<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>

class Solution {
public:
    int secondMinimum(int n, vector<vector<int>>& edges, int time, int change) {
        int edgesSize = edges.size();
        
        // Create the adjacency list
        uint16_t **graph = (uint16_t **)malloc((n + 1) * sizeof(uint16_t *));
        int *connectionCount = (int *)calloc(n + 1, sizeof(int));

        // First pass to count connections
        for (int i = 0; i < edgesSize; i++) {
            connectionCount[edges[i][0]]++;
            connectionCount[edges[i][1]]++;
        }

        // Allocate space for each node's connections
        for (uint16_t i = 0; i <= n; i++) {
            graph[i] = (uint16_t *)malloc((connectionCount[i] + 1) * sizeof(uint16_t));
            graph[i][0] = 0; // Initialize the number of connections to 0
        }

        // Second pass to fill connections
        for (int i = 0; i < edgesSize; i++) {
            uint16_t u = edges[i][0];
            uint16_t v = edges[i][1];
            graph[u][++graph[u][0]] = v;
            graph[v][++graph[v][0]] = u;
        }

        free(connectionCount);

        // Distance array, initialized to INF
        int (*dist)[2] = (int (*)[2])malloc((n + 1) * sizeof(int[2]));
        for (uint16_t i = 0; i <= n; i++) {
            dist[i][0] = INT_MAX;
            dist[i][1] = INT_MAX;
        }

        // Priority queue implemented with a simple array
        uint16_t *queueNode = (uint16_t *)malloc(2 * n * sizeof(uint16_t));
        int *queueTime = (int *)malloc(2 * n * sizeof(int));
        int front = 0, rear = 0;

        queueNode[rear] = 1;
        queueTime[rear++] = 0;

        dist[1][0] = 0;

        while (front < rear) {
            uint16_t u = queueNode[front];
            int t = queueTime[front++];

            int timeTaken = dist[u][t];

            // Calculate the new time considering traffic signals
            int signal = timeTaken / change;
            if (signal % 2 == 1) {
                timeTaken = change * (signal + 1);
            }
            timeTaken += time;

            for (uint16_t i = 1; i <= graph[u][0]; i++) {
                uint16_t v = graph[u][i];

                if (dist[v][0] > timeTaken) {
                    dist[v][1] = dist[v][0];
                    dist[v][0] = timeTaken;
                    queueNode[rear] = v;
                    queueTime[rear++] = 0;
                } else if (dist[v][1] > timeTaken && dist[v][0] != timeTaken) {
                    dist[v][1] = timeTaken;
                    queueNode[rear] = v;
                    queueTime[rear++] = 1;
                }
            }
        }

        int result = (dist[n][1] == INT_MAX) ? -1 : dist[n][1];

        // Free allocated memory
        for (uint16_t i = 0; i <= n; i++) {
            free(graph[i]);
        }
        free(graph);
        free(dist);
        free(queueNode);
        free(queueTime);

        return result;
    }
};
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