# Word Ladder Problem

Problem:

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.

Each intermediate word must exist in the dictionary For example, Given: start = “hit” end = “cog” dict = ["hot","dot","dog","lot","log"] As one shortest transformation is “hit” -> “hot” -> “dot” -> “dog” -> “cog”, return its length 5.

Note: Return 0 if there is no such transformation sequence. All words have the same length. All words contain only lowercase alphabetic characters.

Solution:

We can use BFS to solve this problem. It’s a breadth first search. Put the first word in a queue.
Then, while the queue is not empty, check all possible combinations of the word changing one letter.
If the word exists in the dictionary, check if it is the word you’re looking for, if its not, put it in the queue with the count of the word you’re checking right now plus 1.
If it was the word you’re looking for, return the current word’s count.

Implementation:

```public class CrazyForCode {

public int ladderLength(String start, String end, Set<String> dict) {
if (start == null || end == null |
| dict == null || dict.size() == 0) {
return 0;
}

Queue<String> queue = new LinkedList<String>();
queue.offer(start);
dict.remove(start);
int length = 1;

while (!queue.isEmpty())
{
int count = queue.size();

for (int i = 0; i < count; i++)
{
String current = queue.poll();
for (int j = 0; j < current.length(); j++)
{
for (char c = 'a'; c <= 'z'; c++)
{
if (c == current.charAt(j)) {
continue;
}

String temp = replace(current, j, c);

if (temp.equals(end)) {
return length + 1;
}

if (dict.contains(temp)) {
queue.offer(temp);
dict.remove(temp);
}
}
}
}
length++;
}
return 0;
}

private String replace(String s, int index, char c) {
char[] chars = s.toCharArray();
chars[index] = c;
return new String(chars);

}
}
```

### One Thought on “Word Ladder Problem”

1. jay permanent on June 6, 2016 at 5:52 am said:

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