<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:

    /*
    Intuition :
    See contraints.
    Learn from failed Test Cases (edges cases).

    T.C : O(n^2 * logn)
    */

    // djisktra {O(nlogn)}
    int solve(int src, int dest, int n, vector<pair<int,int>>adj[], int val){
        vector<int> dist(n, INT_MAX);

        set<pair<int,int>> st;
        st.insert({0, src});
        dist[src] = 0;

        while(!st.empty()){
            int node = st.begin()->second;
            st.erase(st.begin());

            for(auto [child, cost] : adj[node]){
                if(cost==-1){
                    if(val==-1){continue;}
                    else{cost = val;}
                }

                int newDist = dist[node] + cost;

                if(newDist < dist[child]){
                    dist[child] = newDist;
                    st.insert({newDist, child});
                }
            }
        }

        return dist[dest];
    }

    vector<vector<int>> modifiedGraphEdges(int n, vector<vector<int>>& edges, int src, int dest, int tar) {
        vector<pair<int,int>> adj[n];
        vector<int> wildEdges;

        for(int i=0; i<edges.size(); i++){
            adj[edges[i][0]].push_back({edges[i][1], edges[i][2]});
            adj[edges[i][1]].push_back({edges[i][0], edges[i][2]});
            if(edges[i][2]==-1){
                wildEdges.push_back(i);
            }
        }

        int dist = solve(src, dest, n, adj, -1);

        if(dist<tar){return {};}    //i.e already more shortest path exits.
        if(dist==tar){              //i.e path with cost=target already exists. 
            vector<vector<int>> res = edges;
            for(int i=0; i<res.size(); i++){
                if(res[i][2]==-1){res[i][2] = tar+1;}
            }
            return res;
        }

        dist = solve(src, dest, n, adj, 1); //by considering smallest-value(i.e 1) for wildEdges,
        if(dist>tar){return {};}            // shortest path is still greater than target.
                                      
        // try djisktra after making each wildEdges val=1, One at a time,
        for(int idx : wildEdges){
            int u = edges[idx][0], v = edges[idx][1];

            edges[idx][2] = 1;

            for(auto &[child, wt] : adj[u]){
                if(child==v){wt = 1; break;}
            }
            for(auto &[child, wt] : adj[v]){
                if(child==u){wt = 1; break;}
            }

            dist = solve(src, dest, n, adj, -1);

            if(dist<=tar){
                edges[idx][2] = (tar-dist+1);

                for(int idx : wildEdges){
                    if(edges[idx][2]==-1){
                        edges[idx][2] = tar + 1;
                    }
                }

                return edges;
            }
        }

        return {};
    }
};
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