<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) {
        vector<double>ans;
        unordered_map<int,int>mp;//to account for late deletions
        priority_queue<double>left; //maxHeap
        priority_queue<double,vector<double>,greater<double>>right; //minHeap
        for(int i=0;i<k;i++) {
            left.push(nums[i]);
        }

        for(int i=0;i<k/2;i++) {
            right.push(left.top());
            left.pop();
        }

        for(int i=k;i<nums.size();i++) {
            if(k%2 == 1) {
                ans.push_back(left.top());
            }else{
                ans.push_back((left.top()+right.top())/2);
            }

            int toDelete=nums[i-k];
            int toAdd=nums[i];

            //firstly we will delete the first element from the window;
            int balance=0;
            if(toDelete <= left.top()) {
                balance--;

                if(toDelete == left.top()) {    //delete it immediately
                    left.pop();
                }else{
                    mp[toDelete]++; //schedule for later deletion
                }
            }
            else{
                balance++;
                 if(toDelete == right.top()) {    //delete it immediately
                    right.pop();
                }else{
                    mp[toDelete]++; //schedule for later deletion
                }
            }

            //add the new element into the window
            if(!left.empty() && toAdd <= left.top()) {
                left.push(toAdd);
                balance++;
            }else{
                right.push(toAdd);
                balance--;
            }

            //balance the heap
            if(balance < 0) {   //means left heap has less elements
                left.push(right.top());
                right.pop();
              
            }else if(balance>0){
                right.push(left.top());
                left.pop();
                
            }

            //delete the elements scheduled for later deletion
            while(!left.empty() && mp[left.top()] > 0) {
         
                mp[left.top()]--;
                left.pop();
            }
            while(!right.empty() && mp[right.top()] > 0) {
                
                mp[right.top()]--;
                right.pop();
            }

        }
         if(k%2 == 1) {
                ans.push_back(left.top());
            }else{
                ans.push_back((left.top()+right.top())/2);
            }

        return ans;

    }
};
41670