|
| 1 | +class Solution { |
| 2 | + /* |
| 3 | + We will do a BFS to find neighbours for a word and add which level it is |
| 4 | + on wrt to start word. Once we have the map, we can do a dfs from start word |
| 5 | + to end word by looping over the neighbours for each word and ensuring they |
| 6 | + are one level away till we find the endword. Once endword is found we backtrack. |
| 7 | + */ |
| 8 | + HashMap<String, Integer> distances; |
| 9 | + HashMap<String, List<String>> neighbours; |
| 10 | + List<List<String>> res; |
| 11 | + HashSet<String> words; |
| 12 | + public List<List<String>> findLadders(String beginWord, String endWord, List<String> wordList) { |
| 13 | + distances = new HashMap<>(); |
| 14 | + neighbours = new HashMap<>(); |
| 15 | + res = new ArrayList<>(); |
| 16 | + words = new HashSet<>(wordList); |
| 17 | + if (wordList == null || wordList.size() == 0 || !words.contains(endWord)) { |
| 18 | + return res; |
| 19 | + } |
| 20 | + bfs(beginWord, endWord, words); |
| 21 | + List<String> path = new ArrayList<>(); |
| 22 | + path.add(beginWord); // adding beginword to the path |
| 23 | + dfs(beginWord, endWord, path); |
| 24 | + return res; |
| 25 | + } |
| 26 | + |
| 27 | + public void bfs(String beginWord, String endWord, HashSet<String> words) { |
| 28 | + if (beginWord.equals(endWord)) return; |
| 29 | + Queue<String> q = new LinkedList<>(); |
| 30 | + q.add(beginWord); |
| 31 | + int level = 0; |
| 32 | + distances.put(beginWord, level); |
| 33 | + while(!q.isEmpty()) { |
| 34 | + int s = q.size(); |
| 35 | + for (int i=0; i<s; i++) { |
| 36 | + String curWord = q.poll(); |
| 37 | + if (curWord.equals(endWord)) return; |
| 38 | + List<String> neighbors = getNeighbors(curWord, words); |
| 39 | + for(String n: neighbors) { |
| 40 | + if (!distances.containsKey(n)) { |
| 41 | + distances.put(n, level+1); |
| 42 | + q.add(n); |
| 43 | + } |
| 44 | + } |
| 45 | + } |
| 46 | + level++; |
| 47 | + } |
| 48 | + } |
| 49 | + public void dfs(String currWord, String endWord, List<String> path) { |
| 50 | + if (currWord.equals(endWord)) { |
| 51 | + res.add(new ArrayList<>(path)); |
| 52 | + return; |
| 53 | + } |
| 54 | + List<String> neighb = getNeighbors(currWord, words); |
| 55 | + for(String n: neighb) { |
| 56 | + if(distances.containsKey(n)) { |
| 57 | + // ensuring we are only moving forward and not going into cycle |
| 58 | + // adding the correct next word which is one level ahead |
| 59 | + if (distances.get(n) == distances.get(currWord) + 1) { |
| 60 | + path.add(n); |
| 61 | + dfs(n, endWord, path); |
| 62 | + path.remove(path.size() - 1); // backtracking |
| 63 | + } |
| 64 | + } |
| 65 | + } |
| 66 | + } |
| 67 | + |
| 68 | + public List<String> getNeighbors(String word, HashSet<String> words) { |
| 69 | + char[] wordArr = word.toCharArray(); |
| 70 | + List<String> neig = new ArrayList<>(); |
| 71 | + for(int i=0; i<wordArr.length; i++) { |
| 72 | + for (char c ='a'; c<'z'; c++) { |
| 73 | + char temp = wordArr[i]; |
| 74 | + wordArr[i] = c; |
| 75 | + String new_string = new String(wordArr); |
| 76 | + if (words.contains(new_string)) { |
| 77 | + neig.add(new_string); |
| 78 | + } |
| 79 | + wordArr[i] = temp; |
| 80 | + } |
| 81 | + } |
| 82 | + neighbours.put(word, neig); |
| 83 | + return neig; |
| 84 | + } |
| 85 | +} |
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