Lattice path from Project Euler with Python solution

Question

I was trying to solve this problem called lattice path from Project Euler:

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...

https://projecteuler.net/project/images/p015.gif
_{(source: projecteuler.net)}

 # 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.

I currently have 1 solution, and I am not sure how much better if we memoize it. But I 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)]

• Daniel Lenz: That's more of a mathematics and algorithm problem than a coding one. Check out this excellent blog post

Answer 1

All of your approaches are very brute force. The answer can be computed with a bit of mathematical reasoning. For an m ×ばつ n lattice, all paths will have m + n segments, of which m segments will go down and n segments will go right. So, out of m + n slots, count the ways to pick m downward moves; the rest of them will be rightward moves. The number of paths is:

$${m + n \choose m} = {m + n \choose n} = \frac{(m+n)!}{m! ,円 n!}$$

So, the Python solution is simply:

from math import factorial
def lattice_paths(m, n):
 return factorial(m + n) // factorial(m) // factorial(n)

Even though the calculation involves large factorials, Python supports big integers seamlessly, and arrives at the answer nearly instantaneously.

Answer 2

here's a solution that uses cache to memoize the solution

def lattice_paths(m, n):
 cache = [1]
 larger = max(m, n)
 smaller = min(m, n)
 while(len(cache) < larger + 1):
 for i in range(1, len(cache)):
 cache[i] += cache[i - 1]
 cache.append(2 * cache[len(cache) - 1])
 return cache[smaller]
# Time Complexity: O(N)
# Auxiliary Space Complexity: O(N)