<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>

class Solution:
    def modifiedGraphEdges(self, n: int, edges: List[List[int]], source: int, destination: int, target: int) -> List[List[int]]:

        def matrixToEdgeList(matrix):
            ans = []

            for u,v,w in edges:
                ans.append([u,v,original[u][v]])

            return ans

        

        def dijkstra(graph, source, destination, n=0):
            # print(graph)
            nodes = max(len(graph), n)
            dist = [float("inf") for _ in range(nodes)]
            parent = [i for i in range(nodes)]
            visited = set()

            q = []
            q.append([0, source])
            dist[source] = 0
            parent[source] = None

            while q:

                distance, node = heapq.heappop(q)
                if node in visited:
                    continue
                
                visited.add(node)

                for neigh, wt in graph[node]:

                    if neigh not in visited and dist[neigh] > dist[node] + wt:
                        dist[neigh] = dist[node] + wt
                        parent[neigh] = node
                        heapq.heappush(q, [dist[neigh], neigh])

            return dist, parent


        graphIgnoredNegative = defaultdict(list)
        graph = defaultdict(list)
        original = [[0 for _ in range(n)] for __ in range(n)]

        for u, v, w in edges:
            original[u][v] = w
            original[v][u] = w
            if w == -1:
                weightIgnore = float("inf")
            else:
                weightIgnore = w
            if w == -1:
                w = 1
            graph[u].append([v, w])
            graph[v].append([u, w])
            graphIgnoredNegative[u].append([v, weightIgnore])
            graphIgnoredNegative[v].append([u, weightIgnore])
            
        dist, parent = dijkstra(graphIgnoredNegative, source, destination)
        if dist[destination] < target:
            return []
        
        dist, parent = dijkstra(graph, source, destination)
        # print(dist)
        # print(parent)

        if dist[destination] > target:
            return []
        
        if dist[destination] == target:
            for i in range(len(original)):
                for j in range(len(original)):
                    if original[i][j] == -1:
                        original[i][j] = 1
                        original[j][i] = 1

            return matrixToEdgeList(original)

        originalNegativeCount = 0
        currentNode = destination
        while currentNode != None:

            parentNode = parent[currentNode]

            # print(currentNode, parentNode)

            if parentNode != None and original[currentNode][parentNode] == -1:
                originalNegativeCount += 1
            
            currentNode = parentNode

        # print(originalNegativeCount)

        if originalNegativeCount == 0:
            return []

        currentNode = destination
        while currentNode != None:

            parentNode = parent[currentNode]

            # print(currentNode, parentNode)

            if parentNode != None and original[currentNode][parentNode] == -1:
                lastSet = [currentNode, parentNode]
                original[currentNode][parentNode] = 1
                original[parentNode][currentNode] = 1

                tempGraph = defaultdict(list)
                for i in range(n):
                    for j in range(i, n):
                        if original[i][j] > 0:
                            tempGraph[i].append([j, original[i][j]])
                            tempGraph[j].append([i, original[j][i]])
                # print("Dfr", source)
                d,p = dijkstra(tempGraph, source, destination,n=n)
                # print(d, p)
                if d[destination] < target:
                    break
            
            currentNode = parentNode

        # last value handling
        remaining = target - d[destination]
        # print(remaining)
        u, v = lastSet
        original[u][v] += remaining
        original[v][u] += remaining

        for i in range(len(original)):
            for j in range(len(original)):

                if original[i][j] == -1:
                    original[i][j] = 2000000000
                    original[j][i] = 2000000000



        return matrixToEdgeList(original)
21896