Array-1 Solution

[shakthinandana]

No description provided.

[super30admin]

Product Except Self (productExceptSelf.py)

1. Strengths:

   • The solution correctly implements an O(n) time algorithm without using division.
   • The space complexity is O(1) excluding the output, which meets the follow-up requirement.
   • The code is concise and uses two passes efficiently.

2. Areas for improvement:

   • Variable naming: Use more descriptive names. For example, lprod could be renamed to result or output, and rp could be running_product to enhance readability.
   • Commenting: Add comments to explain the two passes. For instance, in the first loop, you are computing the left cumulative product for each index. In the second loop, you are computing the right cumulative product and multiplying it with the left product.
   • Initialization: Instead of building lprod by appending (which is O(n) but amortized), you can initialize an array of ones of length n. This is more efficient and standard.
   • Edge cases: The code handles the case when there are zeros? Actually, the problem states that the product is guaranteed to fit in a 32-bit integer, but the code should work with zeros. Your code does work with zeros because it doesn't use division.

3. Code structure: The code is well-structured with two loops. However, consider initializing the result array with ones and then updating it. For example:
   n = len(nums)
   result = [1] * n
   left_running = 1
   for i in range(1, n):
   left_running *= nums[i-1]
   result[i] = left_running
   right_running = 1
   for i in range(n-2, -1, -1):
   right_running *= nums[i+1]
   result[i] *= right_running
   This is similar to your code but initializes the result array to ones. Your code initializes lprod as [1] and then appends, which is equivalent.

4. Minor issue: In the right pass, you are updating lprod[i] by multiplying with rp. This is correct. But note that the last element (index n-1) is not updated in the right pass because the loop goes from n-2 to 0. This is correct because the right product for the last element is 1, and the left product is already set.

Overall, the solution is correct and efficient. The main improvements are in code readability and documentation.

VERDICT: PASS

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Diagonal Traverse (diagonalTraverse.py)

Your solution is well-written and efficient. You have correctly implemented the diagonal traversal algorithm with optimal time and space complexity. The code is clean and easy to follow.

One minor improvement: you can use boolean values for dir (True/False) instead of 1/0 to make it more explicit. Also, consider adding a docstring to explain the algorithm briefly, though it's not strictly necessary.

Additionally, in Python, it's common to use snake_case for variable names. So reslen could be res_len, but this is a minor style point.

Here's a slightly refined version of your code with these suggestions:

class Solution(object):
    def findDiagonalOrder(self, mat):
        """
        :type mat: List[List[int]]
        :rtype: List[int]
        """
        m = len(mat)
        n = len(mat[0])
        total_elements = m * n
        res = []
        # Use True for upward direction, False for downward
        is_upward = True
        r, c = 0, 0
        
        for i in range(total_elements):
            res.append(mat[r][c])
            
            if is_upward:
                if c == n - 1:
                    r += 1
                    is_upward = False
                elif r == 0:
                    c += 1
                    is_upward = False
                else:
                    r -= 1
                    c += 1
            else:
                if r == m - 1:
                    c += 1
                    is_upward = True
                elif c == 0:
                    r += 1
                    is_upward = True
                else:
                    r += 1
                    c -= 1
        
        return res

Overall, great job!

VERDICT: PASS

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Spiral Matrix (sprialMatrix.py)

Your solution is correct and efficient. You have correctly implemented the boundary conditions and checks to avoid duplicate traversals. The code is clean and well-commented. One minor improvement could be to use more descriptive variable names for clarity, such as using 'row' and 'col' instead of 'r' and 'c' in loops, but this is a small stylistic point. Additionally, you might consider adding a check for an empty matrix at the beginning to handle edge cases, although the problem constraints state that m and n are at least 1. Overall, great job!

VERDICT: PASS