Word Ladder
A hard Graph problem included in Love Babbar 450, Striver A2Z. Below: the roles whose interviews prioritise this topic, and how to practise it.
- Topic
- Graph
- Sheets
- 2
- Core for
- 11 roles
- Platform
- LeetCode
The problem
Given a beginWord, endWord, and a wordList, find the length of the shortest transformation sequence from beginWord to endWord such that only one letter can change at each step and each transformed word must exist in the wordList. Return 0 if no such sequence exists.
Example 1
- Input
- beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"]
- Output
- 5
- Why
- One shortest path is hit -> hot -> dot -> dog -> cog, which has 5 words.
Example 2
- Input
- beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log"]
- Output
- 0
- Why
- The endWord cog is not in the wordList, so no transformation is possible.
Constraints
- 1 <= beginWord.length <= 10
- endWord.length == beginWord.length
- 1 <= wordList.length <= 5000
How to think about it
Updated 2026-09-09Every valid one-letter mutation is an unweighted edge in a word graph. The shortest path in an unweighted graph belongs to breadth-first search. Generating all 26 possible substitutions for each character index checks whether candidates exist in the dictionary in O(26 * L) per word, far faster than comparing pairs against the full dictionary.
Approaches, worst first
All-pairs edge construction BFS
time O(N^2 * L) · space O(N^2)
Precompute an adjacency list by comparing every word against every other word to check if they differ by 1 character, then run BFS. Quadratic comparisons across 5000 words time out.
Single-ended BFS with hash set lookup
time O(N * 26 * L^2) · space O(N * L)
Store wordList in a hash set. For the current word, replace each character with 'a' through 'z' and check the set. If present, enqueue the mutated word and erase it from the set so it is never revisited.
Bidirectional BFSWrite this one
time O(N * 26 * L^2) · space O(N * L)
Expand two search frontiers simultaneously: one from beginWord and one from endWord, always expanding the smaller frontier. Halves the effective search depth and slashes visited branch counts.
Where people lose marks · 3
- If endWord is not in wordList, return 0 immediately; beginWord is not required to be in wordList, but endWord is strictly required.
- Forgetting to return sequence length (number of words including beginWord) rather than edge count (number of transitions); the answer is edges + 1.
- Not removing words from the dictionary set upon enqueuing them, leading to exponential cycles.
The theory behind it
Graph — the ground this problem stands on. All Graph problems
What Graph is
A graph is a network of individual points, called vertices or nodes, connected by lines called edges. Think of a subway transit map, an electrical circuit, or a web of social friends. Unlike a tree, a graph has no designated top node and no parent-child hierarchy. Connections can run one-way or both ways, and paths can loop back on themselves to form closed cycles.
When to reach for it
Reach for graph algorithms when inputs describe relationships, networks, flights between cities, course prerequisites, or clone networks. Signals include finding the shortest route across unweighted connections, ordering tasks that depend on earlier tasks, counting isolated clusters, or checking whether a path contains an infinite loop. Whenever problems present pairs of related entities and ask for reachability, distances, or dependencies, graph representations apply.
How the pattern works
First convert edge lists into an adjacency list, mapping each node to an array of its neighbors. Choose your exploration strategy based on the goal: use a queue and breadth-first search to find the shortest path in unweighted networks, or use recursion and depth-first search to explore full paths and detect cycles. Because graphs can have loops, always track visited nodes in a set or boolean array. Add nodes to the visited set at the moment they enter the queue so they are never visited twice.
What each operation costs
| Operation | Time |
|---|---|
| visit all nodes and edges via search | O(v + e) |
| topological sort using in-degree counts | O(v + e) |
| shortest path using dijkstra with a min-heap | O((v + e) log v) |
What usually goes wrong with Graph
- Adding a node to the visited set when popping from the queue instead of when pushing, which lets neighboring nodes enqueue duplicate entries and wastes memory.
- Failing to check for cycles in directed graphs when finding prerequisite orders, causing topological sort routines to hang or return incomplete lists.
- Assuming an input graph is fully connected and scanning from only a single starting node, missing disconnected islands and isolated components.
Which roles need this problem
Graph is a core topic for these 11 roles — if you're targeting one of them, this problem is early in your path, not optional.
Secondary for 6 more roles, including Performance Engineer, Search Engineer, Information Retrieval Engineer.
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