1079. Letter Tile Possibilities
Read the full problem statement on LeetCode.
Difficulty: medium Acceptance: 84% Topics: Hash Table, String, Backtracking, Counting
View full problem on LeetCode Reading material
Reference solution (spoiler · python)
# Time: O(n^2)
# Space: O(n)
import collections
class Solution(object):
def numTilePossibilities(self, tiles):
"""
:type tiles: str
:rtype: int
"""
fact = [0.0]*(len(tiles)+1)
fact[0] = 1.0
for i in xrange(1, len(tiles)+1):
fact[i] = fact[i-1]*i
count = collections.Counter(tiles)
# 1. we can represent each alphabet 1..26 as generating functions:
# G1(x) = 1 + x^1/1! + x^2/2! + x^3/3! + ... + x^count1/count1!
# G2(x) = 1 + x^1/1! + x^2/2! + x^3/3! + ... + x^count2/count2!
# ...
# G26(x) = 1 + x^1/1! + x^2/2! + x^3/3! + ... + x^count26/count26!
#
# 2. let G1(x)*G2(x)*...*G26(x) = c0 + c1*x1 + ... + ck*x^k, k is the max number s.t. ck != 0
# => ci (1 <= i <= k) is the number we need to divide when permuting i letters
# => the answer will be : c1*1! + c2*2! + ... + ck*k!
coeff = [0.0]*(len(tiles)+1)
coeff[0] = 1.0
for i in count.itervalues():
new_coeff = [0.0]*(len(tiles)+1)
for j in xrange(len(coeff)):
for k in xrange(i+1):
if k+j >= len(new_coeff):
break
new_coeff[j+k] += coeff[j]*1.0/fact[k]
coeff = new_coeff
result = 0
for i in xrange(1, len(coeff)):
result += int(round(coeff[i]*fact[i]))
return result
# Time: O(r), r is the value of result
# Space: O(n)
class Solution2(object):
def numTilePossibilities(self, tiles):
"""
:type tiles: str
:rtype: int
"""
def backtracking(counter):
total = 0
for k, v in counter.iteritems():
if not v:
continue
counter[k] -= 1
total += 1+backtracking(counter)
counter[k] += 1
return total
return backtracking(collections.Counter(tiles))
Solution from kamyu104/LeetCode-Solutions · MIT