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NinjaG
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lattice path from project Euler with python solution

I was trying to solve this problem called lattice path from Project Euler that "Count the number of unique paths to travel from the top left to the bottom right of a lattice of squares".

How many such routes are there through a ×ばつ20 grid? It takes me forever to get an answer...

Image https://projecteuler.net/project/images/p015.gif

 # Input: An interger N (which is the size of the lattice)
 # Output: An interger (which represents the number of unique paths)
 #
 # Example: input: 2
 #
 # (2 x 2 lattice of squares)
 # __ __
 # |__|__|
 # |__|__|
 #

When moving through the lattice, you can only move either down or to the right.

Currently, I got 1 solution, and I am not sure how much better if we memoize it. But wanted to ask for code review for my solution below:

def lattice_paths(m, n):
 if m == 0 and n == 0:
 return 1
 if m < 0 or n < 0:
 return 0
 return lattice_paths(m - 1, n) + lattice_paths(m, n - 1)
Here is the time complexity and space complexity of my first solution.
# Time Complexity: O(2^(M+N))
# Auxiliary Space Complexity: O(M+N)

dynamic approach with the memorized solution.

def lattice_paths(m, n): # m rows and n columns
 def helper(x, y):
 if x > helper.m or y > helper.n:
 return 0
 if helper.cache.get((x, y)) is not None: 
 return helper.cache[(x, y)]
 helper.cache[(x, y)] = helper(x + 1, y) + helper(x, y + 1)
 return helper.cache[(x, y)] 
 helper.cache = {}
 helper.m = m
 helper.n = n
 helper(0,0)
 if m == 0 and n == 0:
 return 1
 elif m < 0 or n < 0:
 return 0
 else:
 return helper.cache[(0, 0)]
NinjaG
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lang-py

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