Sometime back, a colleague of mine asked me about the word ladder problem. She was looking for a change. So, I believe she stumbled across this while preparing for data structures and algorithms.

Problem Statement

Typically, the puzzle shared is a flavor of below:

Find the smallest number of transformations needed to change an initial word to a target word of same length. In every transformation, change only one character and make sure word exists in the given dictionary.

Explanation

Assuming all these 4 letter words are there in the dictionary provided, it takes minimum 4 transitions to convert word from SAIL to RUIN, i.e.
SAIL -> MAIL -> MAIN -> RAIN -> RUIN

Intent here is to know about Graph algorithm. So, what are graphs in context of algorithms and how do we apply them to solve such problems?

Graph Data Structure

Graphs are flow structure that represents entities connection with each other. Visually, they are represented with help of a Node (Vertex) & an Edge (Connector).

A tree is an undirected graph in which any two nodes are connected by only one path. In it, each node (except the root node) comprises exactly one parent node.

Most common way to represent a graph is using an Adjacency matrix. In it, Element A[i][j] is 1 if there is an edge from node i to node j or else it is 0. For example, adjacency matrix of above unidirected graph is:

  | 1 2 3 4
------------
1 | 0 1 0 1
2 | 1 0 1 0
3 | 0 1 0 1
4 | 1 0 1 0

Another common way is via Adjacency list. (List format of the data instead of a matrix.)

Related Algorithms

Graphs are applied in search algorithms. Traversing the nodes and edges in a defined order helps in optimizing search. There are two specific approaches to traverse graph:

Breadth First Search (BFS)

Given a graph G and a starting node s, search proceeds by exploring edges in the graph to find all the nodes in G for which there is a path from s. With this approach, it finds all the nodes that are at a distance k from s before it finds any nodes that are at a distance k+1.

For easy visualization, think of it as, in a tree, finding all the child nodes for a parent node as first step. Post it, find all the grandchildren and hence forth.

Depth First Search (DFS)

Given a graph G and a starting node s, search proceeds by exploring edges in the graph to find all the nodes in G traversed from s through it’s edges. With this approach, we go deep in graph connecting as many nodes in the graph as possible and branch where necessary.

For easy visualization, think of it as, in a tree, finding all the family nodes for a parent node. With this, for a given node, we connect its children, grand children, grand grand children and so on before moving to next node of same level.

Thus, with DFS approach, we can have multiple deduced trees.

Knight’s tour is a classic example that leverages Depth First Search algorithm.

Shortest Path First OR Dijkstra’s Algorithm (SPF)

Given a graph G and a starting node s, search the shortest path to reach node d. It uses a concept of weights. It’s an iterative algorithm similar to results of BFS.

Many real world example fits in here, e.g. what would be shortest path from home to office.

With BFS (a simple queue), we visit one node at a time whereas in SPF (a priority queue), we visit a node at any level with lowest cost. In a sense, BFS follows Dijkstra's algorithm, a step at a time with all edge weights equal to 1. The process for exploring the graph is structurally the same in both cases. at times, BFS is preferred with equal weight graphs. This is because, operations on a priority queue are O(log n) compared to operations on a regular queue which is O(1).

Code

I will be using a breadth first graph algorithm here based on the problem need:

import collections
from collections import deque 

class Solution(object):
    # method that will help find the path
    def ladderLength(self, beginWord, 
                        endWord, wordList):
        """
        :type beginWord: str
        :type endWord: str
        :type wordList: Set[str]
        :returntype: int
        """

        # Queue for BFS
        queue = deque()

        # start by adding begin word
        queue.append((beginWord, [beginWord]))

        while queue:
            # let's keep a watch at active queue
            print('Current queue:',queue)

            # get the current node and 
            # path how it came
            node, path = queue.popleft()

            # let's keep track of path length 
            # traversed so far
            print('Current transformation count:',
                                        len(path))

            # find possible next set of 
            # child nodes, 1 diff
            for next in self.next_nodes(node, 
                            wordList) - set(path):
                # traversing through all child nodes
                # if any of the child matches, 
                # we are good               
                if next == endWord:
                    print('found endword at path:',
                                            path)
                    return len(path)
                else:
                    # keep record of next 
                    # possible paths
                    queue.append((next, 
                                path + [next]))
        return 0

    def next_nodes(self, word, word_list):
        # start with empty collection
        possiblenodes = set()

        # all the words are of fixed length
        wl_word_length = len(word)

        # loop through all the words in 
        # the word list
        for wl_word in word_list:
            mismatch_count = 0

            # find all the words that are 
            # only a letter different from 
            # current word those are the 
            # possible next child nodes
            for i in range(wl_word_length):
                if wl_word[i] != word[i]:
                    mismatch_count += 1
            if mismatch_count == 1:
                # only one alphabet different-yes
                possiblenodes.add(wl_word)
        
        # lets see the set of next possible nodes 
        print('possible next nodes:',possiblenodes)
        return possiblenodes

# Setup
beginWord = "SAIL"
endWord = "RUIN"
wordList = ["SAIL","RAIN","REST","BAIL","MAIL",
                                    "MAIN","RUIN"]

# Call
print('Transformations needed: ',
    Solution().ladderLength(beginWord, 
                            endWord, wordList))

# Transformation expected == 4
# One possible shortes path with 4 transformation:
# SAIL -> MAIL -> MAIN -> RAIN -> RUIN

Used deque (doubly ended queue) of Python

deque helps with quicker append and pop operations from both the ends. It has O(1) time complexity for append and pop operations. In comparison, list provides it in O(n) time complexity.

A quick look at the code workflow to validate if all nodes at a particular distance was traversed first and then moved to next level:

Current queue: deque([('SAIL', ['SAIL'])])

Current transformation count: 1
possible next nodes: {'BAIL', 'MAIL'}
Current queue: deque([('BAIL', ['SAIL', 'BAIL']), 
                      ('MAIL', ['SAIL', 'MAIL'])])

Current transformation count: 2
possible next nodes: {'SAIL', 'MAIL'}
Current queue: deque([('MAIL', ['SAIL', 'MAIL']), 
                      ('MAIL', ['SAIL', 'BAIL', 
                       'MAIL'])])

Current transformation count: 2
possible next nodes: {'BAIL', 'MAIN', 'SAIL'}
Current queue: deque([('MAIL', ['SAIL', 'BAIL', 
                                'MAIL']), 
                      ('BAIL', ['SAIL', 'MAIL', 
                                'BAIL']), 
                      ('MAIN', ['SAIL', 'MAIL', 
                                'MAIN'])])

Current transformation count: 3
possible next nodes: {'BAIL', 'MAIN', 'SAIL'}
Current queue: deque([('BAIL', ['SAIL', 'MAIL', 
                                'BAIL']), 
                      ('MAIN', ['SAIL', 'MAIL', 
                                'MAIN']), 
                      ('MAIN', ['SAIL', 'BAIL', 
                                'MAIL', 'MAIN'])])

Current transformation count: 3
possible next nodes: {'SAIL', 'MAIL'}
Current queue: deque([('MAIN', ['SAIL', 'MAIL', 
                                'MAIN']), 
                      ('MAIN', ['SAIL', 'BAIL', 
                                'MAIL', 'MAIN'])])

Current transformation count: 3
possible next nodes: {'RAIN', 'MAIL'}
Current queue: deque([('MAIN', ['SAIL', 'BAIL', 
                                'MAIL', 'MAIN']), 
                      ('RAIN', ['SAIL', 'MAIL', 
                                'MAIN', 'RAIN'])])

Current transformation count: 4
possible next nodes: {'RAIN', 'MAIL'}
Current queue: deque([('RAIN', ['SAIL', 'MAIL', 
                                'MAIN', 'RAIN']), 
                      ('RAIN', ['SAIL', 'BAIL', 
                        'MAIL', 'MAIN', 'RAIN'])])

Current transformation count: 4
possible next nodes: {'MAIN', 'RUIN'}
found endword at path: ['SAIL', 'MAIL', 'MAIN', 
                                        'RAIN']

Transformations needed:  4
Overall path: ['SAIL', 'MAIL', 'MAIN', 
                               'RAIN', 'RUIN']

Complexity

For above code that I used to find the shortest path for transformation:

Time

In next_nodes, for each word in the word list, we iterated over its length to find all the intermediate words corresponding to it. Thus we did M×N iterations, where M is the length of each word and N is the total number of words in the input word list. Further, to form an intermediate word, it takes O(M) time. This adds up to O(M2×N).

In ladderLength, BFS can go to each of the N words and for each word, we need to examine M possible intermediate words. This adds up to O(M2×N).

Overall, it adds up to O2(M2×N) which would be called O(M2×N).

Space

In next_nodes, each word in the word list would have M intermediate combinations. For every word we need a space of M2 to save all the transformations corresponding to it. Thus, it would need a total space of O(M2×N).

In ladderLength, BFS queue would need a space of O(M×N)

Overall, it adds up to O(M2×N) + O(M×N) which would be called O(M2×N)

Wrap Up

It could be little tricky and thus would need some practice to visualize the graph as well to write code for it.

Great, so now we know how to solve problems like word ladder problem. It also touch based other related common graph algorithms that we can refer to.

I had a read of the following reference and it has much more details if needed.

Keep problem solving!

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Linear-time partition is a divide & conquer based selection algorithm. With it, data is split into three groups using a pivot.
.

An integral part of Quick Sort algorithm which uses this partitioning logic recursively. All the elements smaller than the pivot are put on one side and all the larger ones on the other side of the pivot.

Similar to discussion of Dynamic Programming, this algorithm plays on solving sub-problems to solve complex problem.

Algorithm

Post selecting the pivot, Linear-time partition routine separates the data into three groups with values:

• less than the pivot
• equal to the pivot
• greater than the pivot

Generally, this algorithm is done in place. This results in partially sorting the data. There are handful of problems that makes use of this fact, like:

• Sort an array that contains only 0’s, 1’s & 2’s
• Dutch national flag problem
• Print all negative integers followed by positive for an array full of them
• Print all 0’s first and then 1’s or vice-versa for an array with only 0’s & 1’s
• Move all the 0’s to the end maintaining relative order of other elements for an array of integers

If done out of place, (i.e. not changing the original data), it would cost O(n) additional space

Example

Let’s take an example of: sort a array that contains only 0’s, 1’s & 2’s

First thought for such problem is to perform a count of 0’s, 1’s and 2’s. Once we have the counts, reset the array with them. Though it has time complexity O(n), it takes two traversal of the array or uses an extra array.

Below is an attempt to solve using Linear-time partition algorithm to avoid that extra traversal/space.

def threeWayPartition(A):
    start = mid = 0
    end = len(A)-1
    
    # define a Pivot
    pivot = 1
    
    while (mid <= end):
        # mid element is less than pivot
        # current element is 0
        
        # so lets move it to start
        # current start is good. 
        # move start to next element
        # move mid to next element to move forward
        if (A[mid] < pivot) :
            swap(A, start, mid)
            start = start + 1
            mid = mid + 1
            
        # mid element is more than pivot
        # current element is 2
        
        # so lets move it to end
        # current end is good. 
        # move end to previous element
        elif (A[mid] > pivot) :
            swap(A, mid, end)
            end = end - 1
        
        # mid element is same as pivot
        # current element is 1
        
        # just move forward: 
        # mid to next element
        else :
            mid = mid + 1
            
# Swap two elements A[i] and A[j] in the list
def swap(A, i, j):
    A[i], A[j] = A[j], A[i]


# Define an array
inputArray = [0, 1, 2, 2, 1, 0, 0, 2]

# Call the Linear-time partition routine
threeWayPartition(inputArray)

# print the final result
print(inputArray)

# Outputs
# [0, 0, 0, 1, 1, 2, 2, 2]

With a defined pivot, we segregated the data on the either side which resulted in desired output. Dutch nation flag problem or printing all negative first and then positive, or printing all 0s first follows the same code.

For moving all 0’s to end maintaining other elements order, we do a tweak in swap index to maintain order:

def threeWayPartition(A):
    current = 0
    nonzero = 0
    end = len(A)-1
    
    # define a Pivot
    pivot = 0
    
    while (current <= end):
        if (A[current] != pivot) :
            swap(A, current, nonzero)
            nonzero = nonzero + 1
        current = current + 1
            
# Swap two elements A[i] and A[j] in the list
def swap(A, i, j):
    A[i], A[j] = A[j], A[i]


# Define an array
inputArray = [7,0,5,1,2,0,2,0,6]

# Call the Linear-time partition routine
threeWayPartition(inputArray)

# print the final result
print(inputArray)

# Output
# [7, 5, 1, 2, 2, 6, 0, 0, 0]

Complexity

With above algorithm approach, we solved our problem with Time complexity O(n) & Space complexity O(1) (with single traversal of the array)

It was fun solving!

samples GitHub Profile Readme
Learn Python – Beginners step by step – Basics and Examples
Sandeep Mewara Github
Sandeep Mewara Learn By Insight
Matplotlib plot samples
Sandeep Mewara Github Repositories