<p>There is an infrastructure of <code>n</code> cities with some number of <code>roads</code> connecting these cities. Each <code>roads[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> indicates that there is a bidirectional road between cities <code>a<sub>i</sub></code> and <code>b<sub>i</sub></code>.</p>

<p>The <strong>network rank</strong><em> </em>of <strong>two different cities</strong> is defined as the total number of&nbsp;<strong>directly</strong> connected roads to <strong>either</strong> city. If a road is directly connected to both cities, it is only counted <strong>once</strong>.</p>

<p>The <strong>maximal network rank </strong>of the infrastructure is the <strong>maximum network rank</strong> of all pairs of different cities.</p>

<p>Given the integer <code>n</code> and the array <code>roads</code>, return <em>the <strong>maximal network rank</strong> of the entire infrastructure</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2020/09/21/ex1.png" style="width: 292px; height: 172px;" /></strong></p>

<pre>
<strong>Input:</strong> n = 4, roads = [[0,1],[0,3],[1,2],[1,3]]
<strong>Output:</strong> 4
<strong>Explanation:</strong> The network rank of cities 0 and 1 is 4 as there are 4 roads that are connected to either 0 or 1. The road between 0 and 1 is only counted once.
</pre>

<p><strong class="example">Example 2:</strong></p>

<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2020/09/21/ex2.png" style="width: 292px; height: 172px;" /></strong></p>

<pre>
<strong>Input:</strong> n = 5, roads = [[0,1],[0,3],[1,2],[1,3],[2,3],[2,4]]
<strong>Output:</strong> 5
<strong>Explanation:</strong> There are 5 roads that are connected to cities 1 or 2.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> n = 8, roads = [[0,1],[1,2],[2,3],[2,4],[5,6],[5,7]]
<strong>Output:</strong> 5
<strong>Explanation:</strong> The network rank of 2 and 5 is 5. Notice that all the cities do not have to be connected.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= n &lt;= 100</code></li>
	<li><code>0 &lt;= roads.length &lt;= n * (n - 1) / 2</code></li>
	<li><code>roads[i].length == 2</code></li>
	<li><code>0 &lt;= a<sub>i</sub>, b<sub>i</sub>&nbsp;&lt;= n-1</code></li>
	<li><code>a<sub>i</sub>&nbsp;!=&nbsp;b<sub>i</sub></code></li>
	<li>Each&nbsp;pair of cities has <strong>at most one</strong> road connecting them.</li>
</ul>

class Solution {
public:
    int maximalNetworkRank(int n, vector<vector<int>>& roads) {
        //calculate sum  indegree of each pair that will be answer
        unordered_map<int,list<int>>adj;
        vector<int>indegree(n,0);
        for(auto it:roads){
            int u=it[0];
            int v=it[1];
            indegree[u]++;
             indegree[v]++;
            adj[u].push_back(v);
            adj[v].push_back(u);
        }

        int ans=0;
    //     for(auto it:roads){
    //         int u=it[0];
    //         int v=it[1];
    //         int a=indegree[u];
    //         int b=indegree[v];
    //         //if(roads.find({}))
    //         int sum=a+b;
    //         ans=max(sum,ans);
    //     }
    //     return ans;
    // }
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            int sum = indegree[i] + indegree[j];
            // If the two cities are directly connected, subtract 1
            if (find(adj[i].begin(), adj[i].end(), j) != adj[i].end()) {
                sum--;
            }
            ans = max(sum, ans);
        }
    }

    return ans;
    }
};
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