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unique-binary-search-trees.py
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unique-binary-search-trees.py
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"""
96. Unique Binary Search Trees
Medium
Given an integer n, return the number of structurally unique BST's (binary search trees) which has exactly n nodes of unique values from 1 to n.
Example 1:
Input: n = 3
Output: 5
Example 2:
Input: n = 1
Output: 1
Constraints:
1 <= n <= 19
"""
# V0
# V1
# IDEA : DP
# https://leetcode.com/problems/unique-binary-search-trees/solution/
class Solution:
def numTrees(self, n):
"""
:type n: int
:rtype: int
"""
G = [0]*(n+1)
G[0], G[1] = 1, 1
for i in range(2, n+1):
for j in range(1, i+1):
G[i] += G[j-1] * G[i-j]
return G[n]
# V1'
# IDEA : MATH
# https://leetcode.com/problems/unique-binary-search-trees/solution/
class Solution(object):
def numTrees(self, n):
"""
:type n: int
:rtype: int
"""
C = 1
for i in range(0, n):
C = C * 2*(2*i+1)/(i+2)
return int(C)
# V1'
# IDEA : recursion (TLE)
# https://leetcode.com/problems/unique-binary-search-trees/discuss/164915/Python-solution
class Solution:
def numTrees(self, n):
"""
:type n: int
:rtype: int
"""
if n == 0:
return 1
res = 0
for i in range(1, n+1):
res += self.numTrees(i-1)*self.numTrees(n-i)
return res
# V1'''
# IDEA : DP
# https://leetcode.com/problems/unique-binary-search-trees/discuss/164915/Python-solution
class Solution:
def numTrees(self, n):
"""
:type n: int
:rtype: int
"""
arr = [0]*(n+1)
arr[0] = 1
for i in range(1, n+1):
for j in range(1, i+1):
arr[i] += arr[j-1] * arr[i-j]
return arr[-1]
# V1'''''
# https://blog.csdn.net/fuxuemingzhu/article/details/79367789
class Solution(object):
def __init__(self):
self.dp = dict()
def numTrees(self, n):
"""
:type n: int
:rtype: int
"""
if n in self.dp:
return self.dp[n]
if n == 0 or n == 1:
return 1
ans = 0
for i in range(1, n + 1):
ans += self.numTrees(i - 1) * self.numTrees(n - i)
self.dp[n] = ans
return ans
# V1'''''''
class Solution(object):
def numTrees(self, n):
"""
:type n: int
:rtype: int
"""
dp = [1, 1]
for i in range(2, n + 1):
count = 0
for j in range(i):
count += dp[j] * dp[i - j - 1]
dp.append(count)
return dp.pop()
# V2
# Time: O(n)
# Space: O(1)
class Solution(object):
def numTrees(self, n):
"""
:type n: int
:rtype: int
"""
if n == 0:
return 1
def combination(n, k):
count = 1
# C(n, k) = (n) / 1 * (n - 1) / 2 ... * (n - k + 1) / k
for i in range(1, k + 1):
count = count * (n - i + 1) / i
return count
return combination(2 * n, n) - combination(2 * n, n - 1)
# Time: O(n^2)
# Space: O(n)
# DP solution.
class Solution2(object):
# @return an integer
def numTrees(self, n):
counts = [1, 1]
for i in range(2, n + 1):
count = 0
for j in range(i):
count += counts[j] * counts[i - j - 1]
counts.append(count)
return counts[-1]