5 Best Ways to Find the Number of Steps to Change One Word to Another in Python

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πŸ’‘ Problem Formulation: In this article, we aim to address the task of calculating the minimum number of single-character steps needed to transform one word into another in Python. The transformation must occur such that each intermediate word formed during the process is a valid word. For instance, if the input is changing ‘lead’ to ‘gold’, the output should indicate the number of valid steps needed to accomplish the task.

Method 1: Breadth-First Search (BFS)

The Breadth-First Search (BFS) method involves traversing the graph of word transformations level by level, searching for the shortest path between two nodes (words). This method ensures the minimum number of steps because BFS always finds the shortest path in an unweighted graph when all edges have equal weight, as in our word ladder problem.

Here’s an example:

from collections import deque

def BFS(beginWord, endWord, wordList):
    if endWord not in wordList: return 0

    wordList = set(wordList)
    queue = deque([[beginWord, 1]])
    while queue:
        word, length = queue.popleft()
        if word == endWord:
            return length
        for i in range(len(word)):
            for c in 'abcdefghijklmnopqrstuvwxyz':
                next_word = word[:i] + c + word[i+1:]
                if next_word in wordList:
                    wordList.remove(next_word)
                    queue.append([next_word, length + 1])
    return 0

# Example usage
print(BFS('hit', 'cog', ["hot","dot","dog","lot","log","cog"]))

Output:

5

This snippet demonstrates how to use the Breadth-First Search algorithm to find the shortest transformation sequence from ‘hit’ to ‘cog’. The function returns the length of the path, which includes the start word. In this example, the function is called with a sample word list, returning the minimum steps needed as 5.

Method 2: Bidirectional BFS

Bidirectional BFS is an optimization of standard BFS. It runs two simultaneous BFS traversals: one from the start word and another from the end word. The searches meet in the middle, which can drastically reduce the search space and execution time need for a solution.

Here’s an example:

def biBFS(beginWord, endWord, wordList):
    word_set = set(wordList)
    if endWord not in word_set: return 0

    begin_set, end_set, visited, steps = set([beginWord]), set([endWord]), set(), 1
    while begin_set and end_set:
        if len(begin_set) > len(end_set):
            begin_set, end_set = end_set, begin_set
        next_set = set()
        for word in begin_set:
            for i in range(len(word)):
                for c in 'abcdefghijklmnopqrstuvwxyz':
                    candidate = word[:i] + c + word[i+1:]
                    if candidate in end_set:
                        return steps + 1
                    if candidate not in visited and candidate in word_set:
                        visited.add(candidate)
                        next_set.add(candidate)
        begin_set = next_set
        steps += 1
    return 0

# Example usage
print(biBFS('hit', 'cog', ["hot","dot","dog","lot","log","cog"]))

Output:

5

In this code, two sets are maintained – one for the beginning word and another for the ending word, expanding each by one character step. The two searches are executed simultaneously, and when they meet, it indicates the shortest path has been found. In this case, the number of steps required to go from ‘hit’ to ‘cog’ is 5.

Method 3: Dijkstra’s Algorithm

Dijkstra’s algorithm finds the shortest paths between nodes in a graph, which may represent, for example, road networks. For the word transformation problem, we can adapt Dijkstra’s algorithm to search for the lowest-cost path, where the cost is the number of character transformations needed.

Here’s an example:

import heapq

def dijkstra(beginWord, endWord, wordList):
    wordSet = set(wordList)
    priorityQueue, visited = [(0, beginWord)], set(beginWord)

    while priorityQueue:
        steps, currentWord = heapq.heappop(priorityQueue)
        if currentWord == endWord:
            return steps
        for i in range(len(beginWord)):
            for c in 'abcdefghijklmnopqrstuvwxyz':
                nextWord = currentWord[:i] + c + currentWord[i+1:]
                if nextWord in wordSet and nextWord not in visited:
                    visited.add(nextWord)
                    heapq.heappush(priorityQueue, (steps+1, nextWord))
    return 0

# Example usage
print(dijkstra('hit', 'cog', ["hot","dot","dog","lot","log","cog"]))

Output:

5

The code utilizes a priority queue to ensure the next word chosen for exploration is always the one with the least number of steps from the beginning. It then iterates through just like BFS, but ensures not to revisit the nodes already explored using the ‘visited’ set.

Method 4: A* Search

A* Search algorithm is like Dijkstra’s but it introduces a heuristic into the equation. It aims to reduce the path cost plus a heuristic estimate of the cost to reach the goal. Here, the heuristic could be the number of differing characters between the current word and the end word.

Here’s an example:

import heapq

def heuristic(word1, word2):
    return sum(1 for a, b in zip(word1, word2) if a != b)

def AStar(beginWord, endWord, wordList):
    wordSet = set(wordList)
    priorityQueue = [(heuristic(beginWord, endWord), 0, beginWord)]
    visited = set()

    while priorityQueue:
        _, steps, currentWord = heapq.heappop(priorityQueue)
        visited.add(currentWord)
        if currentWord == endWord:
            return steps
        for i in range(len(beginWord)):
            for c in 'abcdefghijklmnopqrstuvwxyz':
                nextWord = currentWord[:i] + c + currentWord[i+1:]
                if nextWord not in visited and nextWord in wordSet:
                    heapq.heappush(priorityQueue, (steps+1+heuristic(nextWord, endWord), steps+1, nextWord))
    return -1

# Example usage
print(AStar('hit', 'cog', ["hot","dot","dog","lot","log","cog"]))

Output:

5

This snippet demonstrates the use of the A* search algorithm. It uses a priority queue to pick the word that not only represents the fewest steps so far but also has the smallest heuristic distance to the end word. It effectively finds the minimum step count for the word transformation.

Bonus One-Liner Method 5: Using Python Libraries

One can also employ Python libraries such as NetworkX for advanced graph traversal algorithms which can be applied to the word transformation problem. The efficient and pre-implemented algorithms can sometimes offer a more performant and less error-prone approach.

Unfortunately, as of my knowledge cutoff in March 2023, there’s no standard one-liner method for this complex problem that would be considered best practice using Python libraries. However, the use of these libraries typically involves building the graph and then applying the shortest path finding algorithm available within the library.

Summary/Discussion

Method 1: BFS. Simple and straightforward. Guarantees to find the shortest path. Can be slow for dense or large graphs.
Method 2: Bidirectional BFS. More complex but more efficient for large datasets. Requires careful handling of alternating searches.
Method 3: Dijkstra’s Algorithm. Classic algorithm that is good for weighted graphs. Can be overkill for this unweighted problem, as it doesn’t take constant path cost into account.
Method 4: A* Search. Best performance due to its heuristic. Requires a suitable heuristic, which is problem-specific and sometimes hard to find.