Add implementation for leetcode problems 740, 931

[rishigoswamy]

No description provided.

[super30admin]

Strengths:

• The solution uses dynamic programming with optimal time and space complexity.
• The approach is correct and handles the boundaries appropriately.

Areas for improvement:

• The code could be made more robust by explicitly handling the case when n=1. Currently, it works, but it might be better to add a check at the beginning to return immediately if n==1. However, it is not necessary.
• The variable names temp and prev could be improved to current_row and previous_row for better clarity.
• Adding comments to explain the steps would make the code more understandable.
• The code for the first and last columns could be written in a unified way to avoid duplication. For example, you could initialize the current row as an array of zeros and then iterate over all columns, and for each column, consider the three neighbors with boundary checks.

Example of a more generalized inner loop:

n = len(matrix)
prev = matrix[0]
for i in range(1, n):
curr = [0] * n
for j in range(n):
# Consider left, middle, right from the previous row
options = []
if j-1 >= 0:
options.append(prev[j-1])
options.append(prev[j])
if j+1 < n:
options.append(prev[j+1])
curr[j] = matrix[i][j] + min(options)
prev = curr
return min(prev)

This avoids separate handling for the first and last columns and is more readable.