# LeetCode – Word Ladder

Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that only one letter can be changed at a time and each intermediate word must exist in the dictionary. For example, given:

```start = "hit"
end = "cog"
dict = ["hot","dot","dog","lot","log"]
```

One shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog", the program should return its length 5.

Analysis

UPDATED on 06/07/2015.

So we quickly realize that this is a search problem, and breath-first search guarantees the optimal solution.

Java Solution

```class WordNode{ String word; int numSteps;   public WordNode(String word, int numSteps){ this.word = word; this.numSteps = numSteps; } }   public class Solution { public int ladderLength(String beginWord, String endWord, Set<String> wordDict) { LinkedList<WordNode> queue = new LinkedList<WordNode>(); queue.add(new WordNode(beginWord, 1));   wordDict.add(endWord);   while(!queue.isEmpty()){ WordNode top = queue.remove(); String word = top.word;   if(word.equals(endWord)){ return top.numSteps; }   char[] arr = word.toCharArray(); for(int i=0; i<arr.length; i++){ for(char c='a'; c<='z'; c++){ char temp = arr[i]; if(arr[i]!=c){ arr[i]=c; }   String newWord = new String(arr); if(wordDict.contains(newWord)){ queue.add(new WordNode(newWord, top.numSteps+1)); wordDict.remove(newWord); }   arr[i]=temp; } } }   return 0; } }```
Category >> Algorithms
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1. Csnerds on 2014-1-12

The second algo has a bug. What the problem says is that “Each intermediate word must exist in the dictionary”. However, this solution assumes the last one(“cog”) also need to be in the dict, which is not necessary..

2. Csnerds on 2014-1-12

Also, “int result = 0;” This sentence seems redundant..

3. 74s0nf on 2014-1-13

hey guess for the “a”, “c” dict = {a,b,c} example the correct ans should be 1 😉

4. elvalord on 2014-3-7

That’s right. A sentence should be added “dict.add(end)”.

5. jason on 2014-3-13

This should be a shortest path problem: if two words has common character, they are connected by edge. The weight of the edge is the number of character in common. So the problem becomes shortest path problem.

6. Kae Pajunar on 2014-5-18

How can I print the world ladder answers, after getting the distance.? Thanks for the answers.

7. Dun Liu on 2014-5-20

Nice!!!

8. hbrong on 2014-5-26

The algorithm can’t guarantee find the shortest transformation sequence.

9. ajay on 2014-6-30

how it will accepted?? as it will not return as cog is not inside in dict, so currWord.equals(end) will false and return 0.

10. Tex on 2014-8-3

Priest solution is obviously wrong even if don’t consider if it’s optimally

11. Tex on 2014-8-3

I mean first

12. AlgorithmFreak on 2014-8-8

The second solution doesn’t work. Try with input: “hit”, “lag”, [“hot”,”hog”,”dot”,”dog”,”lot”,”log”]
Example sequence: hit > hot > hog > log > lag The answer should be 5. But, it shows 0

13. Sole on 2014-8-28

I think there should be a check on the currWord and the end to stop the search in the dict:

``````
/*if(currWord.equals(end)){
return currDistance;
}*/
if(isLastWord(currWord,end)){
break;
}
where isLastWord is something like this:

private boolean isLastWord(String current, String end){
int count = 0;
boolean isLastWord = true;
for(int i=0; i1){
isLastWord = false;
break;
}
}
}
return isLastWord;
}
``````
14. Yifan Peng on 2014-9-20

I think this solution is totally wrong.

15. Shaochen Huang on 2014-11-7

Agree with most of the comments here:

First: the dictionary here needs to include the end word

Second: the second solution does not find the optimal shortest path, it only finds the first valid path.

I have come up with my own solution which does the following:

1. Instead of trying to iterate through all chars from a to z, it parses the dictionary and maintains a map between index and possible chars for that index in the dictionary, so flipping a character only flipped to possible dictionary words.

2. recursive Graph depth first search which keeps tracks of all sub path size and choose the shortest one.

Please kindly review and comment to help improve the algo

16. ryanlr on 2014-11-10

Why? And how does a correct one look like?

17. Guanting on 2014-11-11

Hi! I think you were oversimplifying the problem. the edge cannot be weighted just based on the number of characters in common. And an edge cannot even be determined between two words just because they have common characters. You have to factor in what are included the dictionary. One simple example, ‘good’ and ‘doom’ have two common characters and suppose the dictionary only contains ‘good’ and ‘doom’…Apparently no edge of path could be established between the two words.

18. dfgd on 2014-11-15

yeah second algo is wrong

19. BK on 2015-1-3

We can add a condition that if (newword.equals(target)) return currDistance +1;

20. BK on 2015-1-3

we stop at the first time we find the target word and return “length”. All the words that will be added in the stack after that will be > length as we push words in stack after length +1 .

21. hdante on 2015-1-10

shortest path problem with dictionary extended to include start and end and weight equals to 1 if distance(a, b) == 1, else weight equals infinite

22. Cong on 2015-2-9

Yes, you are right. Surprisingly, the seoncd one can be acceptted by leetcode.

23. Curious Guy on 2015-2-20

You can create a separate list that stores words when found in the dictionary.

24. zhuoran on 2015-2-27

You just need to add end to dict. Then it will be correct.
Btw, thanks a lot for ur code:)

25. ryanlr on 2015-2-27

Thanks! Changed. And you are welcome:)

26. ryanlr on 2015-2-27

The problem is fixed.

27. Nooby on 2015-3-1

Thanks a ton!! You rock!! Keep it up 🙂

28. dwelo on 2015-3-12

You also don’t necessarily need two queues. You could create a WordNode class inside the method and have a word field and a distance field. Then you can create a queue of “WordNode”s. Also, I think there is no need to compare the distance to previously recorded one, since breadth search will reach the end word in the shortest path. Whenever you reach the end word, that is the shortest distance.

29. Stephen Boesch on 2015-4-11

suboptimal solutions. Better would be 1-grams.

30. Stephen Boesch on 2015-4-11

That’s nice idea to keep only the actual existing letters at each index.

31. Dado on 2015-4-15

Hi
I have solved this problem using Graph of word and then run on it BFS.

package Questions;

import java.util.HashMap;
import java.util.Queue;
import java.util.HashSet;

public class WordLadder {

//solution function
public static int ladderLength(String start, String end,
HashSet dict) {
// saving the graph nodes on hash map.
HashMap nodes = new HashMap();
// adding the start and the end words to the dict
for (String word : dict) {
nodes.put(word, new GraphNode(word));
}
// update each node’s adjacents according to one character different relation
Object[] dictArray = dict.toArray();
for (int i = 0; i < dictArray.length; i++) {
for (int j = i + 1; j < dictArray.length; j++) {
if (isNeighbor((String) dictArray[i], (String) dictArray[j])) {
.get((String) dictArray[j]));
.get((String) dictArray[i]));
}
}
}

// Run BFS on the Graph and take the dist generated as result
HashMap result = BFS(nodes, start);

// Return the distance of the end word node from the start word node
return result.get(end);

}

// BFS function
public static HashMap BFS(
HashMap nodes, String start) {

HashMap visited = new HashMap();
HashMap dist = new HashMap();
for (String key : nodes.keySet()) {
visited.put(key, 0);
dist.put(key, 0);
}
Queue q = new LinkedList();
visited.put(start, 1);
while (!q.isEmpty()) {
String dequeued = q.remove();
GraphNode curNode = nodes.get(dequeued);
for (int i = 0; i < currAdjs.size(); i++) {
if (visited.get(adj.word) == 0) {
dist.put(adj.word, dist.get(dequeued) + 1);
}

}
}
return dist;

}

// check if two words differ by one character
public static boolean isNeighbor(String a, String b) {
assert a.length() == b.length();
int differ = 0;
for (int i = 0; i 1)
return false;
}
return true;
}

public static void main(String[] args) {
// dict = [“hot”,”dot”,”dog”,”lot”,”log”] result 5;
HashSet dict = new HashSet();
}

}

class GraphNode {
String word;

public GraphNode(String word) {
this.word = word;
childs = new LinkedList();
}
}
I would like to have feedback on my code ?
Thanks

32. Alibek Datbayev on 2015-9-10

BFS is a really good way of solving this problem

33. Leonardo Campos on 2015-9-25

Too much code for an interview or competition problem.
Try to keep it as simple and small as possible 😉

34. Jayesh on 2015-10-19
35. peng li on 2015-11-1

Can be solved by Dijkstra.

36. CRH on 2015-12-6

Without using a queue. Please let me know if you see any issues with this?

``` public int getCount(String start, String end, Set dict){ // Basic edge cases if(start==null || end==null || start.length()!=end.length()) return -1; if(dict.isEmpty()) return -1; dict.add(end); int count = 0; int i = 0; while(i<start.length()){ boolean found = false; char[] arr = start.toCharArray(); char tmp = arr[i]; for(int j=97;j<=122;j++){ arr[i] = (char)j; String checker = new String(arr); if(dict.contains(checker)){ count++; dict.remove(checker); if(checker.equals(end)) { return count; } else { start = checker; found = true; i = 0; break; } } } if(!found){ arr[i] = tmp; i++; } } return -1; } ```

37. Roman on 2015-12-10

Solution posted in the task is not correct.

Here is mine:

Test.java file:
``` import java.util.*; public class Test { public static void main(String argv[]){ String start = "hit"; String end = "cog"; String[] dict = {"cot","hot","dot","cit","dog","lot","log"}; // "hit" -> "cit" -> "cot" -> "cog" // length = 4 int minDistance = getDistance(start, end); List nodesToCheck = new ArrayList(); List newNodes = new ArrayList(); List winners = new ArrayList(); nodesToCheck.add(new TreeElem(start, null, minDistance, 1)); do { for (TreeElem node : nodesToCheck) { for (int i = 0; i 0); if (winners.size() == 0) { System.out.println("No solution found."); return; } TreeElem winner = null; int minLength = -1;```

``` for (TreeElem e : winners) { if (winner == null) { winner = e; minLength = e.distanceFromTop; } ```

``` if (e.distanceFromTop = 0; i--) { System.out.print("[" + steps[i] + "] -> "); } System.out.println("[" + end + "]"); } public static int getDistance(String str1, String str2) { int dist = 0; for (int i = 0; i < str1.length(); i++) { if (str1.charAt(i) != str2.charAt(i)) { dist++; } } return dist; } public static boolean isUsed(TreeElem node, String value) { while (node.parent != null) { if (node.value.equals(value)) { return true; } node = node.parent; } return false; } } ```

TreeElem.java file:

``` import java.util.*; public class TreeElem{ public String value; public TreeElem parent; List childs; public int distanceToEnd; public int distanceFromTop; public TreeElem(String value, TreeElem parent, int distanceToEnd, int distanceFromTop) { this.value = value; this.parent = parent; this.childs = new ArrayList(); this.distanceToEnd = distanceToEnd; this.distanceFromTop = distanceFromTop; } public void addChild(TreeElem child) { childs.add(child); } } ```

38. subash sethy on 2015-12-11

can you tell me what is the complexity of this method

39. Surbhi Motghare on 2016-1-17

Can you please tell me the complexity for both the methods?

40. Dhanaraj D on 2016-2-4

Does this solution give length of shortest transformation sequence or length of first sequence found ?

41. Eugene Arnatovich on 2016-3-4

Solution just using 2 HashMaps and 1 ArrayList
` `

``` package leetcode; import java.util.ArrayList; import java.util.HashMap; import java.util.List; import java.util.Map; public class Solution { private static int min = Integer.MAX_VALUE; private static List visitedkeys = new ArrayList(); public void wordladder(String start, String end, String[] dict) { //Word ladder Map map2 = new HashMap(); Map<String, List> map = new HashMap(); int i=0; map2.put(i++, start); map.put(start, new ArrayList()); for (String d : dict) { map2.put(i++, d); map.put(d, new ArrayList()); } map2.put(i, end); map.put(end, new ArrayList()); int n = map2.size(); for (i=0; i < n; i++) { for (int j=0; j < n; j++) { if (isConvertible(map2.get(i), map2.get(j))) { map.get(map2.get(i)).add(map2.get(j)); } } } length(map, start, end, 1); if (min != Integer.MAX_VALUE){ System.out.println(min); } else { System.out.println("no path"); } } private boolean isConvertible(String s1, String s2) { if (s1.length() != s2.length()) { return false; } int diff = 0; for (int i = 0; i < s1.length(); i++) { if (s1.charAt(i) != s2.charAt(i)) { diff += 1; } } return (diff == 1) ? true : false; } private void length(Map<String, List> map, String key, String end, int len) { if (key == end) { if (len < min) min = len; return; } List lst = map.get(key); for (String v : lst) { if (!visitedkeys.contains(v)){ visitedkeys.add(v); length(map, v, end, len+1); visitedkeys.remove(visitedkeys.size()-1); } } } } ```

42. Sohrab Ahmad on 2016-6-16

for the same input as yours with dictionary changed to below:

dict = [“hot”,”dot”,”dog”,”lot”,”aog”,”log”]

answer will change like : “hit” -> “hot” -> “dot” -> “dog” -> “aog”->”cog”,

but from dog to cog also we can go directly, how will you avoid that?

43. Steve Dyson on 2016-9-16

this is not working, returns 0

44. Steve Dyson on 2016-9-17

any c# code?

45. lekzeey on 2016-10-9

This can definitely be better. You should include visited words and make sure you are not adding visited words back to the graph.

46. Bismoy Murasing on 2016-11-29

I don’t understand why this code is posted, it has time limit exceeded issue on Leetcode.

47. indiver kumar on 2017-1-25

I have Solved this problem in my way.Can please anyone locate the use case in which it will break

package com.cpa.examples;

import java.util.ArrayList;
import java.util.HashSet;
import java.util.List;

public class WordLadder {
private static String start = “hit”;
private static String end = “cog”;
private static List Dictionary = null;
private static ArrayList availablePathCntList = new ArrayList();

public static void main(String[] args) {

Dictionary = new ArrayList();

HashSet used = new HashSet();

int count = 0;

find(Dictionary, start, end, used, count);

System.out.println(availablePathCntList);

}

private static int find(List dictionary, String start, String end, HashSet used, int count) {

if (start == end) {
return 0;

}
if (difference(start, end) == 1) {
return 0;
}

for (String word : dictionary) {

int diff = difference(start, word);
if (difference(end, word) == 0) {
continue;
}
if (diff == 0) {
continue;
}
if (diff <= 1 && used.add(word)) {
System.out.println("Your word is: " + start + " and matching word is " + word);

if (diff <= 1) {
if (difference(end, word) == 1) {
} else {
count++;
find(Dictionary, new String(word), new String(end), (HashSet) used.clone(), count);
}

}
}

}
return 0;

}

private static int difference(String first, String second) {

char[] firstArr = first.toCharArray();
char[] secondArr = second.toCharArray();
int mismatch = 0;

for (int i = 0; i < firstArr.length; i++) {

if (firstArr[i] != secondArr[i]) {
mismatch++;
}
}

return mismatch;
}

}

48. Howard Wang on 2017-5-17

Uhhh… It’s breadth-first search

49. Hoc Ngo on 2017-5-19

// “hit” -> “cit” -> “cot” -> “cog” => every intermediate word must exist in the dictionary, but “cit” doesn’t.

50. Mohammad Tbeishat on 2017-6-9

You have to check the newWord with the endWord
``` String newWord = new String(arr); if (dict.contains(newWord)) { System.out.println(newWord); queue.add(new WordNode(newWord, top.numSteps + 1)); dict.remove(newWord); // you have to add the next lines. if (newWord.equals(end)) { // System.out.println("equals!"); return top.numSteps; } } ```