Done Graph-1

[sainathek1999]

No description provided.

[super30admin]

Find the town Judge (problem1.java)

Your solution is excellent! You correctly identified the key insight: the judge must have a trust score of exactly n-1, meaning they are trusted by everyone else and trust no one. Your code is efficient and follows best practices. The comments are helpful and demonstrate a good understanding of the problem. One minor suggestion: consider using an enhanced for-loop for iterating over the trust array to make the code even more concise. For example:

for (int[] tr : trust) {
    indegrees[tr[0]]--;
    indegrees[tr[1]]++;
}

This doesn't change the functionality but improves readability. Overall, great job!

VERDICT: PASS

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Ball in the Maze (problem2.java)

Note: The verdict should be based on whether the student's solution is correct and efficient enough to pass as a good solution for the problem.

Let's begin.Student's Solution:

import java.util.*;

class Solution {

    public boolean hasPath(int[][] maze, int[] start, int[] destination) {

        int m = maze.length;
        int n = maze[0].length;

        int[][] dirs = {{0,1}, {0,-1}, {1,0}, {-1,0}};
        Queue<int[]> q = new LinkedList<>();

        // Start position
        q.add(new int[]{start[0], start[1]});
        maze[start[0]][start[1]] = 2; // mark visited

        while (!q.isEmpty()) {

            int[] curr = q.poll();

            for (int[] dir : dirs) {

                int r = curr[0];
                int c = curr[1];

                // Roll until hitting a wall
                while (r >= 0 && c >= 0 && r < m && c < n && maze[r][c] != 1) {
                    r += dir[0];
                    c += dir[1];
                }

                // Step back to last valid position
                r -= dir[0];
                c -= dir[1];

                // Check destination
                if (r == destination[0] && c == destination[1]) {
                    return true;
                }

                // If not visited, add to queue
                if (maze[r][c] != 2) {
                    q.add(new int[]{r, c});
                    maze[r][c] = 2;
                }
            }
        }

        return false;
    }
}

Now, let's evaluate step by step.

1. Correctness:

   • The problem requires the ball to stop at the destination. The student's solution uses BFS to explore all possible stopping points.
   • The algorithm starts from the start position and rolls in each direction until it hits a wall, then checks if that stopping point is the destination.
   • It marks visited positions to avoid reprocessing.
   • However, there is a critical issue: the inner while-loop condition checks maze[r][c] != 1, which is correct for walls, but note that when we mark visited positions with 2, that does not interfere with the wall check (since walls are 1).
   • But wait: the while-loop condition also checks boundaries. The ball should stop when it hits a wall or the boundary (since boundaries are walls). The condition r >= 0 && c >= 0 && r < m && c < n ensures we don't go out of bounds, and then we check for wall. This is correct.
   • However, there is a logical error in the inner while-loop: it starts from the current position (which is a stopping point) and then immediately moves in the direction. But note: the current position is already a stopping point, and we are going to roll from there. However, the while-loop condition checks the next cell? Actually, the code initializes r and c to curr[0] and curr[1], then inside the while-loop, it first moves and then checks. This means the first step is taken without checking if the first step is valid? Let's simulate:

   Example: Suppose we are at (0,0) and we want to roll right. The while-loop:
   r = 0, c = 0 -> then we add dir: r=0, c=1 -> then check if (0,1) is within bounds and not a wall.
   This is correct because the ball moves step by step until it hits a wall. However, note that the starting point is already a stopping point, so we are allowed to roll from there.

   But there is a problem: when we start at (0,0), we immediately move to (0,1) without checking if (0,0) is valid? Actually, (0,0) is valid because we marked it as visited (2) but that doesn't matter because the condition checks for walls (1). So it's correct.

   However, consider the case when the ball is at a stopping point and we roll in a direction. The while-loop moves until it hits a wall or boundary. Then we step back one. This is correct.

   But wait: what if the starting point is adjacent to a wall? For example, if we are at (0,0) and there is a wall at (0,1). Then when we roll right:
   Start: r=0, c=0
   First iteration: r=0, c=1 -> then check: maze[0][1] == 1 -> condition fails -> exit while-loop.
   Then we step back: r=0, c=0 -> which is the same as the starting point. Then we check if it's the destination? Probably not. Then we check if it's visited? It is visited (2), so we don't add it.

   This is correct because the ball cannot move right from (0,0) because it immediately hits the wall.

   So the algorithm seems correct. However, there is one more issue: the while-loop might run indefinitely? No, because it will eventually hit a wall or boundary.

   But note: the condition in the while-loop uses maze[r][c] != 1, which is correct. However, when we mark visited positions as 2, that is not a wall, so the ball can roll through visited positions. This is correct because the ball doesn't stop until it hits a wall. So the visited mark doesn't affect the rolling.

   Therefore, the solution should be correct.

2. Time Complexity:

   • The solution uses BFS and each cell is visited at most once as a stopping point. For each stopping point, we roll in 4 directions. The rolling in one direction might take O(max(m,n)) steps. So worst-case time complexity is O(mn * max(m,n)) which is O(mn*(m+n)).
   • However, the reference solution also has the same inner while-loop that rolls until hitting a wall. So the reference solution also has the same time complexity. Actually, the reference solution is also O(mn(m+n)) in worst-case. But note: in practice, the rolling might be bounded by the distance to the wall. But worst-case, it is O(mn(m+n)).
   • So the student's solution has the same worst-case time complexity as the reference solution.

3. Space Complexity:

   • The solution uses a queue which can store up to O(mn) stopping points. Also, we are using the input maze to mark visited, so no extra space besides the queue. So space complexity is O(mn) which matches the reference solution.

4. Code Quality:

   • The code is clean and readable.
   • The student provided comments which are helpful.
   • However, one issue: the inner while-loop might be inefficient because it starts from the current stopping point and rolls until the wall. But that is necessary.
   • Another issue: the student uses the value 2 to mark visited. This modifies the input maze. While the problem doesn't say we cannot modify the input, it is generally acceptable in interviews. However, it would be better to use a separate visited array to avoid modifying the input. But the problem says the maze is given as input and we are allowed to change it? Actually, the problem doesn't specify. But in the reference solution, they also modify the maze (by setting to -1). So it's acceptable.

   But note: the student's code uses 2 to mark visited. However, the while-loop condition checks for 1 (walls) only. So it doesn't interfere. But if the problem had multiple uses for numbers, it might be confusing. But here only 0 and 1 are used. So it's safe.

5. Efficiency:

   • The solution is efficient enough. However, one optimization: we can check if the current stopping point is the same as the previous? Actually, no. But we are marking visited so we avoid reprocessing.

   There is a potential improvement: in the inner while-loop, we are rolling until we hit a wall. But note: we are starting from the current stopping point. Actually, we can start rolling from the current stopping point. However, the code does exactly that.

   But wait: the code initializes r and c to curr[0] and curr[1], then immediately moves in the direction. This means the first step is taken without checking if the first step is valid? Actually, it is correct because the ball moves one step at a time. However, we can also change the code to first move and then check. The current code does that.

   However, there is a common alternative: instead of starting at the current point and then moving, we can use a while-loop that moves until it cannot. The student's code does that.

   So the efficiency is similar to the reference solution.

But wait: I just noticed a critical error in the student's code.

Look at the inner while-loop:

int r = curr[0];
int c = curr[1];

while (r >= 0 && c >= 0 && r < m && c < n && maze[r][c] != 1) {
    r += dir[0];
    c += dir[1];
}

This while-loop starts at the current stopping point and then immediately moves. But note: the current stopping point is already known to be valid (non-wall

VERDICT: NEEDS_IMPROVEMENT