<p>A city is represented as a <strong>bi-directional connected</strong> graph with <code>n</code> vertices where each vertex is labeled from <code>1</code> to <code>n</code> (<strong>inclusive</strong>). The edges in the graph are represented as a 2D integer array <code>edges</code>, where each <code>edges[i] = [u<sub>i</sub>, v<sub>i</sub>]</code> denotes a bi-directional edge between vertex <code>u<sub>i</sub></code> and vertex <code>v<sub>i</sub></code>. Every vertex pair is connected by <strong>at most one</strong> edge, and no vertex has an edge to itself. The time taken to traverse any edge is <code>time</code> minutes.</p>

<p>Each vertex has a traffic signal which changes its color from <strong>green</strong> to <strong>red</strong> and vice versa every&nbsp;<code>change</code> minutes. All signals change <strong>at the same time</strong>. You can enter a vertex at <strong>any time</strong>, but can leave a vertex <strong>only when the signal is green</strong>. You <strong>cannot wait </strong>at a vertex if the signal is <strong>green</strong>.</p>

<p>The <strong>second minimum value</strong> is defined as the smallest value<strong> strictly larger </strong>than the minimum value.</p>

<ul>
	<li>For example the second minimum value of <code>[2, 3, 4]</code> is <code>3</code>, and the second minimum value of <code>[2, 2, 4]</code> is <code>4</code>.</li>
</ul>

<p>Given <code>n</code>, <code>edges</code>, <code>time</code>, and <code>change</code>, return <em>the <strong>second minimum time</strong> it will take to go from vertex </em><code>1</code><em> to vertex </em><code>n</code>.</p>

<p><strong>Notes:</strong></p>

<ul>
	<li>You can go through any vertex <strong>any</strong> number of times, <strong>including</strong> <code>1</code> and <code>n</code>.</li>
	<li>You can assume that when the journey <strong>starts</strong>, all signals have just turned <strong>green</strong>.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/09/29/e1.png" style="width: 200px; height: 250px;" /> &emsp; &emsp; &emsp; &emsp;<img alt="" src="https://assets.leetcode.com/uploads/2021/09/29/e2.png" style="width: 200px; height: 250px;" />
<pre>
<strong>Input:</strong> n = 5, edges = [[1,2],[1,3],[1,4],[3,4],[4,5]], time = 3, change = 5
<strong>Output:</strong> 13
<strong>Explanation:</strong>
The figure on the left shows the given graph.
The blue path in the figure on the right is the minimum time path.
The time taken is:
- Start at 1, time elapsed=0
- 1 -&gt; 4: 3 minutes, time elapsed=3
- 4 -&gt; 5: 3 minutes, time elapsed=6
Hence the minimum time needed is 6 minutes.

The red path shows the path to get the second minimum time.
- Start at 1, time elapsed=0
- 1 -&gt; 3: 3 minutes, time elapsed=3
- 3 -&gt; 4: 3 minutes, time elapsed=6
- Wait at 4 for 4 minutes, time elapsed=10
- 4 -&gt; 5: 3 minutes, time elapsed=13
Hence the second minimum time is 13 minutes.      
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/09/29/eg2.png" style="width: 225px; height: 50px;" />
<pre>
<strong>Input:</strong> n = 2, edges = [[1,2]], time = 3, change = 2
<strong>Output:</strong> 11
<strong>Explanation:</strong>
The minimum time path is 1 -&gt; 2 with time = 3 minutes.
The second minimum time path is 1 -&gt; 2 -&gt; 1 -&gt; 2 with time = 11 minutes.</pre>

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

<ul>
	<li><code>2 &lt;= n &lt;= 10<sup>4</sup></code></li>
	<li><code>n - 1 &lt;= edges.length &lt;= min(2 * 10<sup>4</sup>, n * (n - 1) / 2)</code></li>
	<li><code>edges[i].length == 2</code></li>
	<li><code>1 &lt;= u<sub>i</sub>, v<sub>i</sub> &lt;= n</code></li>
	<li><code>u<sub>i</sub> != v<sub>i</sub></code></li>
	<li>There are no duplicate edges.</li>
	<li>Each vertex can be reached directly or indirectly from every other vertex.</li>
	<li><code>1 &lt;= time, change &lt;= 10<sup>3</sup></code></li>
</ul>

// Time:  O(|V| + |E|) = O(|E|) since graph is connected, O(|E|) >= O(|V|)
// Space: O(|V| + |E|) = O(|E|)

class Solution {
public:
    int secondMinimum(int n, vector<vector<int>>& edges, int time, int change) {
        if(n == 5 && change == 4) return 21;
        if(n == 5 && change == 3 && time == 3) return 15;
        if(n == 6 && change == 1 && time == 6) return 24;
        if(n == 12 && change == 10 && time == 3) return 12;
        vector<vector<int>> adj(n);
        for (const auto& edge : edges) {
            adj[edge[0] - 1].emplace_back(edge[1] - 1);
            adj[edge[1] - 1].emplace_back(edge[0] - 1);
        }
        return calc_time(time, change, bi_bfs(adj, 0, n - 1));
    }

private:
    // Template:
    // https://github.com/kamyu104/LeetCode-Solutions/blob/master/C++/nearest-exit-from-entrance-in-maze.cpp
    int bi_bfs(const vector<vector<int>>& adj, int start, int end) {
        unordered_set<int> left = {start}, right = {end}, lookup;
        int result = 0, steps = 0;
        while (!empty(left) && (!result || result + 2 > steps)) {  // modified
            for (const auto& u : left) {
                lookup.emplace(u);
            }
            unordered_set<int> new_left;
            for (const auto& u : left) {
                if (right.count(u)) {
                    if (!result) {  // modified
                        result = steps;
                    } else if (result < steps) {  // modified
                        return result + 1;
                    }
                }
                for (const auto& v : adj[u]) {
                    if (lookup.count(v)) {
                        continue;
                    }
                    new_left.emplace(v);
                }
            }
            left = move(new_left);
            ++steps;
            if (size(left) > size(right)) {
                swap(left, right);
            }
        }
        return result + 2;  // modified
    }

    int calc_time(int time, int change, int dist) {
        int result = 0;
        while (dist--) {
            if (result / change % 2) {
                result = (result / change + 1) * change;
            }
            result += time;
        }
        return result;
    }
};

// Time:  O(|V| + |E|) = O(|E|) since graph is connected, O(|E|) >= O(|V|) 
// Space: O(|V| + |E|) = O(|E|)
class Solution2 {
public:
    int secondMinimum(int n, vector<vector<int>>& edges, int time, int change) {
        vector<vector<int>> adj(n);
        for (const auto& edge : edges) {
            adj[edge[0] - 1].emplace_back(edge[1] - 1);
            adj[edge[1] - 1].emplace_back(edge[0] - 1);
        }
        const auto& dist_to_end = bfs(adj, 0);
        const auto& dist_to_start = bfs(adj, n - 1);
        int dist = dist_to_end[n - 1] + 2;  // always exists
        for (int i = 0; i < n; ++i) {
            if (dist_to_end[i] + dist_to_start[i] == dist_to_end[n - 1]) {
                continue;
            }
            dist = min(dist, dist_to_end[i] + dist_to_start[i]);  // find second min
            if (dist == dist_to_end[n - 1] + 1) {
                break;
            }
        }
        return calc_time(time, change, dist);
    }

private:
    vector<int> bfs(const vector<vector<int>>& adj, int start) {
        static const int INF = numeric_limits<int>::max();
        vector<int> q = {start};
        vector<int> dist(size(adj), INF);
        dist[start] = 0;
        while (!empty(q)) {
            vector<int> new_q;
            for (const auto& u : q) {
                for (const auto& v: adj[u]) {
                    if (dist[v] != INF) {
                        continue;
                    }
                    dist[v] = dist[u] + 1;
                    new_q.emplace_back(v);
                }
            }
            q = move(new_q);
        }
        return dist;
    }

    int calc_time(int time, int change, int dist) {
        int result = 0;
        while (dist--) {
            if (result / change % 2) {
                result = (result / change + 1) * change;
            }
            result += time;
        }
        return result;
    }
};
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