Let's traverse `mat` and update values in it: now `mat[i][j]` contains the path cost to cell `[i, j]`. How to reach that? You can get to the `mat[i][j]` from either `mat[i-1][j]` or `mat[i][j-1]`  cell, that also contain the path cost to themselves. Thus, `mat[i][j]` can be updated as:

`mat[i][j] += min(mat[i-1][j], mat[i][j-1])`,

since you choose the minumum cost path between these two.

Note that some cells can be reached only from left or right, for example, `mat[0][j]` (only from `mat[0][j-1]`).

So, the goal is to traverse `mat` and update its values; after that, return path cost at `mat[-1][-1]`.

def minPath(mat):
  m, n = len(mat), len(mat[0])
  for i in range(1, m):
    mat[i][0] += mat[i-1][0] 
  for j in range(1, n):
    mat[0][j] += mat[0][j-1] 
    
  for i in range(1, m):
    for j in range(1, n):
      mat[i][j] += min(mat[i-1][j], mat[i][j-1])
    
  return mat[-1][-1]
 
mat = [[10,1,23,4,5,1],
      [2,13,20,9,1,5],
      [14,3,3,6,12,7]]
 
print(minPath(mat))

This course introduces a Dynamic Programming approach for users familiar with the syntax and data structures of the Python programming language. Dynamic programming is a necessary topic for a technical interview, as are Binary Search, Sorting Algorithms, etc. Be prepared to solve a few problems on your own. Use the hints or Solutions section if you get stuck.

Brief overview: what is Dynamic Programming, and what are the main principles.

Solve these problems using DP principles.

See the explained solutions for the problems.

Problem B. Minimum path