<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>

bool G[10001];

class Solution {
public:
    int maximalNetworkRank(int n, vector<vector<int>>& roads) {
        for (int i = 0; i < n * n; ++i) {
            G[i] = 0;
        }
        int counts[101] = {};
        for (auto & road : roads) {
            counts[road[0]] += 1;
            counts[road[1]] += 1;
            G[road[0] * n + road[1]] = 1;
            G[road[1] * n + road[0]] = 1;
        }
        int a = 0;
        int b = 0;
        for (int i = 0; i < n; ++i) {
            const int c = counts[i];
            if (c > a) {
                b = a;
                a = c;
            } else if (c > b) {
                b = c;
            }
        }
        int result = 0;
        for (int i = 0; i < n; ++i) {
            //if (counts[i] < b) {
            //    continue;
            //}
            for (int j = i + 1; j < n; ++j) {
                //if (counts[j] < b) {
                //    continue;
                //}
                //cout << i << " - " << j << ", " << G[i * 100 + j] << endl;
                result = std::max(counts[i] + counts[j] - G[i * n + j], result);
            }
        }
        return result;
    }
};
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