Category: dsa Date: 2026-03-10
System Design Discussion: Word Ladder
Problem Statement: Given two words, find a sequence of words where each word is formed by changing one letter from the previous word, with the goal of transforming the first word into the second word.
Requirements (Functional + Non-functional)
Functional Requirements:
start and endstart to end with one-letter differences between consecutive wordsNon-functional Requirements:
High-Level Architecture:
The system will consist of a single microservice written in a language like Java or Python, using a framework like Spring Boot or Flask. The service will be deployed on a cloud provider like AWS or GCP, using a containerization platform like Docker.
Database Design:
We will use a NoSQL database like Redis or Cassandra to store the word ladder graph. The graph will be represented as an adjacency list, where each node represents a word and each edge represents a one-letter difference between two words.
Scaling Strategy:
To scale the system, we will use the following strategies:
Bottlenecks:
Potential bottlenecks in the system include:
Trade-offs:
Trade-offs in the system include:
Solution using the 1st Principle of System Design:
The 1st principle of system design is to “Keep it Simple, Stupid” (KISS). To apply this principle to the Word Ladder problem, we can simplify the system by:
By keeping the system simple, we can reduce the complexity and increase the performance of the system.
Learning Links:
Code Snippet (Python):
from collections import deque
def word_ladder(start, end):
graph = {}
# build the word ladder graph using a dictionary
for word in words:
graph[word] = []
for other_word in words:
if len(word) == len(other_word) and sum(c1 != c2 for c1, c2 in zip(word, other_word)) == 1:
graph[word].append(other_word)
queue = deque([(start, [start])])
visited = set()
while queue:
word, path = queue.popleft()
if word == end:
return path
for neighbor in graph[word]:
if neighbor not in visited:
visited.add(neighbor)
queue.append((neighbor, path + [neighbor]))
return []
# usage
start_word = "cat"
end_word = "dog"
print(word_ladder(start_word, end_word))