<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 {
private:
const int INF = 1e9+1;
private:
int getShortestPath(int n, int src, int dest,vector<vector<int>>&edges) {
    // create the graph
    vector<vector<pair<int,int>>> graph(n);
    for(auto &e: edges) {
        graph[e[0]].push_back({e[1],e[2]});
        graph[e[1]].push_back({e[0],e[2]});
    }
    priority_queue<pair<int,int>> pq;
    vector<int> d(n,INF);
    d[src] = 0;
    pq.push({0,src});
    while(!pq.empty()) {
        auto top = pq.top();
        pq.pop();
        int u = top.second, dist = -top.first;
        for(auto &e: graph[u]) {
            int v = e.first;
            if(e.second== -1) continue;
            if(d[v]>dist+e.second) {
                d[v] = dist+e.second;
                pq.push({-d[v],v});
            }
        }class Solution {
public:
    vector<vector<int>> modifiedGraphEdges(int n, vector<vector<int>>& edges, int source, int destination, int target) {
        vector<vector<pair<int, int>>> adjacencyList(n);
        for (int i = 0; i < edges.size(); i++) {
            int nodeA = edges[i][0], nodeB = edges[i][1];
            adjacencyList[nodeA].emplace_back(nodeB, i);
            adjacencyList[nodeB].emplace_back(nodeA, i);
        }

        vector<vector<int>> distances(n, vector<int>(2, INT_MAX));
        distances[source][0] = distances[source][1] = 0;

        runDijkstra(adjacencyList, edges, distances, source, 0, 0);
        int difference = target - distances[destination][0];
        if (difference < 0) return {}; 
        runDijkstra(adjacencyList, edges, distances, source, difference, 1);
        if (distances[destination][1] < target) return {}; 

        for (auto& edge : edges) {
            if (edge[2] == -1) edge[2] = 1;
        }
        return edges;
    }

private:
    void runDijkstra(const vector<vector<pair<int, int>>>& adjacencyList, vector<vector<int>>& edges, vector<vector<int>>& distances, int source, int difference, int run) {
        int n = adjacencyList.size();
        priority_queue<pair<int, int>, vector<pair<int, int>>, greater<>> priorityQueue;
        priorityQueue.push({0, source});
        distances[source][run] = 0;

        while (!priorityQueue.empty()) {
            auto [currentDistance, currentNode] = priorityQueue.top();
            priorityQueue.pop();

            if (currentDistance > distances[currentNode][run]) continue;

            for (auto& neighbor : adjacencyList[currentNode]) {
                int nextNode = neighbor.first, edgeIndex = neighbor.second;
                int weight = edges[edgeIndex][2];

                if (weight == -1) weight = 1; 

                if (run == 1 && edges[edgeIndex][2] == -1) {
                    int newWeight = difference + distances[nextNode][0] - distances[currentNode][1];
                    if (newWeight > weight) {
                        edges[edgeIndex][2] = weight = newWeight;
                    }
                }

                if (distances[nextNode][run] > distances[currentNode][run] + weight) {
                    distances[nextNode][run] = distances[currentNode][run] + weight;
                    priorityQueue.push({distances[nextNode][run], nextNode});
                }
            }
        }
    }
};
        static const auto speedup = []() {
            std::ios::sync_with_stdio(false);
            std::cin.tie(nullptr);
            std::cout.tie(nullptr);
            return 0;
        }();
    }
    return d[dest];
}
public:
    vector<vector<int>> modifiedGraphEdges(int n, vector<vector<int>>& edges, int source, int destination, int target) {

        int minDistance = getShortestPath(n,source,destination,edges);
        bool matchesTarget = minDistance==target;
        // will not be possible if shortest path without -1 is already < target
        
        if(minDistance < target) return {};
       
        vector<vector<int>> res = edges;
        // if a path is at most d' = target, you can replace 
        // one of the -1's with a 1+(target-d')
        // if target distance is reached already then a -1 edge can be made >=target
        // to keep shortest path == target
        for(auto &e: res) {
            if(e[2]!=-1) continue;

            e[2] = matchesTarget? target:1;

            if(!matchesTarget) {
                int minDistance = getShortestPath(n,source,destination,res);
                if(minDistance <= target) {
                    e[2] += target-minDistance;
                    matchesTarget = true;
                }
            }

        }

        return matchesTarget? res:vector<vector<int>>{};
        
    }
};
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