"Graph-1 done"

[Sahithipsl470]

No description provided.

[super30admin]

Find the town Judge (Problem-1.py)

Your solution is excellent and correctly solves the problem. You have clearly understood the requirements and implemented an efficient algorithm. Here are a few points to consider:

1. While using two arrays is perfectly fine and clear, you could also use a single array to track the net trust count. For example, for each trust relationship [a, b], you could decrement the count for a (since a trusts someone) and increment for b (since b is trusted). Then, the judge would be the person with a net trust count of n-1. This approach is used in the reference solution and reduces the space usage by one array. However, your approach with two arrays is equally valid and might be more intuitive for some.

2. Your code is well-commented and follows good practices. Keep up the good work!

3. One minor suggestion: In the loop, you are iterating from 1 to n inclusive, which is correct. However, note that the arrays are indexed from 0 to n, so the indices 0 are unused. This is acceptable since the people are labeled from 1 to n.

Overall, your solution is correct, efficient, and well-written. Great job!

VERDICT: PASS

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Ball in the Maze (Problem-2.py)

1. Correctness Issue: The while loop condition in the rolling phase is incorrect. It checks nx + dx and ny + dy to be within bounds, but if the current (nx, ny) is already at the border, the next step would be out of bounds, and the loop should not proceed. However, the condition correctly prevents this by checking the next step. But note: the condition maze[nx + dx][ny + dy] == 0 might cause an index error if nx + dx or ny + dy is out of bounds. Actually, the condition first checks the bounds, so it should be safe. However, the condition should be written to avoid any index error. The current condition is safe because it checks bounds first. But there is a logical issue: the while loop moves until it hits a wall, but it should stop when the next step is a wall or out of bounds. The condition is correct in that regard.

2. Efficiency Issue: The DFS approach might lead to deep recursion for large mazes (up to 100x100 = 10000 cells). This could cause a stack overflow in Python. It is better to use an iterative DFS with a stack or a BFS (like the reference solution) to avoid recursion depth issues.

3. Code Quality: The code is readable and well-structured. The use of a visited matrix is appropriate. However, the variable names are clear, and the logic is straightforward.

4. Potential Bug: The DFS function might not explore all paths correctly because it marks the current cell as visited before exploring directions. This is correct because we only need to visit each stopping point once. However, the rolling logic might not always stop at the correct cell. For example, when rolling, the code moves until the next step is invalid, then uses the current (nx, ny) as the stopping point. This is correct because the while loop moves until the next step is invalid, so the current (nx, ny) is the last valid cell before hitting a wall or boundary. But note: the condition in the while loop allows the ball to move as long as the next cell is within bounds and is empty. This is correct.

5. Comparison with Reference: The reference solution uses BFS with a queue and marks visited by modifying the maze (setting to -1). This is efficient and avoids recursion. The student's DFS approach is conceptually similar but uses recursion which might be problematic for large inputs.

VERDICT: NEEDS_IMPROVEMENT