<p>You are given a <strong>0-indexed</strong> 2D integer array <code>grid</code> of size <code>m x n</code> that represents a map of the items in a shop. The integers in the grid represent the following:</p>

<ul>
	<li><code>0</code> represents a wall that you cannot pass through.</li>
	<li><code>1</code> represents an empty cell that you can freely move to and from.</li>
	<li>All other positive integers represent the price of an item in that cell. You may also freely move to and from these item cells.</li>
</ul>

<p>It takes <code>1</code> step to travel between adjacent grid cells.</p>

<p>You are also given integer arrays <code>pricing</code> and <code>start</code> where <code>pricing = [low, high]</code> and <code>start = [row, col]</code> indicates that you start at the position <code>(row, col)</code> and are interested only in items with a price in the range of <code>[low, high]</code> (<strong>inclusive</strong>). You are further given an integer <code>k</code>.</p>

<p>You are interested in the <strong>positions</strong> of the <code>k</code> <strong>highest-ranked</strong> items whose prices are <strong>within</strong> the given price range. The rank is determined by the <strong>first</strong> of these criteria that is different:</p>

<ol>
	<li>Distance, defined as the length of the shortest path from the <code>start</code> (<strong>shorter</strong> distance has a higher rank).</li>
	<li>Price (<strong>lower</strong> price has a higher rank, but it must be <strong>in the price range</strong>).</li>
	<li>The row number (<strong>smaller</strong> row number has a higher rank).</li>
	<li>The column number (<strong>smaller</strong> column number has a higher rank).</li>
</ol>

<p>Return <em>the </em><code>k</code><em> highest-ranked items within the price range <strong>sorted</strong> by their rank (highest to lowest)</em>. If there are fewer than <code>k</code> reachable items within the price range, return <em><strong>all</strong> of them</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/12/16/example1drawio.png" style="width: 200px; height: 151px;" />
<pre>
<strong>Input:</strong> grid = [[1,2,0,1],[1,3,0,1],[0,2,5,1]], pricing = [2,5], start = [0,0], k = 3
<strong>Output:</strong> [[0,1],[1,1],[2,1]]
<strong>Explanation:</strong> You start at (0,0).
With a price range of [2,5], we can take items from (0,1), (1,1), (2,1) and (2,2).
The ranks of these items are:
- (0,1) with distance 1
- (1,1) with distance 2
- (2,1) with distance 3
- (2,2) with distance 4
Thus, the 3 highest ranked items in the price range are (0,1), (1,1), and (2,1).
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/12/16/example2drawio1.png" style="width: 200px; height: 151px;" />
<pre>
<strong>Input:</strong> grid = [[1,2,0,1],[1,3,3,1],[0,2,5,1]], pricing = [2,3], start = [2,3], k = 2
<strong>Output:</strong> [[2,1],[1,2]]
<strong>Explanation:</strong> You start at (2,3).
With a price range of [2,3], we can take items from (0,1), (1,1), (1,2) and (2,1).
The ranks of these items are:
- (2,1) with distance 2, price 2
- (1,2) with distance 2, price 3
- (1,1) with distance 3
- (0,1) with distance 4
Thus, the 2 highest ranked items in the price range are (2,1) and (1,2).
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/12/30/example3.png" style="width: 149px; height: 150px;" />
<pre>
<strong>Input:</strong> grid = [[1,1,1],[0,0,1],[2,3,4]], pricing = [2,3], start = [0,0], k = 3
<strong>Output:</strong> [[2,1],[2,0]]
<strong>Explanation:</strong> You start at (0,0).
With a price range of [2,3], we can take items from (2,0) and (2,1). 
The ranks of these items are: 
- (2,1) with distance 5
- (2,0) with distance 6
Thus, the 2 highest ranked items in the price range are (2,1) and (2,0). 
Note that k = 3 but there are only 2 reachable items within the price range.
</pre>

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

<ul>
	<li><code>m == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= m * n &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= grid[i][j] &lt;= 10<sup>5</sup></code></li>
	<li><code>pricing.length == 2</code></li>
	<li><code>2 &lt;= low &lt;= high &lt;= 10<sup>5</sup></code></li>
	<li><code>start.length == 2</code></li>
	<li><code>0 &lt;= row &lt;= m - 1</code></li>
	<li><code>0 &lt;= col &lt;= n - 1</code></li>
	<li><code>grid[row][col] &gt; 0</code></li>
	<li><code>1 &lt;= k &lt;= m * n</code></li>
</ul>

class Solution:
    def highestRankedKItems(self, grid: List[List[int]], pricing: List[int], start: List[int], k: int) -> List[List[int]]:
        q = deque()
        m = len(grid)
        n = len(grid[0])

        h = []

        q.append((start[0], start[1]))
        d = 0
        visited = set()
        while q:
            cnt = len(q)
            for _ in range(cnt):
                x, y = q.popleft()
                if (x, y) in visited:
                    continue
                visited.add((x, y))
                if pricing[0] <= grid[x][y] <= pricing[1]:
                    heappush(h, (-1*d, -1*grid[x][y], -1*x, -1*y))
                if len(h) > k:
                    heappop(h)
                for i in [[-1,0], [1,0], [0,-1], [0,1]]:
                    dx = x + i[0]
                    dy = y + i[1]
                    if 0 <= dx < m and 0 <= dy < n and grid[dx][dy] > 0 and (dx, dy) not in visited:
                        q.append((dx, dy))
            d += 1

        
        result = [[] for _ in range(len(h))]
        idx = len(h)-1
        while h:
            d, v, x, y = heapq.heappop(h)
            result[idx] = [-1*x, -1*y]
            idx -= 1
        return result


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