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cracking-the-safe.py
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# Time: O(k^n)
# Space: O(k^n)
class Solution(object):
def crackSafe(self, n, k):
"""
:type n: int
:type k: int
:rtype: str
"""
M = k**(n-1)
P = [q*k+i for i in xrange(k) for q in xrange(M)] # rotate: i*k^(n-1) + q => q*k + i
result = [str(k-1)]*(n-1)
for i in xrange(k**n):
j = i
# concatenation in lexicographic order of Lyndon words
while P[j] >= 0:
result.append(str(j//M))
P[j], j = -1, P[j]
return "".join(result)
# Time: O(k^n)
# Space: O(k^n)
class Solution2(object):
def crackSafe(self, n, k):
"""
:type n: int
:type k: int
:rtype: str
"""
total = k**n
M = total//k
unique_rolling_hash = 0
result = [str(0)]*(n-1)
lookup = set()
while len(lookup) < total:
for i in reversed(xrange(k)): # preorder like traversal relative to initial result to avoid getting stuck, i.e. don't use 0 until there is no other choice
new_unique_rolling_hash = unique_rolling_hash*k + i
if new_unique_rolling_hash not in lookup:
lookup.add(new_unique_rolling_hash)
result.append(str(i))
unique_rolling_hash = new_unique_rolling_hash%M
break
return "".join(result)
# Time: O(k^n)
# Space: O(k^n)
class Solution3(object):
def crackSafe(self, n, k):
"""
:type n: int
:type k: int
:rtype: str
"""
M = k**(n-1)
def dfs(k, unique_rolling_hash, lookup, result):
for i in reversed(xrange(k)): # preorder like traversal relative to initial result to avoid getting stuck, i.e. don't use 0 until there is no other choice
new_unique_rolling_hash = unique_rolling_hash*k + i
if new_unique_rolling_hash not in lookup:
lookup.add(new_unique_rolling_hash)
result.append(str(i))
dfs(k, new_unique_rolling_hash%M, lookup, result)
break
unique_rolling_hash = 0
result = [str(0)]*(n-1)
lookup = set()
dfs(k, unique_rolling_hash, lookup, result)
return "".join(result)
# Time: O(n * k^n)
# Space: O(n * k^n)
class Solution4(object):
def crackSafe(self, n, k):
"""
:type n: int
:type k: int
:rtype: str
"""
result = [str(k-1)]*(n-1)
lookup = set()
total = k**n
while len(lookup) < total:
node = result[len(result)-n+1:]
for i in xrange(k): # preorder like traversal relative to initial result to avoid getting stuck, i.e. don't use k-1 until there is no other choice
neighbor = "".join(node) + str(i)
if neighbor not in lookup:
lookup.add(neighbor)
result.append(str(i))
break
return "".join(result)
# Time: O(n * k^n)
# Space: O(n * k^n)
class Solution5(object):
def crackSafe(self, n, k):
"""
:type n: int
:type k: int
:rtype: str
"""
def dfs(k, node, lookup, result):
for i in xrange(k): # preorder like traversal relative to initial result to avoid getting stuck, i.e. don't use k-1 until there is no other choice
neighbor = node + str(i)
if neighbor not in lookup:
lookup.add(neighbor)
result.append(str(i))
dfs(k, neighbor[1:], lookup, result)
break
result = [str(k-1)]*(n-1)
lookup = set()
dfs(k, "".join(result), lookup, result)
return "".join(result)