[Python3] DP in 4 directions, commented code
\nclass Solution:\n    def maxTrailingZeros(self, A: List[List[int]]) -> int:\n        \n        # Helper method with cache to return number of twos in prime fa
```\nclass Solution:\n    def maxTrailingZeros(self, A: List[List[int]]) -> int:\n        \n        # Helper method with cache to return number of twos in prime factorization of given number\n        @cache\n        def twos(x):\n            res = 0\n            while (x&1)==0:\n                res += 1\n                x >>= 1\n            return res\n        \n        # Helper method with cache to return number of fives in prime factorization of given number\n        @cache\n        def fives(x):\n            res = 0\n            while x%5 == 0:\n                res += 1\n                x //= 5\n            return res\n        \n        m, n = len(A), len(A[0])\n        \n        # Creating 2D arrays for storing number of 2\'s and number of 5\'s in all elements above, below, left or right\n\t\t# of the current element (including current element).The two counts will be stored as a tuple i.e. as pair<int, int>\n        \n        up, down, left, right = [[0]*n for _ in range(m)], [[0]*n for _ in range(m)], [[0]*n for _ in range(m)], [[0]*n for _ in range(m)]\n        \n        # Initializing first row of up array and last row of down array\n        for j in range(n):\n            up[0][j] = (twos(A[0][j]),fives(A[0][j]))\n            down[m-1][j] = (twos(A[m-1][j]), fives(A[m-1][j]))\n        \n        # Initializing leftmost column of left array and rightmost column of right array\n        for i in range(m):\n            left[i][0] = (twos(A[i][0]), fives(A[i][0]))\n            right[i][n-1] = (twos(A[i][n-1]), fives(A[i][n-1]))\n            \n        # Populating all remaining cells of all arrays\n        \n        for i in range(1, m):\n            for j in range(n):\n                up[i][j] = (up[i-1][j][0] + twos(A[i][j]), up[i-1][j][1] + fives(A[i][j]))\n        \n        for i in range(m-2, -1, -1):\n            for j in range(n):\n                down[i][j] = (down[i+1][j][0] + twos(A[i][j]), down[i+1][j][1] + fives(A[i][j]))\n        \n        for i in range(m):\n            for j in range(1, n):\n                left[i][j] = (left[i][j-1][0] + twos(A[i][j]), left[i][j-1][1] + fives(A[i][j]))\n        \n        for i in range(m):\n            for j in range(n-2, -1, -1):\n                right[i][j] = (right[i][j+1][0] + twos(A[i][j]), right[i][j+1][1] + fives(A[i][j]))\n        \n        # Iterating over each cell to first take pairwise sum in two directions of 2\'s and 5\'s then subtract current cell values (because it\'s counted twice) \n\t\t# finally take min of 2\'s and 5\'s for that cell where the horizontal/vertical turn happens\n        res = 0\n        for i in range(m):\n            for j in range(n):\n                res = max(res, min(up[i][j][0] + right[i][j][0] - twos(A[i][j]), up[i][j][1] + right[i][j][1] - fives(A[i][j])))\n                res = max(res, min(up[i][j][0] + left[i][j][0] - twos(A[i][j]), up[i][j][1] + left[i][j][1] - fives(A[i][j])))\n                res = max(res, min(down[i][j][0] + right[i][j][0] - twos(A[i][j]), down[i][j][1] + right[i][j][1] - fives(A[i][j])))\n                res = max(res, min(down[i][j][0] + left[i][j][0] - twos(A[i][j]), down[i][j][1] + left[i][j][1] - fives(A[i][j])))\n        \n        return res\n```
['Dynamic Programming']