Done Graph-1

[samikshm]

No description provided.

[super30admin]

Solution 1: ball_in_maze.java

• Correctness: The DFS approach correctly handles the rolling ball mechanics by simulating movement until hitting a wall. Marking visited positions with -1 prevents infinite loops. The logic for rolling and stopping is sound.
• Time Complexity: O(m*n) is correct since each cell is visited at most once.
• Space Complexity: O(m*n) for the recursion stack in the worst case is accurate.
• Code Quality: Good structure with clear separation of concerns. However, modifying the input maze (setting visited cells to -1) is a side effect that could be problematic if the maze needs to be reused. Using a separate visited array would be cleaner.
• Edge Cases: The solution handles boundary conditions well with the while loop bounds checking. However, there's a potential issue: if start equals destination, the code returns true correctly, but if start is on a wall (value 1), the initial check would cause an ArrayIndexOutOfBoundsException before reaching the visited check.

Solution 2: town_judge.java

• Correctness: The indegree approach is optimal and correctly identifies the town judge. The logic of decrementing for trusting and incrementing for being trusted is correct.
• Time Complexity: O(n + m) is correct.
• Space Complexity: O(n) is correct.
• Code Quality: Clean, concise, and follows best practices. The solution handles the edge case of n=1 (no trust relationships) correctly, returning 1 as the judge.
• Edge Cases: Well-handled. When trust array is empty, all indegrees remain 0, so for n=1 it returns 1 (correct), and for n>1 it returns -1 (correct).

General Observations:

• Both solutions include good comments explaining the approach.
• The time/space complexity analysis is accurate for both.
• Both solutions successfully run on Leetcode as stated.