<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>

// Have to reduce time complexity

/*
T:  O(E * (V+E)logV) 

Make graph without negative edges and find shortest distance from source to destination.

    Case 1: (shortest_dist < target)
            We can never make shortest_distance == target because we can not modify + edges.
    
    Case 2: (shortest_dist == target)
            Make all -ve edges == 2e9 so that they can't interfere. 
    
    Case 3: (shortest_dist > target)
            Add -ve edges one by one with value min_possible(1) and find shortest _distance.

            1. If we get shortest_dist changed to == target.
                    Make rest of the -ve edges 2e9 and return.
            2. If we get shortest_dist changed to <= target.
                    Remodify the last modified edge so that shortest_dist == target.
            3. Even after adding all negative edges with minimum value 1, we get shortest_dist > target.
                    There is no path possible return empty array.
*/


class Solution {
public:

    int dijkstra (vector<vector<pair<int,int>>>& adj,int n, int source, int dest){

        vector<int> distance(n , 2e9);

        priority_queue<vector<int>, vector<vector<int>>, greater<vector<int>> > pq;

        distance[source] = 0;

        pq.push({0, source});

        while(!pq.empty()){
            auto temp = pq.top();
            pq.pop();

            int dist = temp[0], node = temp[1];

            if(dist > distance[dest]) continue;
            // never return shortest distance inside while loop. because of multiple relaxation.

            for(auto it : adj[node]){
                int nbr = it.first, w = it.second;

                if(distance[nbr] > dist + w){
                    distance[nbr] =  dist + w;

                    pq.push({distance[nbr], nbr});
                }
            }

        }

        return distance[dest];
    }


    vector<vector<int>> modifiedGraphEdges(int n, vector<vector<int>>& edges, int source, int destination, int target) {
        
        vector<vector<pair<int,int>>> adj(n);
        
        for(int i = 0; i < edges.size(); i++){

            if(edges[i][2] != -1){
                adj[edges[i][0]].push_back({edges[i][1], edges[i][2]});
                adj[edges[i][1]].push_back({edges[i][0], edges[i][2]});
            }
        }

        int shortest = dijkstra(adj, n, source, destination);

        if(shortest < target) return {};

        else if(shortest == target){
            for(int i = 0; i < edges.size(); i++){
                if(edges[i][2] == -1) edges[i][2] = 2e9;
            }
            return edges;
        }

        // adding edges one by one until we get shortest path <= target
        else if (shortest > target){

            for(int i = 0; i < edges.size(); i++){

                if(edges[i][2] == -1){
                    edges[i][2] = 1;
                    adj[edges[i][0]].push_back({edges[i][1], edges[i][2]});
                    adj[edges[i][1]].push_back({edges[i][0], edges[i][2]});

                    shortest = dijkstra(adj, n, source, destination);

                    if(shortest <= target){

                        edges[i][2] = target-shortest+1;

                        for(int j = i+1; j < edges.size(); j++){
                            if(edges[j][2] == -1) edges[j][2] = 2e9;
                        }

                        return edges;
                    }
                }
            }

            return {};
        }

        return {};
    }
};
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