Lattice path from Project Euler with Python solution 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 20×20 grid? It takes me forever to get an answer...


(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 &lt; 0 or n &lt; 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 &gt; helper.m or y &gt; 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 &lt; 0 or n &lt; 0:
      return 0
    else:
      return helper.cache[(0, 0)]
All of your approaches are very brute force.  The answer can be computed with a bit of mathematical reasoning.  For an m × 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:

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

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.