# Time:  O(n)
# Space: O(k)

# Given an array nums, there is a sliding window of size k
# which is moving from the very left of the array to the
# very right. You can only see the k numbers in the window.
# Each time the sliding window moves right by one position.
#
# For example,
# Given nums = [1,3,-1,-3,5,3,6,7], and k = 3.
#
# Window position                Max
# ---------------               -----
# [1  3  -1] -3  5  3  6  7       3
#  1 [3  -1  -3] 5  3  6  7       3
#  1  3 [-1  -3  5] 3  6  7       5
#  1  3  -1 [-3  5  3] 6  7       5
#  1  3  -1  -3 [5  3  6] 7       6
#  1  3  -1  -3  5 [3  6  7]      7
# Therefore, return the max sliding window as [3,3,5,5,6,7].
#
# Note:
# You may assume k is always valid, ie: 1 <= k <= input array's size for non-empty array.
#
# Follow up:
# Could you solve it in linear time?

from collections import deque

class Solution(object):
    def maxSlidingWindow(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: List[int]
        """
        dq = deque()
        max_numbers = []

        for i in xrange(len(nums)):
            while dq and nums[i] >= nums[dq[-1]]:
                dq.pop()
            dq.append(i)
            if i >= k and dq and dq[0] <= i - k:
                dq.popleft()
            if i >= k - 1:
                max_numbers.append(nums[dq[0]])

        return max_numbers

# Time:  O(n)
# Space: O(k)

# Given an array nums, there is a sliding window of size k
# which is moving from the very left of the array to the
# very right. You can only see the k numbers in the window.
# Each time the sliding window moves right by one position.
#
# For example,
# Given nums = [1,3,-1,-3,5,3,6,7], and k = 3.
#
# Window position                Max
# ---------------               -----
# [1  3  -1] -3  5  3  6  7       3
#  1 [3  -1  -3] 5  3  6  7       3
#  1  3 [-1  -3  5] 3  6  7       5
#  1  3  -1 [-3  5  3] 6  7       5
#  1  3  -1  -3 [5  3  6] 7       6
#  1  3  -1  -3  5 [3  6  7]      7
# Therefore, return the max sliding window as [3,3,5,5,6,7].
#
# Note:
# You may assume k is always valid, ie: 1 <= k <= input array's size for non-empty array.
#
# Follow up:
# Could you solve it in linear time?

from collections import deque

class Solution(object):
    def maxSlidingWindow(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: List[int]
        """
        dq = deque()
        max_numbers = []

        for i in xrange(len(nums)):
            while dq and nums[i] >= nums[dq[-1]]:
                dq.pop()
            dq.append(i)
            if i >= k and dq and dq[0] == i - k:
                dq.popleft()
            if i >= k - 1:
                max_numbers.append(nums[dq[0]])

        return max_numbers

Update sliding-window-maximum.py