Python3 Vectorized Solution
I hope this answer is helpful to those who prefer more formal explanations (like myself).\n___\n# Intuition\n Describe your first thoughts on how to solve this 
I hope this answer is helpful to those who prefer more formal explanations (like myself).\n___\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nI made the following two observations:\n1. Let $g(x)$ denote the number of trailing zeros of a number $x$. Then \n$g(x) = \\min_{p\\in\\{2,5\\}} f(x,p),$ \nwhere $f(x, p)$ is the number of times that prime number $p$ appears in the prime factorization of $x$.\n2. The solution can always be represented by a cornered path which extends all the way to the edges of the grid. This is because the entries are non-negative, and so adding more cells can never harm the solution.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nDefine a _leg_ $(i,j,d)$ to be a path of cells from (_excluding_) $(i,j)$ to the edge of the grid, following direction $d\\in\\{{\\rm up,down,left,right}\\}$. WLOG, we only consider cornered paths which extend all the way to the edges of the grid, so that each cornered path can be decomposed into a corner and two legs. We encode each cornered path by a tuple $(i,j,d_1,d_2)$, where $(i,j)$ is the location of the path\'s corner, and $(d_1,d_2)$ are the directions of the its legs, $(i,j,d_1)$ and $(i,j,d_2)$.\n\nDefine the _gain_ $g(\\pi)$ of $\\pi$ to be the number of trailing zeros in the product of the path, i.e. $g(\\pi) = g(\\Pi_{ij\\in\\pi}x_{ij})$, and let $f(\\pi, p) = \\sum_{ij\\in\\pi} f(x_{ij}, p)$ be the number of times prime number $p$ appears along a path $\\pi$. Then,\n$g(\\pi) = g(\\Pi_{ij\\in\\pi}x_{ij}) \\\\ = \\min_{p\\in\\{2,5\\}} f(\\Pi_{ij\\in\\pi}x_{ij}, p) \\\\ = \\min_{p\\in\\{2,5\\}} \\sum_{ij\\in\\pi} f(x_{ij}, p) \\\\ = \\min_{p\\in\\{2,5\\}} f(\\pi, p).$\nOur objective is to find a cornered path $\\pi$ which maximizes $g(\\pi)$. First, we need to compute $f(\\pi, p)$ for all cornered paths $\\pi$ and $p\\in\\{2,5\\}$.\n\nNotice that, for a cornered path, $\\pi=(i,j,d_1,d_2)$, the quantity $f(\\pi,p)$ can be decomposed over $\\pi$\'s corner and two legs:\n$f((i,j,d_1,d_2), p) = f(x_{ij}, p) + f((i,j,d_1), p) + f((i,j,d_2), p).$\n\nThese leg counts, $f((i,j,d), p)$, are our building blocks, so I computed them for each possible leg $(i,j,d)$. Since a leg may be a sub-leg of a longer leg, I stored and reused intermediate results in a dynamic programming style. I then used these leg counts to compute $f(\\pi, 2)$ and $f(\\pi, 5)$ for each possible cornered path $\\pi$, which together yield $g(\\pi)$. Finally, I computed $\\max_\\pi g(\\pi)$ by searching over all cornered paths $\\pi$. Crucially, I vectorized this search over $(i,j)$ using NumPy; without doing so, my solution exceeded the time limit.\n\n# Complexity\n- Time complexity: $O(mn)$, because $O(1)$ visits per cell\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $O(mn)$, because $O(1)$ values stored per cell\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nimport numpy as np\n\n# directions\nU = 0 # Up\nD = 1 # Down\nL = 2 # Left\nR = 3 # Right\nN_dir = 4\n\n# factors\nT = 0 # Two\nF = 1 # Five\nN_fac = 2\n\ndef f(x, p):\n    \'\'\'returns the highest power of `p` that divides `x`\'\'\'\n    c = 0\n    while x % p == 0:\n        c += 1\n        x //= p\n    return c\n\nclass Solution:\n    def maxTrailingZeros(self, grid: List[List[int]]) -> int:\n        m, n = len(grid), len(grid[0])\n        grid = np.array(grid)\n\n        # y[i, j, p] = \n        #   highest power of `p` that divides `grid[i, j]`\n        y = np.zeros((m, n, N_fac), dtype=int)\n        for i in range(m):\n            for j in range(n):\n                y[i, j, T] = f(grid[i, j], 2)\n                y[i, j, F] = f(grid[i, j], 5)\n\n        # x[i, j, d, p] = \n        #   count of times that `p` appears in leg `(i,j,d)`\n        x = np.zeros((m, n, N_dir, N_fac), dtype=int)\n        for i in range(m):\n            for j in range(n):\n                if i > 0:\n                    x[i, j, U] = y[i-1, j] + x[i-1, j, U]\n                    x[m-1-i, j, D] = y[m-i, j] + x[m-i, j, D]\n                if j > 0:\n                    x[i, j, L] = y[i, j-1] + x[i, j-1, L]\n                    x[i, n-1-j, R] = y[i, n-j] + x[i, n-j, R]\n\n        # the cornered path\'s legs can be oriented in one of four ways\n        path_orientations = [(U,R), (U,L), (D,R), (D,L)]\n\n        # for each orientation `(d1, d2)`, \n        #   for each corner `(i, j)` (vectorized):\n        #       find the the number of trailing zeros in the product\n        #       of the path `(i, j, d1, d2)`.\n        #   max-reduce over corners (vectorized)\n        # max-reduce over orientations.\n        return max(\n            (y + x[:, :, d1] + x[:, :, d2]).min(-1).max()\n            for d1, d2 in path_orientations\n        )\n```
['Python3']