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

Problem B. Minimum path