Maximum Trailing Zeros in a Cornered Path
<p>You are given a 2D integer array <code>grid</code> of size <code>m x n</code>, where each cell contains a positive integer.</p>

<p>A <strong>cornered path</strong> is defined as a set of adjacent cells with <strong>at most</strong> one turn. More specifically, the path should exclusively move either <strong>horizontally</strong> or <strong>vertically</strong> up to the turn (if there is one), without returning to a previously visited cell. After the turn, the path will then move exclusively in the <strong>alternate</strong> direction: move vertically if it moved horizontally, and vice versa, also without returning to a previously visited cell.</p>

<p>The <strong>product</strong> of a path is defined as the product of all the values in the path.</p>

<p>Return <em>the <strong>maximum</strong> number of <strong>trailing zeros</strong> in the product of a cornered path found in </em><code>grid</code>.</p>

<p>Note:</p>

<ul>
	<li><strong>Horizontal</strong> movement means moving in either the left or right direction.</li>
	<li><strong>Vertical</strong> movement means moving in either the up or down direction.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2200-2299/2245.Maximum%20Trailing%20Zeros%20in%20a%20Cornered%20Path/images/ex1new2.jpg" style="width: 577px; height: 190px;" />
<pre>
<strong>Input:</strong> grid = [[23,17,15,3,20],[8,1,20,27,11],[9,4,6,2,21],[40,9,1,10,6],[22,7,4,5,3]]
<strong>Output:</strong> 3
<strong>Explanation:</strong> The grid on the left shows a valid cornered path.
It has a product of 15 * 20 * 6 * 1 * 10 = 18000 which has 3 trailing zeros.
It can be shown that this is the maximum trailing zeros in the product of a cornered path.

The grid in the middle is not a cornered path as it has more than one turn.
The grid on the right is not a cornered path as it requires a return to a previously visited cell.

</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2200-2299/2245.Maximum%20Trailing%20Zeros%20in%20a%20Cornered%20Path/images/ex2.jpg" style="width: 150px; height: 157px;" />
<pre>
<strong>Input:</strong> grid = [[4,3,2],[7,6,1],[8,8,8]]
<strong>Output:</strong> 0
<strong>Explanation:</strong> The grid is shown in the figure above.
There are no cornered paths in the grid that result in a product with a trailing zero.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= m * n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= grid[i][j] &lt;= 1000</code></li>
</ul>
Array; Matrix; Prefix Sum
class Solution {
public:
    int maxTrailingZeros(vector<vector<int>>& grid) {
        int m = grid.size(), n = grid[0].size();
        vector<vector<int>> r2(m + 1, vector<int>(n + 1));
        vector<vector<int>> c2(m + 1, vector<int>(n + 1));
        vector<vector<int>> r5(m + 1, vector<int>(n + 1));
        vector<vector<int>> c5(m + 1, vector<int>(n + 1));
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                int x = grid[i - 1][j - 1];
                int s2 = 0, s5 = 0;
                for (; x % 2 == 0; x /= 2) {
                    ++s2;
                }
                for (; x % 5 == 0; x /= 5) {
                    ++s5;
                }
                r2[i][j] = r2[i][j - 1] + s2;
                c2[i][j] = c2[i - 1][j] + s2;
                r5[i][j] = r5[i][j - 1] + s5;
                c5[i][j] = c5[i - 1][j] + s5;
            }
        }
        int ans = 0;
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                int a = min(r2[i][j] + c2[i - 1][j], r5[i][j] + c5[i - 1][j]);
                int b = min(r2[i][j] + c2[m][j] - c2[i][j], r5[i][j] + c5[m][j] - c5[i][j]);
                int c = min(r2[i][n] - r2[i][j] + c2[i][j], r5[i][n] - r5[i][j] + c5[i][j]);
                int d = min(r2[i][n] - r2[i][j - 1] + c2[m][j] - c2[i][j], r5[i][n] - r5[i][j - 1] + c5[m][j] - c5[i][j]);
                ans = max({ans, a, b, c, d});
            }
        }
        return ans;
    }
};