Understanding the Problem
We are given a startWord
, a targetWord
, and a wordList
. The goal is to transform the startWord
into the targetWord
by changing only one character at a time, with the restriction that each transformed word must be a valid word in the given wordList
.
This problem can be visualized as a graph where each word is a node, and an edge exists between two words if they differ by exactly one character. Our task is to find the shortest path (in terms of number of transformations) from the startWord
to the targetWord
. The most suitable algorithm for this task is Breadth-First Search (BFS), which efficiently finds the shortest path in an unweighted graph.
Step-by-Step Solution with Example
Step 1: Check if transformation is possible
If the targetWord
is not in the wordList
, then there’s no way to reach it via valid transformations. So, we return 0
immediately.
Step 2: Treat words as graph nodes
Each word becomes a node in a graph. Two words are connected by an edge if they differ by exactly one letter. For example, "hot" and "dot" are neighbors, but "hot" and "dog" are not.
Step 3: Use BFS for shortest path
We initialize a queue for BFS with the pair (startWord
, level = 1). The level
variable tracks how many transformations we've made so far. We also convert the wordList
to a set for O(1) lookup and remove the startWord
if it's present.
Step 4: Generate possible transformations
For each word we dequeue, we try changing every character to every letter from 'a' to 'z'. For example, from "hot", we generate words like "aot", "bot", ..., "zot", then "hat", "hbt", ..., "hzt", and finally "hoa", "hob", ..., "hoz".
If a transformed word exists in the wordList
set, it means we’ve found a valid transformation. We enqueue this new word with level + 1
and remove it from the set to avoid revisiting.
Step 5: Stop when targetWord is found
If during BFS we dequeue a word that matches the targetWord
, we return the current level
. This ensures the minimum number of steps due to the nature of BFS (shortest path search).
Step 6: Return 0 if targetWord is unreachable
If the BFS completes and we never encounter the targetWord
, that means there is no valid transformation path, and we return 0
.
Edge Cases
- targetWord not in wordList: Return 0 immediately.
- startWord is same as targetWord: Usually undefined by the problem, but often treated as requiring at least one transformation, so result should be 0 if no path exists.
- wordList is empty: No transformation is possible, return 0.
- All words are the same length but differ significantly: The solution should still handle them with correct transformation checks.
Finally
This problem is a great example of using BFS for shortest path discovery in unweighted graphs. By thinking of each word as a node and transformations as edges, we can apply standard graph traversal techniques. Always remember to handle edge cases like missing target words or cycles due to repeated transformations.
With a clear understanding of BFS and word transformation rules, even beginners can confidently approach and solve the Word Ladder problem.
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