Problem URL: Check if there is a path with equal number of 0s and 1s

We will use top-down dynamic programming to solve this problem. We will create a dfs function to find if there is a path from the current node to the end node. We will return the result of the dfs function. We will use memoization to cache the results of the dfs function.

class Solution:
    def isThereAPath(self, grid: List[List[int]]) -> bool:
        ROWS, COLS = len(grid), len(grid[0])

        if (ROWS+COLS-1) % 2 == 1: 
            return False

        @cache
        def dfs(r, c, cur):
            if r >= ROWS or c >= COLS:
                return False

            cur += (grid[r][c]*2 - 1)

            if r == ROWS-1 and c == COLS-1:
                return cur == 0

            return dfs(r+1, c, cur) or dfs(r, c+1, cur)

        return dfs(0, 0, 0)

Time complexity: O(mn)
Space complexity: O(mn)