System-Design-Question

Word Ladder

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:

Non-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))