Skip to main content

Word Ladders - Medium


Word ladder is a word game invented by Lewis Carroll. A word ladder puzzle begins with two words, and to solve the puzzle one must find a chain of other words to link the two, in which two adjacent words (that is, words in successive steps) differ by one letter. Eg. SPORT <-> SHORT
Source: Wikipedia

If you've followed my previous Dijkstra's Shortest Path Algorithm Tutorial, you'll be able to solve this challenge with minimum effort.

Essentially, we create a graph where each word is a vertex and there exists an edge if two words differ by a single character. To find the word ladder is basically running any shortest path algorithm. You could also use a Breadth First Search.

Some important observations

  • The length of the source and destination words must be the same for a path to exist.
  • The path from source to destination  can only contain words of the same length. So we can prune off the rest of the dictionary.
  • The graph is undirected.
Here's my implementation, it uses the modules from the tutorial.

import sys,graphs,itertools
def prepare_graph():
diff_by_1 = lambda (x,y): len(x) == len(y) and sum(1 for a,b in zip(x,y) if a != b) == 1
graph = {}
for x,y in filter(diff_by_1,itertools.product(words,repeat=2)):
if x not in graph: graph[x] = []
if y not in graph: graph[y] = []
graph[x].append( ( y , 1) )
graph[y].append( ( x , 1) )
return graph
if __name__ == '__main__':
if len(sys.argv) >= 3 and len(sys.argv[1]) == len(sys.argv[2]):
src,dst = sys.argv[1:]
words = filter(lambda w : len(w) == len(src), sys.stdin.read().split()) #only consider those words with same length as parameter
if src in words and dst in words:
print graphs.dijkstra(prepare_graph(),src,dst)
else:
print "Both words should be present in the dictionary."
else:
print "Missing command line parameters."
view raw Word Ladder hosted with ❤ by GitHub


To run the program, "wordladder.py SOURCE DESTINATION < dict.txt"
For eg.

C:\>wordladder.py BOX TEN < words.txt
(4, ['BOX', 'BOD', 'BED', 'BEN', 'TEN'])

Labels

Labels:

Comments

Post a Comment

Popular posts from this blog

Find Increasing Triplet Subsequence - Medium

Problem - Given an integer array A[1..n], find an instance of i,j,k where 0 < i < j < k <= n and A[i] < A[j] < A[k]. Let's start with the obvious solution, bruteforce every three element combination until we find a solution. Let's try to reduce this by searching left and right for each element, we search left for an element smaller than the current one, towards the right an element greater than the current one. When we find an element that has both such elements, we found the solution. The complexity of this solution is O(n^2). To reduce this even further, One can simply apply the longest increasing subsequence algorithm and break when we get an LIS of length 3. But the best algorithm that can find an LIS is O(nlogn) with O( n ) space . An O(nlogn) algorithm seems like an overkill! Can this be done in linear time? The Algorithm: We iterate over the array and keep track of two things. The minimum value iterated over (min) The minimum increa...
2 comments
Read more

Merge k-sorted lists - Medium

Problem - Given k-sorted lists, merge them into a single sorted list. A daft way of doing this would be to copy all the list into a new array and sorting the new array. ie O(n log(n)) The naive method would be to simply perform k-way merge similar to the auxiliary method in Merge Sort. But that is reduces the problem to a minimum selection from a list of k-elements. The Complexity of this algorithm is an abysmal O(nk). Here's how it looks in Python. We maintain an additional array called Index[1..k] to maintain the head of each list. We improve upon this by optimizing the minimum selection process by using a familiar data structure, the Heap! Using a MinHeap, we extract the minimum element from a list and then push the next element from the same list into the heap, until all the list get exhausted. This reduces the Time complexity to O(nlogk) since for each element we perform O(logk) operations on the heap. An important implementation detail is we need to keep track ...
Post a Comment
Read more

3SUM - Hard

Problem - Given an Array of integers, A. Find out if there exists a triple (i,j,k) such that A[i] + A[j] + A[k] == 0. The 3SUM  problem is very similar to the 2SUM  problem in many aspects. The solutions I'll be discussing are also very similar. I highly recommend you read the previous post first, since I'll explain only the differences in the algorithm from the previous post. Let's begin, We start with the naive algorithm. An O(n^3) solution with 3 nested loops each checking if the sum of the triple is 0. Since O(n^3) is the higher order term, we can sort the array in O(nlogn) and add a guard at the nested loops to prune of parts of the arrays. But the complexity still remains O(n^3). The code is pretty simple and similar to the naive algorithm of 2SUM. Moving on, we'll do the same thing we did in 2SUM, replace the inner-most linear search with a binary search. The Complexity now drops to O(n^2 logn) Now, the hash table method, this is strictly not ...
Post a Comment
Read more