<p>You are given an array of integers&nbsp;<code>nums</code>, there is a sliding window of size <code>k</code> which is moving from the very left of the array to the very right. You can only see the <code>k</code> numbers in the window. Each time the sliding window moves right by one position.</p>

<p>Return <em>the max sliding window</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,3,-1,-3,5,3,6,7], k = 3
<strong>Output:</strong> [3,3,5,5,6,7]
<strong>Explanation:</strong> 
Window position                Max
---------------               -----
[1  3  -1] -3  5  3  6  7       <strong>3</strong>
 1 [3  -1  -3] 5  3  6  7       <strong>3</strong>
 1  3 [-1  -3  5] 3  6  7      <strong> 5</strong>
 1  3  -1 [-3  5  3] 6  7       <strong>5</strong>
 1  3  -1  -3 [5  3  6] 7       <strong>6</strong>
 1  3  -1  -3  5 [3  6  7]      <strong>7</strong>
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1], k = 1
<strong>Output:</strong> [1]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
	<li><code>1 &lt;= k &lt;= nums.length</code></li>
</ul>

class Solution {
public:
    // Solution 2-Monotonous queue: O(n) and O(n)
    vector<int> maxSlidingWindow(vector<int>& nums, int k) {
        vector<int> ret;
        for(int i = 0; i < nums.size(); i++){
            pushIntoMonotoneQ(nums[i]);
            if(i >= k-1) {
                ret.push_back(monoQ.front().first);
                popFromMonotoneQ();
            }
        }
        return ret;
    }
private:
    deque<pair<int, int>> monoQ;// pair.first = actual value
                                // pair.second = count of the number of elements popped before placing a number into the back of the deque.
    void pushIntoMonotoneQ(int val){
        int count = 0;
        while(!monoQ.empty() and val > monoQ.back().first){
            count += monoQ.back().second + 1;
            monoQ.pop_back();
        }
        monoQ.emplace_back(val, count);
            
    }
    void popFromMonotoneQ(){
        if(monoQ.front().second > 0){
            monoQ.front().second--;
            return;
        }
        monoQ.pop_front();
    }
};

/*
class Solution {
public:
    // Solution 1- Brute force: O(nk) and O(1) | TLE
    vector<int> maxSlidingWindow(vector<int>& nums, int k) {
        vector<int> ret;
        for(int i = 0; i < nums.size() - k + 1; i++){
            int maxx = INT_MIN;
            for(int j = i; j < i + k; j++){
                maxx = max(maxx, nums[j]);
            }
            ret.push_back(maxx);
        }
        
        return ret;
    }
};
*/
140756