<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>

from math import inf
from collections import deque
def bfs(g, d, par):
    q = deque([1])
    d[1] = 0
    while q:
        x = q.popleft()
        for i in g[x]:
            if d[i]>d[x]+1:
                d[i] = d[x]+1
                q.append(i)
                par[i] = x
class Solution:
    def secondMinimum(self, n: int, edges: List[List[int]], time: int, change: int) -> int:
        g = [[] for i in range(n+1)]
        for u, v in edges:
            g[u].append(v)
            g[v].append(u)
        d = [inf]*(n+1)
        par = [-1]*(n+1)
        bfs(g, d, par)
        tt = d[n]+2
        q = deque([n])
        f = 1
        while q and f:
            x = q.popleft()
            for i in g[x]:
                if d[i]==d[x]-1:
                    q.append(i)
                elif d[i]==d[x]:
                    tt -=1
                    f = 0
                    break
        #print(tt)
        t = 0
        for i in range(tt):
            y = t//change
            if y%2:
                t += (change-t%change)
            t += time
        return t
992