<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>

#include <bits/stdc++.h>
using namespace std; 

#define ll long long
#define pii pair<int, int>

#pragma GCC optimize("Ofast")
#pragma GCC optimize ("unroll-loops")

class Solution {
public:
    vector<vector<int>> modifiedGraphEdges(int n, vector<vector<int>>& E, int st, int en, int targ) {
        ios_base::sync_with_stdio(0); cin.tie(0);  

        vector<int> unknown;
        vector<vector<int>> adj(n, vector<int>(n, 2e9 + 5));

        for (int i = 0; i < E.size(); i++){
            if (E[i][2] == -1)
                unknown.push_back(i);
        }
        auto sp = [&](){
            vector<int> dist(n, 2e9 + 5);
            vector<bool> vis(n, false);

            for (auto e : E){
                int a = e[0];
                int b = e[1];
                int w = e[2];
                adj[a][b] = adj[b][a] = w;
            }

            dist[st] = 0;

            for (int i = 0; i < n; i++){
                int best = -1;

                for (int j = 0; j < n; j++)
                    if (!vis[j] and (best == -1 or dist[j] < dist[best]))
                        best = j;
                
                if (dist[best] == 2e9 + 5)
                    break;
                
                vis[best] = true;

                for (int j = 0; j < n; j++){
                    dist[j] = min((ll)dist[j], (ll)dist[best] + adj[best][j]);
                }
            }
            return dist[en];
        };

        int same = 0;

        auto getVal = [&](int val){
            for (int i : unknown)
                E[i][2] = val;
            
            return sp();
        };
        auto chk1 = [&](int val){
            return getVal(val) <= targ;
        };

        auto chk2 = [&](int pref){
            for (int i = 0; i < unknown.size(); i++){
                E[unknown[i]][2] = same + (i < pref);
            }
            return sp() <= targ;
        };
        
        // Figure out what to set everything to
        {
            if (getVal(1e9 + 5) < targ or getVal(0) > targ)
                return vector<vector<int>>();
            
            int L = 0, H = 1e9 + 5;

            while (L < H){
                int M = (L + H + 1) / 2;
                chk1(M) ? L = M : H = M - 1;
            }

            same = L;                
        }

        // Figure out which prefix to increase by 1 (<pref)
        
        int L = 0, H = unknown.size();

        while (L < H){
            int M = (L + H + 1) / 2;
            chk2(M) ? L = M : H = M - 1;
        }
        chk2(L);

        if (sp() != targ)
            return vector<vector<int>>();
        
        return E;
    }
};
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