Leetcode #127: Word Ladder
In this guide, we solve Leetcode #127 Word Ladder in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
A transformation sequence from word beginWord to word endWord using a dictionary wordList is a sequence of words beginWord -> s1 -> s2 -> ... -> sk such that: Every adjacent pair of words differs by a single letter.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Breadth-First Search, Hash Table, String
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"]
Output: 5
Explanation: One shortest transformation sequence is "hit" -> "hot" -> "dot" -> "dog" -> cog", which is 5 words long.
Python Solution
while q1 and q2:
if len(q1) <= len(q2):
# Prioritize the queue with fewer elements for expansion
extend(m1, m2, q1)
else:
extend(m2, m1, q2)
def extend(m1, m2, q):
# New round of expansion
for _ in range(len(q)):
p = q.popleft()
step = m1[p]
for t in next(p):
if t in m1:
# Already visited before
continue
if t in m2:
# The other direction has been searched, indicating that a shortest path has been found
return step + 1 + m2[t]
q.append(t)
m1[t] = step + 1
Complexity
The time complexity is O(n). The space complexity is O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.