<p>The <strong>median</strong> is the middle value in an ordered integer list. If the size of the list is even, there is no middle value. So the median is the mean of the two middle values.</p>

<ul>
	<li>For examples, if <code>arr = [2,<u>3</u>,4]</code>, the median is <code>3</code>.</li>
	<li>For examples, if <code>arr = [1,<u>2,3</u>,4]</code>, the median is <code>(2 + 3) / 2 = 2.5</code>.</li>
</ul>

<p>You are given an integer array <code>nums</code> and an integer <code>k</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 median array for each window in the original array</em>. Answers within <code>10<sup>-5</sup></code> of the actual value will be accepted.</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> [1.00000,-1.00000,-1.00000,3.00000,5.00000,6.00000]
<strong>Explanation:</strong> 
Window position                Median
---------------                -----
[<strong>1  3  -1</strong>] -3  5  3  6  7        1
 1 [<strong>3  -1  -3</strong>] 5  3  6  7       -1
 1  3 [<strong>-1  -3  5</strong>] 3  6  7       -1
 1  3  -1 [<strong>-3  5  3</strong>] 6  7        3
 1  3  -1  -3 [<strong>5  3  6</strong>] 7        5
 1  3  -1  -3  5 [<strong>3  6  7</strong>]       6
</pre>

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

<pre>
<strong>Input:</strong> nums = [1,2,3,4,2,3,1,4,2], k = 3
<strong>Output:</strong> [2.00000,3.00000,3.00000,3.00000,2.00000,3.00000,2.00000]
</pre>

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

<ul>
	<li><code>1 &lt;= k &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>-2<sup>31</sup> &lt;= nums[i] &lt;= 2<sup>31</sup> - 1</code></li>
</ul>

class Solution {
public:
    vector<double> medianSlidingWindow(vector<int>& nums, int k) {
        set<pair<int, int>> maxh; // get max element using *(maxh.rbegin());
        set<pair<int, int>> minh; // get min element using *(minh.begin());
        
        for(int i = 0; i < k; i++) {
            maxh.insert({nums[i], i}); // a {number, index} pair will always be unique (for handling duplicates)
        }
        
        for(int i = 0; i < k/2; i++) {
            minh.insert(*maxh.rbegin());
            maxh.erase(*maxh.rbegin());
        }
        
        vector<double> res;
        
        int n = nums.size();
        
        for(int i = k; i < n; i++) {
            if(k % 2 == 0) {
                res.push_back(((double)(maxh.rbegin())->first + (double)(minh.begin())->first) / 2.0);
            } else {
                res.push_back((double) (maxh.rbegin())->first);
            }
            
            // delete the element that is out of window, deletion in O(log k)
            
            if(maxh.count({nums[i-k], i-k})) {
                maxh.erase({nums[i-k], i-k});
            }
            
            if(minh.count({nums[i-k], i-k})) {
                minh.erase({nums[i-k], i-k});
            }
            
            // insert new element
            
            if(maxh.size() == 0 || nums[i] <= (maxh.rbegin())->first) {
                maxh.insert({nums[i], i});
            } else {
                minh.insert({nums[i], i});
            }
            
            // balance the sets/heaps
            
            if(minh.size() > maxh.size()) {
                pair<int, int> x = *(minh.begin());
                minh.erase(x);
                maxh.insert(x);
            }
            
            // maxh can have at max one more element than minh
            
            if(maxh.size() > minh.size()+1) {
                pair<int, int> x = *(maxh.rbegin());
                maxh.erase(x);
                minh.insert(x);
            }
        }
        
        if(k % 2 == 0) {
            res.push_back(((double)(maxh.rbegin())->first + (double)(minh.begin())->first) / 2.0);
        } else {
            res.push_back((double) (maxh.rbegin())->first);
        }
        
        return res;
    }
};
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