Array 1 done

[pranjay01]

No description provided.

[super30admin]

Strengths:

• You have correctly identified the need to alternate directions and handle boundary conditions.
• The time and space complexity are optimal.
• You provided a test case to validate your code.

Areas for Improvement:

• The boundary adjustment logic after each while loop in direction 1 and -1 is complex and may not handle all cases correctly. For instance, when moving upward and hitting the top row and rightmost column simultaneously, your adjustment sets row += 2 and column -= 1. This might work for some cases but can lead to index errors or incorrect traversals.
• Consider a simpler approach like the reference solution: instead of having separate while loops for each direction, you can use a single for loop and adjust the direction and indices based on current position. This reduces the complexity of boundary checks.
• Use more descriptive variable names (e.g., num_rows and num_cols instead of rows and columns).
• Avoid using magic numbers (like 1 and -1 for directions); instead, use constants or enums for better readability.
• Test with more edge cases, such as single-row matrices, single-column matrices, and non-square matrices.

Suggested Revision:
You can simplify your code by following the reference solution's structure. Here's a Python version of the reference solution for your reference:

class Solution:
    def findDiagonalOrder(self, mat: List[List[int]]) -> List[int]:
        m, n = len(mat), len(mat[0])
        result = []
        r, c = 0, 0
        direction = 1  # 1 for upward, -1 for downward

        for i in range(m * n):
            result.append(mat[r][c])
            if direction == 1:
                if c == n - 1:
                    r += 1
                    direction = -1
                elif r == 0:
                    c += 1
                    direction = -1
                else:
                    r -= 1
                    c += 1
            else:
                if r == m - 1:
                    c += 1
                    direction = 1
                elif c == 0:
                    r += 1
                    direction = 1
                else:
                    r += 1
                    c -= 1
        return result

This approach is simpler and handles all boundary conditions elegantly.