Two solutions completed

[dbkuppagiri]

No description provided.

[super30admin]

Least Falling Path Sum (minFallingPathSum.js)

Strengths:

• The solution is efficient with O(n^2) time and O(n) space.
• The code is clean, readable, and well-commented.
• The approach correctly handles boundary conditions (j-1 and j+1) by checking indices.

Areas for improvement:

• Since the matrix is n x n, the number of rows and columns are the same. Using 'n' for both dimensions would improve consistency and readability. Instead of defining 'm = matrix[0].length', you can simply use 'n' throughout.
• The variable name 'minAbove' might be slightly misleading because it represents the minimum value from the previous row for the three possible directions (left, right, center). Consider renaming it to 'minPrev' or something similar to indicate it's the minimum from the previous row's relevant cells.
• Although not necessary, you could add a check for an empty matrix at the beginning to handle edge cases (e.g., if matrix is empty, return 0 or appropriate value). However, the constraints state n>=1, so it's optional.

VERDICT: PASS

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Delete and Earn (deleteAndEarn.js)

Your second solution is excellent: it efficiently transforms the problem into a House Robber-style DP and solves it iteratively. This is optimal in both time and space. Your code is clean and well-commented.

For your first solution, while the DFS approach is conceptually correct, the memoization implementation has a flaw: when memo[idx] is 0 (which is a valid cached value), the condition if (memo[idx]) returns false, leading to recomputation. You should initialize memo as an array of undefined (or use a different marker) and check explicitly for undefined (e.g., if (memo[idx] !== undefined)). Also, recursion might lead to stack overflow for large inputs (maxLength up to 10^4), so the iterative approach is preferred.

You might consider using a more robust memoization technique in the future, such as:

if (memo[idx] !== undefined) return memo[idx];

Additionally, for problems with large state spaces, iterative DP is usually more efficient and avoids recursion limits.

Both solutions demonstrate a good understanding of the problem. Keep up the good work!

VERDICT: PASS