Count unique paths in matrix from top left to bottom right (Memoization)

Given m x n matrix, count of all unique possible paths from top left to the bottom right. From each cell, only move right or down is allowed.

Hint

This algorithm uses memoization to improve performance. It's same as the recursive version of Count unique paths in matrix from top left to bottom right, but with a special 'cache' to store results. The 'cache key' is crucial. It's how we identify and store past results. Sometimes we need one key, sometimes multiple keys. We add code to:

• Initialize cache to store results.
• Using the key to check cached result. If the result is not found in the cache, do the calculation and save the result in cache.
• Return the cached result.

• Python
• Javascript
•  Run

cache = None

def count(m, n):
  if m == 1 or n == 1:
    return 1
  
  global cache
  if not cache:
    cache = [[None for _ in range(n + 1)] for _ in range(m + 1)]

  if not cache[m][n]:
    cache[m][n] = count(m - 1, n) + count(m, n - 1)

  return cache[m][n]

m = 2
n = 2

print(count(m, n))

let cache = null;

function count(m, n) {
  if (m == 1 || n == 1) {
    return 1;
  }

  if (!cache) {
    cache = Array.from(Array(n + 1), () => Array(m + 1));
  }

  if (cache[m][n] === undefined) {
    cache[m][n] = count(m - 1, n) + count(m, n - 1);
  }

  return cache[m][n];
}

const m = 2;
const n = 2;

console.log(count(m, n));