2045 - Second Minimum Time to Reach Destination

 JAVA

• class Solution {
      public int secondMinimum(int n, int[][] edges, int time, int change) {
          List<Integer>[] g = new List[n + 1];
          Arrays.setAll(g, k -> new ArrayList<>());
          for (int[] e : edges) {
              int u = e[0], v = e[1];
              g[u].add(v);
              g[v].add(u);
          }
          Deque<int[]> q = new LinkedList<>();
          q.offerLast(new int[] {1, 0});
          int[][] dist = new int[n + 1][2];
          for (int i = 0; i < n + 1; ++i) {
              Arrays.fill(dist[i], Integer.MAX_VALUE);
          }
          dist[1][1] = 0;
          while (!q.isEmpty()) {
              int[] e = q.pollFirst();
              int u = e[0], d = e[1];
              for (int v : g[u]) {
                  if (d + 1 < dist[v][0]) {
                      dist[v][0] = d + 1;
                      q.offerLast(new int[] {v, d + 1});
                  } else if (dist[v][0] < d + 1 && d + 1 < dist[v][1]) {
                      dist[v][1] = d + 1;
                      if (v == n) {
                          break;
                      }
                      q.offerLast(new int[] {v, d + 1});
                  }
              }
          }
          int ans = 0;
          for (int i = 0; i < dist[n][1]; ++i) {
              ans += time;
              if (i < dist[n][1] - 1 && (ans / change) % 2 == 1) {
                  ans = (ans + change) / change * change;
              }
          }
          return ans;
      }
  }

C++