<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>

INFINITY = 2**31
DIR_FWD = 0
DIR_RVS = 1
DIR_BREAK = 2

class Solution:
    def modifiedGraphEdges(self, n: int, edges: List[List[int]], source: int, destination: int, target: int) -> List[List[int]]:
        class Node:
            def __init__(self, id):
                self.id = id
                self.distance = [INFINITY, INFINITY, INFINITY]
                self.wildcards = 0
                self.neighbors = {}
                self.processed = -1

            @staticmethod
            def link(nodeA, nodeB, metric):
                nodeA.neighbors[nodeB] = metric
                nodeB.neighbors[nodeA] = metric

            def __lt__(self, other):
                return self.id < other.id

            def __hash__(self):
                return self.id

        def edgeHash(nidA, nidB):
            return min(nidA, nidB) * 1000 + max(nidA, nidB)
        
        def doDijkstra(nodes, source, dest, direction, target):
            nodes[source].distance[direction] = 0
            heap = [(0, nodes[source])]
            bestA, bestB, bestVal = None, None, target + 1
            while len(heap) > 0:
                (_, node) = heappop(heap)
                if node.processed == direction:
                    continue
                if node.distance[direction] == INFINITY or node.id == dest:
                    break
                for (nbr, metric) in node.neighbors.items():
                    if nbr.processed == direction:
                        continue
                    if direction == DIR_BREAK and metric == -1:
                        continue
                    if nbr.distance[direction] > node.distance[direction] + abs(metric):
                        nbr.distance[direction] = node.distance[direction] + abs(metric)
                        nbr.wildcards = node.wildcards
                        if metric == -1:
                            nbr.wildcards += 1
                            if direction == DIR_RVS:
                                val = target - (nbr.distance[DIR_FWD] + node.distance[DIR_RVS])
                                if val < bestVal:
                                    bestA, bestB, bestVal = nbr.id, node.id, val
                        heappush(heap, (nbr.distance[direction], nbr))
                node.processed = direction
            return (bestA, bestB, bestVal)

        nodes = [Node(i) for i in range(n)]
        edgeMap = {}
        modifiableEdges = set()
        for i in range(len(edges)):
            eH = edgeHash(edges[i][0], edges[i][1])
            edgeMap[eH] = edges[i][2]
            Node.link(nodes[edges[i][0]], nodes[edges[i][1]], edges[i][2])
            if edges[i][2] == -1:
                modifiableEdges.add(eH)

        doDijkstra(nodes, source, destination, DIR_BREAK, target)
        if nodes[destination].distance[DIR_BREAK] < target:
            return [] # Immutable shortest path exists under target
            
        while True:
            for node in nodes:
                node.distance = [INFINITY, INFINITY]
                node.processed = -1
            doDijkstra(nodes, source, destination, DIR_FWD, target)
            if nodes[destination].distance[DIR_FWD] == target:                
                break
            if nodes[destination].distance[DIR_FWD] == INFINITY:
                return [] # Disconnected
            if nodes[destination].distance[DIR_FWD] != target and nodes[destination].wildcards == 0:
                return [] # Shortest path immutable and wrong
            if nodes[destination].distance[DIR_FWD] > target:
                return [] # Can't improve to reach target

            (A, B, val) = doDijkstra(nodes, destination, source, DIR_RVS, target)
            if A is None:
                return [] # No viable modifications

            eH = edgeHash(A, B)
            modifiableEdges.remove(eH)
            edgeMap[eH] = val
            Node.link(nodes[A], nodes[B], val)

        return [[eH // 1000, eH % 1000, abs(edgeMap[eH])] for eH in edgeMap.keys()]

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