<p>You are given an <strong>undirected weighted</strong> <strong>connected</strong> graph containing <code>n</code> nodes labeled from <code>0</code> to <code>n - 1</code>, and an integer array <code>edges</code> where <code>edges[i] = [a<sub>i</sub>, b<sub>i</sub>, w<sub>i</sub>]</code> indicates that there is an edge between nodes <code>a<sub>i</sub></code> and <code>b<sub>i</sub></code> with weight <code>w<sub>i</sub></code>.</p>

<p>Some edges have a weight of <code>-1</code> (<code>w<sub>i</sub> = -1</code>), while others have a <strong>positive</strong> weight (<code>w<sub>i</sub> &gt; 0</code>).</p>

<p>Your task is to modify <strong>all edges</strong> with a weight of <code>-1</code> by assigning them <strong>positive integer values </strong>in the range <code>[1, 2 * 10<sup>9</sup>]</code> so that the <strong>shortest distance</strong> between the nodes <code>source</code> and <code>destination</code> becomes equal to an integer <code>target</code>. If there are <strong>multiple</strong> <strong>modifications</strong> that make the shortest distance between <code>source</code> and <code>destination</code> equal to <code>target</code>, any of them will be considered correct.</p>

<p>Return <em>an array containing all edges (even unmodified ones) in any order if it is possible to make the shortest distance from </em><code>source</code><em> to </em><code>destination</code><em> equal to </em><code>target</code><em>, or an <strong>empty array</strong> if it&#39;s impossible.</em></p>

<p><strong>Note:</strong> You are not allowed to modify the weights of edges with initial positive weights.</p>

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

<p><strong class="example"><img alt="" src="https://assets.leetcode.com/uploads/2023/04/18/graph.png" style="width: 300px; height: 300px;" /></strong></p>

<pre>
<strong>Input:</strong> n = 5, edges = [[4,1,-1],[2,0,-1],[0,3,-1],[4,3,-1]], source = 0, destination = 1, target = 5
<strong>Output:</strong> [[4,1,1],[2,0,1],[0,3,3],[4,3,1]]
<strong>Explanation:</strong> The graph above shows a possible modification to the edges, making the distance from 0 to 1 equal to 5.
</pre>

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

<p><strong class="example"><img alt="" src="https://assets.leetcode.com/uploads/2023/04/18/graph-2.png" style="width: 300px; height: 300px;" /></strong></p>

<pre>
<strong>Input:</strong> n = 3, edges = [[0,1,-1],[0,2,5]], source = 0, destination = 2, target = 6
<strong>Output:</strong> []
<strong>Explanation:</strong> The graph above contains the initial edges. It is not possible to make the distance from 0 to 2 equal to 6 by modifying the edge with weight -1. So, an empty array is returned.
</pre>

<p><strong class="example">Example 3:</strong></p>

<p><strong class="example"><img alt="" src="https://assets.leetcode.com/uploads/2023/04/19/graph-3.png" style="width: 300px; height: 300px;" /></strong></p>

<pre>
<strong>Input:</strong> n = 4, edges = [[1,0,4],[1,2,3],[2,3,5],[0,3,-1]], source = 0, destination = 2, target = 6
<strong>Output:</strong> [[1,0,4],[1,2,3],[2,3,5],[0,3,1]]
<strong>Explanation:</strong> The graph above shows a modified graph having the shortest distance from 0 to 2 as 6.
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 100</code></li>
	<li><code><font face="monospace">1 &lt;= edges.length &lt;= n * (n - 1) / 2</font></code></li>
	<li><code>edges[i].length == 3</code></li>
	<li><code>0 &lt;= a<sub>i</sub>, b<sub>i&nbsp;</sub>&lt;&nbsp;n</code></li>
	<li><code><font face="monospace">w<sub>i</sub>&nbsp;= -1&nbsp;</font></code>or <code><font face="monospace">1 &lt;= w<sub>i&nbsp;</sub>&lt;= 10<sup><span style="font-size: 10.8333px;">7</span></sup></font></code></li>
	<li><code>a<sub>i&nbsp;</sub>!=&nbsp;b<sub>i</sub></code></li>
	<li><code>0 &lt;= source, destination &lt; n</code></li>
	<li><code>source != destination</code></li>
	<li><code><font face="monospace">1 &lt;= target &lt;= 10<sup>9</sup></font></code></li>
	<li>The graph is connected, and there are no self-loops or repeated edges</li>
</ul>

from queue import PriorityQueue

class Solution:
    def modifiedGraphEdges(self, n: int, edges: List[List[int]], source: int, destination: int, target: int) -> List[List[int]]:
        edges_adj = [[] for _ in range(n)]
        distances = [[] for _ in range(n)]

        empty_edges_dest = {}

        for (a, b, w) in edges:
            edges_adj[a].append((b,w))
            edges_adj[b].append((a,w))
        
        queue = PriorityQueue()

        queue.put(((0, 0), (source, [])))
        while not queue.empty():
            (num_empty, dist), (vertex, empty_edges) = queue.get()
            if (not distances[vertex]) or (distances[vertex][-1][1] > dist):
                distances[vertex].append((num_empty, dist))
                if vertex == destination:
                    empty_edges_dest[num_empty] = empty_edges

                for neighbour, weight in edges_adj[vertex]:
                    if weight == -1:
                        queue.put(((num_empty+1, dist), (neighbour, empty_edges + [(vertex, neighbour)])))
                    else:
                        queue.put(((num_empty, dist+weight), (neighbour, empty_edges)))

        for num_empty, dist in distances[destination]:
            if (num_empty == 0) and (dist < target):
                return []

            elif (num_empty == 0) and (dist == target):
                final_edges = []
                for (a,b,w) in edges:
                    if w == -1:
                        final_edges.append((a,b,target))
                    else:
                        final_edges.append((a,b,w))
                return final_edges

            elif num_empty + dist > target:
                continue

            empty_edges = empty_edges_dest[num_empty]

            final_edges = []
            for (a,b,w) in edges:
                index = None
                if (a,b) in empty_edges:
                    index = empty_edges.index((a,b))
                elif (b,a) in empty_edges:
                    index = empty_edges.index((b,a))
                
                if index is None:
                    if w == -1:
                        final_edges.append((a,b,target))
                    else:
                        final_edges.append((a,b,w))
                elif index == 0:
                    final_edges.append((a,b,target-dist-num_empty+1))
                else:
                    final_edges.append((a,b,1))

            return final_edges
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