def findWords(firstWord, lastWord):
global dictionary
#---------------Enter your source code here----------------#
#Print a shortest word ladder for the firstWord and lastWord and return 1 if such a ladder can be found
#Return -1 if there exists no word ladder for the firstWord and lastWord
queue = MyQueue() #declare a new queue
existingwordlist = [] #to keep track of used words
firtFound = False
lastFound = False
# if both words exist in dictionary, then posible to have ladder
for i in range(len(dictionary)):
if firstWord == dictionary[i]:
firstFound = True
if lastWord == dictionary[i]:
lastFound = True
if lastFound and firstFound:
if firstWord == lastWord:
print[firstWord]
return 1
elif len(firstWord) != len(lastWord):
return -1
else:
existingwordlist.append(firstWord)
refStack = MyStack()
refStack.push(firstWord)
whileCounter = 0
curword = firstWord
while lastWord not in existingwordlist and whileCounter < 100:
queuedCount = 0
for i in range(len(dictionary)): #go through dictionary list
if len(curword) == len(
dictionary[i]): #if word length matches
counter = 0
for k in range(len(curword)):
a = dictionary[i]
if a[k] == curword[
k]: #go through each character and match if they are the same
counter += 1
if counter == (len(dictionary[i]) - 1) and dictionary[
i] not in existingwordlist: #do checking if only one letter different
existingwordlist.append(dictionary[i])
curStack = MyStack()
curStack.copyFrom(refStack)
curStack.push(dictionary[i])
queue.enqueue(
curStack) #enqueue the stack into the queue
queuedCount += 1
if queuedCount == 0 and queue.size() > 0:
refStack = MyStack()
firstStack = queue.dequeue()
refStack.copyFrom(firstStack)
curword = firstStack.pop()
whileCounter += 1 # can't think of any smarter way to prevent infinite loop
outputlist = [] #declare a list to store the word ladder
if lastWord in existingwordlist:
for k in range(queue.size()):
outputlist.append(queue.dequeue().toString())
print(outputlist[-1])
return 1
else:
return -1
else:
return -1
def findWords(firstWord, lastWord):
global dictionary
#---------------Enter your source code here----------------#
#Print a shortest word ladder for the firstWord and lastWord and return 1 if such a ladder can be found
#Return -1 if there exists no word ladder for the firstWord and lastWord
usedWords = []
usedWords.append(firstWord)
#step 1
count = 0
breakFirstWordIntoChars = list(firstWord)
myQueue = MyQueue()
for i in range(len(dictionary)):
if len(dictionary[i]) == len(breakFirstWordIntoChars):
breakDicObjectIntoChars = list(dictionary[i])
for j in range(len(breakFirstWordIntoChars)):
if breakFirstWordIntoChars[j] != breakDicObjectIntoChars[j]:
count += 1
if count == 1 and dictionary[i] not in usedWords:
myStack = MyStack()
myStack.push(firstWord)
myStack.push(dictionary[i])
usedWords.append(dictionary[i])
myQueue.enqueue(myStack)
count = 0
#step 2
#print myQueue.size()
if myQueue.size() > 0:
count = 0
while (True):
if myQueue.isEmpty() is not True:
firstStack = myQueue.dequeue()
firstStackWord = firstStack.peek()
print firstStack.toString()
if firstStackWord == lastWord:
#return firstStack.toString()
break
else:
firstStackWordIntoChars = list(firstStackWord)
for m in range(len(dictionary)):
if len(dictionary[m]) == len(firstStackWordIntoChars):
breakDicObjectIntoChars = list(dictionary[m])
for n in range(len(firstStackWordIntoChars)):
if firstStackWordIntoChars[
n] != breakDicObjectIntoChars[n]:
count += 1
if count == 1 and dictionary[m] not in usedWords:
myStack2 = MyStack()
myStack2.copyFrom(firstStack)
myStack2.push(dictionary[m])
myQueue.enqueue(myStack2)
usedWords.append(dictionary[m])
count = 0
else:
return -1
break
else:
myStack2 = MyStack()
myStack2.push(firstWord)
return myStack2.toString()
