I was just curious if there is a way to compare two tries data structure for similarities?
trie1 trie2
root root
/ | / |
m b m b
| | | |
a o a o
| \ | | |
t x b x b
def compare_trie(trie1, trie2):
pass
Output["max","bob"]
Edit: So far I tried to implement a dfs algorithm, but struck on how to manage two stacks of the different tries
Code that I tried still struck by managing two stacks for two different tries:
def compareTrie(trie1, trie2):
dfsStack = []
result = []
stack1 = [x for x in trie1.keys()]
stack2 = [y for y in trie2.keys()]
similar = list(set(stack1) & set(stack2))
dfsStack.append((similar, result))
while (dfsStack):
current, result = dfsStack.pop()
print(current, result)
result.append(current)
for c in current:
trie1 = trie1[c]
trie2 = trie2[c]
st1 = [x for x in trie1.keys()]
st2 = [x for x in trie2.keys()]
simm = list(set(st1) & set(st2))
dfsStack.append((simm, result))
print(result)
Trie Implementation:
def create_trie(words):
trie = {}
for word in words:
curr = trie
for c in word:
if c not in curr:
curr[c] = {}
curr = curr[c]
# Mark the end of a word
curr['#'] = True
return trie
s1 = "mat max bob"
s2 = "max bob"
words1 = s1.split()
words2 = s2.split()
t1 = create_trie(words1)
t2 = create_trie(words2)
Your idea to use dfs was correct; however, you could've opted a simple recusive approach to solve the task at hand. Here's the recursive version:
def create_trie(words):
trie = {}
for word in words:
curr = trie
for c in word:
if c not in curr:
curr[c] = {}
curr = curr[c]
# Mark the end of a word
curr['#'] = True
return trie
def compare(trie1, trie2, curr):
for i in trie1.keys():
if trie2.get(i, None):
if i=="#":
result.append(curr)
else:
compare(trie1[i], trie2[i], curr+i)
s1 = "mat max bob temp2 fg f r"
s2 = "max bob temp fg r c"
words1 = s1.split()
words2 = s2.split()
t1 = create_trie(words1)
t2 = create_trie(words2)
result = []
compare(t1, t2, "")
print(result) #['max', 'bob', 'fg', 'r']