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Problem solution · Java

Word Ladder II

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

Technique
Breadth-first search
Source
walkccc LeetCode Solutions
Length
85 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

Breadth-first search

For Word Ladder II, the implementation explores reachable states in layers, which is the standard shape for unweighted shortest paths and minimum-step transitions.

  1. Model each valid configuration as a state and each legal move as an edge.
  2. Seed the queue with the starting state and mark it immediately.
  3. Expand each state once, recording distance or reachability for unseen neighbours.

Code notes

  • 85 lines of Java from the credited upstream file 126.java.
  • The implementation visibly relies on sequence storage, hash lookup, ordered lookup.
  • 7 loop blocks detected, together with recursive traversal.

Complexity

Verify that each state and transition is processed only a bounded number of times; that determines the traversal cost.

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 Ladder II · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public List<List<String>> findLadders(String beginWord, String endWord, List<String> wordList) {    Set<String> wordSet = new HashSet<>(wordList);    if (!wordSet.contains(endWord))      return new ArrayList<>();     // {"hit": ["hot"], "hot": ["dot", "lot"], ...}    Map<String, List<String>> graph = new HashMap<>();     // Build the graph from the beginWord to the endWord.    if (!bfs(beginWord, endWord, wordSet, graph))      return new ArrayList<>();     List<List<String>> ans = new ArrayList<>();    List<String> path = new ArrayList<>(List.of(beginWord));    dfs(graph, beginWord, endWord, path, ans);    return ans;  }   private boolean bfs(final String beginWord, final String endWord, Set<String> wordSet,                      Map<String, List<String>> graph) {    Set<String> currentLevelWords = new HashSet<>();    currentLevelWords.add(beginWord);    boolean reachEndWord = false;     while (!currentLevelWords.isEmpty()) {      for (final String word : currentLevelWords)        wordSet.remove(word);      Set<String> nextLevelWords = new HashSet<>();      for (final String parent : currentLevelWords) {        graph.putIfAbsent(parent, new ArrayList<>());        for (final String child : getChildren(parent, wordSet)) {          if (wordSet.contains(child)) {            nextLevelWords.add(child);            graph.get(parent).add(child);          }          if (child.equals(endWord))            reachEndWord = true;        }      }      if (reachEndWord)        return true;      currentLevelWords = nextLevelWords;    }     return false;  }   private List<String> getChildren(final String parent, Set<String> wordSet) {    List<String> children = new ArrayList<>();    StringBuilder sb = new StringBuilder(parent);     for (int i = 0; i < sb.length(); ++i) {      final char cache = sb.charAt(i);      for (char c = 'a'; c <= 'z'; ++c) {        if (c == cache)          continue;        sb.setCharAt(i, c);        final String child = sb.toString();        if (wordSet.contains(child))          children.add(child);      }      sb.setCharAt(i, cache);    }     return children;  }   private void dfs(Map<String, List<String>> graph, final String word, final String endWord,                   List<String> path, List<List<String>> ans) {    if (word.equals(endWord)) {      ans.add(new ArrayList<>(path));      return;    }    if (!graph.containsKey(word))      return;     for (final String child : graph.get(word)) {      path.add(child);      dfs(graph, child, endWord, path, ans);      path.remove(path.size() - 1);    }  }}

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