Done DP-3

[pranjay01]

No description provided.

[super30admin]

Strengths:

• You recognized that dynamic programming is suitable for this problem and attempted to optimize space by using a single array to store the previous row.
• Your code includes comments that explain your approach, which is good for readability.

Areas for Improvement:

1. Correctness: The main issue is in the inner loop logic. For each cell (i, j), you need to consider the values from the previous row at positions (j-1), (j), and (j+1). Your current approach does not correctly capture these three values. Specifically:
   • For the first column (j=0), you should compare the top (j=0) and top-right (j=1) values.
   • For the last column (j=columns-1), you should compare the top-left (j=columns-2) and top (j=columns-1) values.
   • For middle columns, compare all three: top-left, top, and top-right.
2. Variable Usage: The use of previousTopVal is confusing and leads to errors. Instead, you should create a temporary copy of the previous row before updating it. This way, you can access the original values when computing the min for the current row.
3. Edge Cases: Ensure your solution handles matrices of size 1x1 correctly.
4. Code Clarity: Use more descriptive variable names. For example, psblValuesAtEachColumn could be named dp or prevRow. Also, avoid using None for initializing variables when numbers are expected.

Suggested Correct Approach:

• Initialize an array dp of length n (number of columns) with the first row of the matrix.
• For each subsequent row i:
  • Create a temporary copy of dp (call it temp) to store the previous row's values.
  • For each column j:
    if j == 0:
    dp[j] = matrix[i][j] + min(temp[j], temp[j+1])
    elif j == n-1:
    dp[j] = matrix[i][j] + min(temp[j-1], temp[j])
    else:
    dp[j] = matrix[i][j] + min(temp[j-1], temp[j], temp[j+1])
• After processing all rows, return the minimum value in dp.

This approach ensures that you are always using the correct values from the previous row without overwriting them during the inner loop.