Medium
Count Partitions With Max-Min Difference at Most K — Python
Full explanation · Time O(n) · Space O(n)
# Time: O(n)
# Space: O(n)
import collections
# mono deque, two pointers, sliding window, dp, prefix sum
class Solution(object):
def countPartitions(self, nums, k):
"""
:type nums: List[int]
:type k: int
:rtype: int
"""
MOD = 10**9+7
max_dq, min_dq = collections.deque(), collections.deque()
dp = [0]*(len(nums)+1)
dp[0] = 1
left = suffix = 0
for right in xrange(len(nums)):
suffix = (suffix+dp[right])%MOD
while max_dq and nums[max_dq[-1]] <= nums[right]:
max_dq.pop()
max_dq.append(right)
while min_dq and nums[min_dq[-1]] >= nums[right]:
min_dq.pop()
min_dq.append(right)
while nums[max_dq[0]]-nums[min_dq[0]] > k:
if min_dq[0] == left:
min_dq.popleft()
if max_dq[0] == left:
max_dq.popleft()
suffix = (suffix-dp[left])%MOD
left += 1
dp[right+1] = suffix
return dp[-1]