give sql query

Here’s a straightforward solution that returns the names of all trainers who have conducted at least one training, in the same order they appear in the trainer table (i.e., ascending trainer_id):

sql
SELECT
    t.trainer_name
FROM
    trainer AS t
WHERE
    EXISTS ( SELECT 1
             FROM candidate_appearance AS ca
             WHERE ca.trainer_id = t.trainer_id )
ORDER BY
    t.trainer_id;

• EXISTS (or IN, if you prefer) filters to trainers whose trainer_id appears at least once in candidate_appearance.
• ORDER BY t.trainer_id preserves the original insertion order of the trainer table.

Below is a clean, O(V + E) solution in Java that counts articulation points (a.k.a. cut-vertices) in the undirected power-grid graph.
An articulation point is a “critical” point whose failure would split the grid into two or more disconnected parts, so the count equals the number of fail-safe machines Powa needs to install.

java
import java.io.*;
import java.util.*;

/**
 * Solve the “critical points in a power grid” problem.
 *
 * Input
 *   n               – number of points (0 … n-1)                 1 < n < 100
 *   n lines of n integers (0/1) – adjacency matrix (1 ⇒ edge)
 *
 * Output
 *   single integer – count of articulation points
 *
 * Algo
 *   • Build an adjacency list from the matrix (undirected graph).
 *   • Run Tarjan’s DFS to compute discovery/low times and mark cut-vertices.
 *   • Print the total number of cut-vertices.
 */
public class Main {

    private static List<Integer>[] graph;
    private static boolean[] visited, isAP;
    private static int[] disc, low, parent;
    private static int timer = 0;

    private static void dfs(int u) {
        visited[u] = true;
        disc[u] = low[u] = ++timer;
        int children = 0;

        for (int v : graph[u]) {
            if (!visited[v]) {
                children++;
                parent[v] = u;
                dfs(v);

                // update low-link value
                low[u] = Math.min(low[u], low[v]);

                // articulation-point cases
                if (parent[u] == -1 && children > 1)          // root with >1 child
                    isAP[u] = true;
                if (parent[u] != -1 && low[v] >= disc[u])      // non-root
                    isAP[u] = true;
            } else if (v != parent[u]) {                       // back-edge
                low[u] = Math.min(low[u], disc[v]);
            }
        }
    }

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String line = br.readLine();
        if (line == null || line.isEmpty()) return;
        int n = Integer.parseInt(line.trim());

        /* ---------- build graph ---------- */
        graph = new ArrayList[n];
        for (int i = 0; i < n; i++) graph[i] = new ArrayList<>();

        for (int i = 0; i < n; i++) {
            String[] parts = br.readLine().trim().split("\\s+");
            for (int j = 0; j < n; j++) {
                if (parts[j].charAt(0) == '1' && i != j) {     // ignore self-loops
                    graph[i].add(j);
                }
            }
        }

        /* ---------- Tarjan articulation-point DFS ---------- */
        visited = new boolean[n];
        isAP    = new boolean[n];
        disc    = new int[n];
        low     = new int[n];
        parent  = new int[n];
        Arrays.fill(parent, -1);

        for (int i = 0; i < n; i++) {
            if (!visited[i]) dfs(i);
        }

        /* ---------- count & output ---------- */
        int count = 0;
        for (boolean b : isAP) if (b) count++;
        System.out.println(count);
    }
}

Why it works

• Connectivity model – The matrix describes an undirected graph of power-grid points.
• Articulation point definition – If removing node u increases the number of connected components, u is critical; Tarjan’s algorithm identifies exactly those nodes with one DFS.
• Complexity – Each edge and vertex is processed a constant number of times: O(V + E) time, O(V + E) memory (with V ≤ 99, well within limits).

Compile & run (example):

bash
javac Main.java
echo "5
0 1 0 1 0
1 0 1 0 0
0 1 0 1 1
1 0 1 0 0
0 0 1 0 0" | java Main

Outputs the number of fail-safe machines needed for that grid.

Below is a drop-in replacement for the skeleton shown in your screenshot.
Just copy-paste over the editor in HirePro and click Run Tests / Submit & Continue Coding.

java
import java.util.*;

public class TestClass {

    /* -----------------------------------------------------------
       Count articulation points (critical nodes) in an undirected
       graph represented by an n × n adjacency-matrix `arr`.
       A critical node’s failure disconnects the grid, so the
       answer equals the number of fail-safe machines required.
       ----------------------------------------------------------- */
    static int findNumberOfPowerGrid(int[][] arr, int n) {

        /* ---------- build adjacency list ---------- */
        List<Integer>[] g = new ArrayList[n];
        for (int i = 0; i < n; i++) g[i] = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (arr[i][j] == 1 && i != j) {      // ignore self-loops
                    g[i].add(j);
                }
            }
        }

        /* ---------- Tarjan articulation-point DFS ---------- */
        boolean[] visited = new boolean[n];
        boolean[] isAP    = new boolean[n];
        int[] disc        = new int[n];
        int[] low         = new int[n];
        int[] parent      = new int[n];
        Arrays.fill(parent, -1);
        int[] timer = {0};                           // single-element “ref”

        // DFS lambda (Java 8 doesn’t allow local methods – use array for timer)
        class DFS {
            void run(int u) {
                visited[u] = true;
                disc[u] = low[u] = ++timer[0];
                int children = 0;

                for (int v : g[u]) {
                    if (!visited[v]) {
                        children++;
                        parent[v] = u;
                        run(v);

                        low[u] = Math.min(low[u], low[v]);

                        if (parent[u] == -1 && children > 1)  // root case
                            isAP[u] = true;
                        if (parent[u] != -1 && low[v] >= disc[u]) // bridge case
                            isAP[u] = true;
                    } else if (v != parent[u]) {              // back-edge
                        low[u] = Math.min(low[u], disc[v]);
                    }
                }
            }
        }
        DFS dfs = new DFS();

        for (int i = 0; i < n; i++) {
            if (!visited[i]) dfs.run(i);
        }

        /* ---------- count critical points ---------- */
        int res = 0;
        for (boolean b : isAP) if (b) res++;
        return res;
    }

    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        int n = input.nextInt();
        int[][] arr = new int[n][n];
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                arr[i][j] = input.nextInt();
            }
        }
        System.out.print(findNumberOfPowerGrid(arr, n));
    }
}

How it matches the required format

• Function signature unchanged (findNumberOfPowerGrid(int[][], int)).
• Uses the provided main exactly (only the algorithm body is filled).
• Reads the adjacency matrix exactly as in the sample.
• Outputs a single integer — the number of fail-safe machines (articulation points) — with System.out.print, no extra spaces or newlines.