## # 239. Sliding Window Maximum

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. Return the max sliding window.

Example:

``````Input: nums = [1,3,-1,-3,5,3,6,7], and k = 3
Output: [3,3,5,5,6,7]
Explanation:
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
``````

### # Solution

Approach 1: O(n*k) - Brute force.

Approach 2: O(n) - keep a monotonously decreasing deque with increasing freshness -- new items get appended to the right, and outdated items get kicked out from the left.

### # Code (Python)

Approach 2:

``````from collections import deque

class Solution:
def maxSlidingWindow(self, nums: List[int], k: int) -> List[int]:
# Optimization: to save space we could have stored only the indices instead of (value, index) pairs.
result = []
window = deque()
for i in range(len(nums)):
# append new item
while window and window[-1] <= nums[i]:
window.pop()
window.append((nums[i], i))
# throw away outdated items
while window <= i - k:
window.popleft()
# append max to result after forming the first window
if i >= k - 1:
result.append(window)

return result
``````

### # Code (C++)

Approach 1:

``````// O(n*k).
class Solution {
public:
vector<int> maxSlidingWindow(vector<int>& nums, int k) {
vector<int> res;
if (k <= 0)
return res;
for (int i = 0; i <= nums.size()-k; ++i)
{
int max = nums[i];
for (int j = i+1; j < i + k; ++j)
{
if (nums[j] > max)
max = nums[j];
}
res.push_back(max);
}
return res;
}
};
``````

Approach 2:

``````// O(n).
class Solution {
public:
vector<int> maxSlidingWindow(vector<int>& nums, int k) {
vector<int> res;
vector<int> buf;
for (int tail = 0; tail < nums.size(); ++tail)
{
while (!buf.empty() && buf.back() < nums[tail])
buf.pop_back();
buf.push_back(nums[tail]);
if (tail - head + 1 == k)
{
res.push_back(buf.front());
buf.erase(buf.begin());
}
}
return res;
}
};
class Solution {
public:
vector<int> maxSlidingWindow(vector<int>& nums, int k) {
vector<int> res;
deque<int> buf;
for (int tail = 0; tail < nums.size(); ++tail)
{
while (!buf.empty() && buf.back() < nums[tail])
buf.pop_back();
buf.push_back(nums[tail]);
if (tail - head + 1 == k)
{
res.push_back(buf.front());