<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 {
public:
    vector<vector<int>> modifiedGraphEdges(int n, vector<vector<int>>& edges, int source, int destination, int target) {
        vector<vector<pair<int,int>>> adjList(n);

        for(int i = 0; i < edges.size(); i++){
            vector<int> edge = edges[i];
            int a = edge[0];
            int b = edge[1];
            int w = edge[2];
            if(w == -1) continue;
            adjList[a].push_back({b, w});
            adjList[b].push_back({a, w});
        }

        int d1 = getShortestPath(adjList, source, destination, n);
        if(d1 < target) return {};
        bool found = (d1 == target);

        for(int i = 0; i < edges.size(); i++){
            vector<int> edge = edges[i];
            int a = edge[0];
            int b = edge[1];
            int w = edge[2];
            if(w != -1) continue;
            adjList[a].push_back({b, 1});
            adjList[b].push_back({a, 1});
            edges[i][2] = found ? 2e9 : 1;
            if(!found){
                int d = getShortestPath(adjList, source, destination, n);
                if(d <= target){
                    edges[i][2] += (target-d);
                    found = true;
                }
            }
        }
        if(!found) return {};
        return edges;
    }
private:
    struct Compare {
        bool operator()(const pair<int, int>& a, const pair<int, int>& b){
            return a.second > b.second;
        }
    };

    int getShortestPath(vector<vector<pair<int,int>>> adjList, int source, int destination, int n) {
        vector<pair<int, int>> distances(n, {-1, INT_MAX});
        priority_queue<pair<int, int>, vector<pair<int, int>>, Compare> pq;

        distances[source] = {-1, 0};
        pq.push({source, 0});

        while(pq.size() > 0){
            pair<int, int> top = pq.top(); pq.pop();
            int node = top.first;
            int dist = top.second;

            vector<pair<int,int>> children = adjList[node];

            for(auto& it: children){
                int child = it.first;
                int w = it.second;
                if(w == -1) continue;
                int newDist = w + dist;
                if(newDist < distances[child].second){
                    distances[child] = {node, newDist};
                    pq.push({child, newDist});
                }
            }
        }

        return distances[destination].second;
    }
};

/*
        0 -> 1
        | \  |
        v  \ |
        3 -> 2

    shortest distance from source to destination is target after modification

    get the shortest distance (D) from source to destination by changing all the -1 to 1

    if the D is less than target and there exists an edge between source and

    we can change -1 to [1, 2,000,000,000]

    how do we assign the appropriate weights to the edges to return?

    we can keep track of the parent that leads to the shortest distance, then we do dfs again
*/
740957