArray()) {      final int i = c - 'a';      if (node.children[i] == null)        return new ArrayList<>();      node = node.children[i];    }    return node.startsWith;  }   private TrieNode root = new TrieNode();   private void insert(final String word) {    TrieNode node = root;    for (final char c : word.toCharArray()) {      final int i = c - 'a';      if (node.children[i] == null)        node.children[i] = new TrieNode();      node = node.children[i];      node.startsWith.add(word);    }  }} class Solution {  public List<List<String>> wordSquares(String[] words) {    if (words.length == 0)      return new ArrayList<>();     final int n = words[0].length();    List<List<String>> ans = new ArrayList<>();    List<String> path = new ArrayList<>();    Trie trie = new Trie(words);     for (final String word : words) {      path.add(word);      dfs(trie, n, path, ans);      path.remove(path.size() - 1);    }     return ans;  }   private void dfs(Trie trie, final int n, List<String> path, List<List<String>> ans) {    if (path.size() == n) {      ans.add(new ArrayList<>(path));      return;    }     final String prefix = getPrefix(path);     for (final String s : trie.findBy(prefix)) {      path.add(s);      dfs(trie, n, path, ans);      path.remove(path.size() - 1);    }  }   // e.g. path = ["wall",  //              "area"]  //    prefix =  "le.."  private String getPrefix(List<String> path) {    StringBuilder sb = new StringBuilder();    final int index = path.size();    for (final String s : path)      sb.append(s.charAt(index));    return sb.toString();  }}

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