<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>

import heapq
class Solution:
    def modifiedGraphEdges(self, n: int, edges: List[List[int]], source: int, destination: int, target: int) -> List[List[int]]:
        #instead of recreating adjList everytime,try creating adjMatrix 
        # every time edge weight is modified just change edge weight two cells of the matrix
        #[u][v] , [v][u]
        
        
        def Dijkstra(source,destination,adjMatrix):
            n=len(adjMatrix)
            visited=[False for i in range(n)]
            visited[source]=True
            h=[(0,source,visited)]
            while len(h)!=0:
                out=heapq.heappop(h)
                node=out[1];dist=out[0];visited=out[2]
                if node==destination:
                    return dist
                adj=adjMatrix[node]

                for i in range(n):
                    if adj[i]==None:
                        continue
                    n1=i
                    d1=adj[i]
                    if visited[n1]:
                        continue
                    if d1==-1:
                        continue
                    if dist+d1>target:
                        continue
                    heapq.heappush(h,(dist+d1,n1,visited[:n1]+[True]+visited[n1+1:]))
            return int(1e9)

        
        adjMatrix=[[None for i in range(n)] for j in range(n)]
        for e in edges:
            u=e[0];v=e[1];d=e[2]
            adjMatrix[u][v]=d
            adjMatrix[v][u]=d
        shortest=Dijkstra(source,destination,adjMatrix)
        if shortest<target:
            return []
        elif shortest==target:
            for i in range(len(edges)):
                if edges[i][-1]==-1:
                    edges[i][-1]=target
            return edges
        for i in range(len(edges)):
            if edges[i][-1]==-1:
                edges[i][-1]=1
                u=edges[i][0]
                v=edges[i][1]
                adjMatrix[u][v]=1
                adjMatrix[v][u]=1
                shortest=Dijkstra(source,destination,adjMatrix)
                if shortest<=target:
                    if shortest<target:
                        edges[i][-1]+=target-shortest
                    for j in range(i+1,len(edges)):
                        if edges[j][-1]==-1:
                            edges[j][-1]=target
                    return edges
        return []

3709