Help | Understanding Complexity of Dijkstra's algorithm

I am having hard time in understanding the runtime of Dijkstra’s algorithm. I couldn’t get how the complexity of Dijkstra’s is E Log(V).

For the given graph, vertex 3 gets inserted twice in the priority queue, so how the size of PQ can not grow beyond |V|.

public class Dijkstra {

    public static void main(String args[]) {
        Dijkstra sol = new Dijkstra();

        System.out.println(sol.networkDelayTime(new int[][]{
                {1, 2, 1},
                {1, 5, 3},
                {1, 4, 2},
                {2, 3, 6},
                {5, 3, 1},
                {4, 2, 7},
        }, 5, 1));
    }

    public int[] dijkstra(int[][] times, int N, int K) {
        Graph g = new Graph(N, times);
        PriorityQueue<int[]> pq = new PriorityQueue<>(Comparator.comparingInt(x -> x[1]));
        int distance[] = new int[N + 1];
        boolean visited[] = new boolean[N + 1];
        Arrays.fill(distance, Integer.MAX_VALUE);
        distance[K] = 0;
        pq.offer(new int[]{K, 0});
        int time = 0;
        while (!pq.isEmpty()) {
            int node[] = pq.poll();
            List<int[]> neighbours = g.neighbours(node[0]);
            visited[node[0]] = true;
            for (int[] neighbour : neighbours) {
                int v = neighbour[0];
                int w = neighbour[1];
                if (!visited[v] && distance[v] > node[1] + w) {
                    distance[v] = node[1] + w;
                    pq.offer(new int[]{v, distance[v]});
                }
            }

        }
        return distance;
    }

    private static class Graph {
        List<List<int[]>> adj = new ArrayList<>();

        public Graph(int n, int edges[][]) {
            for (int i = 0; i <= n; i++) {
                adj.add(new ArrayList<>());
            }
            for (int[] edge : edges) {
                adj.get(edge[0]).add(new int[]{edge[1], edge[2]});
            }
        }

        List<int[]> neighbours(int u) {
            return adj.get(u);
        }
}

}