Medium
Partition Array to Minimize XOR — Python
Full explanation · Time O(n^2 * k) · Space O(n)
# Time: O(n^2 * k)
# Space: O(n)
# dp, prefix sum
class Solution(object):
def minXor(self, nums, k):
"""
:type nums: List[int]
:type k: int
:rtype: int
"""
INF = float("inf")
prefix = [0]*(len(nums)+1)
for i in xrange(len(nums)):
prefix[i+1] = prefix[i]^nums[i]
dp = prefix[:]
dp[0] = INF
for l in xrange(2, k+1):
for i in reversed(xrange(l-1, len(dp))):
mn = INF
for j in xrange(l-1, i):
v = prefix[i]^prefix[j]
mx = dp[j] if dp[j] > v else v
if mx < mn:
mn = mx
dp[i] = mn
return dp[-1]
# Time: O(n^2 * k)
# Space: O(n)
# dp, prefix sum
class Solution2(object):
def minXor(self, nums, k):
"""
:type nums: List[int]
:type k: int
:rtype: int
"""
INF = float("inf")
prefix = [0]*(len(nums)+1)
for i in xrange(len(nums)):
prefix[i+1] = prefix[i]^nums[i]
dp = [INF]*(len(nums)+1)
dp[0] = 0
for l in xrange(1, k+1):
for i in reversed(xrange(l-1, len(dp))):
dp[i] = INF
for j in xrange(l-1, i):
dp[i] = min(dp[i], max(dp[j], prefix[i]^prefix[j]))
return dp[-1]