Medium
Maximum Product of Two Integers With No Common Bits — Python
Full explanation · Time O(n + rlogr) · Space O(r)
# Time: O(n + rlogr), r = max(nums)
# Space: O(r)
# dp, bitmasks
class Solution(object):
def maxProduct(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
l = max(nums).bit_length()
dp = [0]*(1<<l)
for x in nums:
dp[x] = x
for i in xrange(l):
for j in xrange(0, 1<<l, 1<<(i+1)):
for k in xrange(j, j+(1<<i)):
if dp[k] > dp[k+(1<<i)]:
dp[k+(1<<i)] = dp[k]
result = 0
for x in nums:
if x*dp[((1<<l)-1)^x] > result:
result = x*dp[((1<<l)-1)^x]
return result
# Time: O(n + rlogr), r = max(nums)
# Space: O(r)
# dp, bitmasks
class Solution2(object):
def maxProduct(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
l = max(nums).bit_length()
dp = [0]*(1<<l)
for x in nums:
dp[x] = x
for i in xrange(l):
for mask in xrange(1<<l):
if mask&(1<<i):
continue
if dp[mask] > dp[mask|(1<<i)]:
dp[mask|(1<<i)] = dp[mask]
result = 0
for x in nums:
if x*dp[((1<<l)-1)^x] > result:
result = x*dp[((1<<l)-1)^x]
return result