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longest-increasing-subsequence.py
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longest-increasing-subsequence.py
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# Time: O(nlogn)
# Space: O(n)
import bisect
class Solution(object):
def lengthOfLIS(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
LIS = []
def insert(target):
left = bisect.bisect_left(LIS, target)
# If not found, append the target.
if left == len(LIS):
LIS.append(target)
else:
LIS[left] = target
for num in nums:
insert(num)
return len(LIS)
# Time: O(nlogn)
# Space: O(n)
class Solution2(object):
def lengthOfLIS(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
LIS = []
def insert(target):
left, right = 0, len(LIS) - 1
# Find the first index "left" which satisfies LIS[left] >= target
while left <= right:
mid = left + (right - left) // 2
if LIS[mid] >= target:
right = mid - 1
else:
left = mid + 1
# If not found, append the target.
if left == len(LIS):
LIS.append(target)
else:
LIS[left] = target
for num in nums:
insert(num)
return len(LIS)
# Range Maximum Query
class SegmentTree(object): # 0-based index
def __init__(self, N,
build_fn=lambda x, y: [y]*(2*x),
query_fn=lambda x, y: y if x is None else max(x, y), # (lambda x, y: y if x is None else min(x, y))
update_fn=lambda x, y: y,
default_val=0):
self.N = N
self.H = (N-1).bit_length()
self.query_fn = query_fn
self.update_fn = update_fn
self.default_val = default_val
self.tree = build_fn(N, default_val)
self.lazy = [None]*N
def __apply(self, x, val):
self.tree[x] = self.update_fn(self.tree[x], val)
if x < self.N:
self.lazy[x] = self.update_fn(self.lazy[x], val)
def update(self, L, R, h): # Time: O(logN), Space: O(N)
def pull(x):
while x > 1:
x //= 2
self.tree[x] = self.query_fn(self.tree[x*2], self.tree[x*2+1])
if self.lazy[x] is not None:
self.tree[x] = self.update_fn(self.tree[x], self.lazy[x])
L += self.N
R += self.N
L0, R0 = L, R
while L <= R:
if L & 1: # is right child
self.__apply(L, h)
L += 1
if R & 1 == 0: # is left child
self.__apply(R, h)
R -= 1
L //= 2
R //= 2
pull(L0)
pull(R0)
def query(self, L, R): # Time: O(logN), Space: O(N)
def push(x):
n = 2**self.H
while n != 1:
y = x // n
if self.lazy[y] is not None:
self.__apply(y*2, self.lazy[y])
self.__apply(y*2 + 1, self.lazy[y])
self.lazy[y] = None
n //= 2
result = None
if L > R:
return result
L += self.N
R += self.N
push(L)
push(R)
while L <= R:
if L & 1: # is right child
result = self.query_fn(result, self.tree[L])
L += 1
if R & 1 == 0: # is left child
result = self.query_fn(result, self.tree[R])
R -= 1
L //= 2
R //= 2
return result
def __str__(self):
showList = []
for i in xrange(self.N):
showList.append(self.query(i, i))
return ",".join(map(str, showList))
# Time: O(nlogn)
# Space: O(n)
# optimized from Solution4
class Solution3(object):
def lengthOfLIS(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
sorted_nums = sorted(set(nums))
lookup = {num:i for i, num in enumerate(sorted_nums)}
segment_tree = SegmentTree(len(lookup))
for num in nums:
segment_tree.update(lookup[num], lookup[num],
segment_tree.query(0, lookup[num]-1)+1 if lookup[num] >= 1 else 1)
return segment_tree.query(0, len(lookup)-1) if len(lookup) >= 1 else 0
# Time: O(n^2)
# Space: O(n)
# Traditional DP solution.
class Solution4(object):
def lengthOfLIS(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
dp = [] # dp[i]: the length of LIS ends with nums[i]
for i in xrange(len(nums)):
dp.append(1)
for j in xrange(i):
if nums[j] < nums[i]:
dp[i] = max(dp[i], dp[j] + 1)
return max(dp) if dp else 0