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feat: add solutions to lc problem: No.2245
No.2245.Maximum Trailing Zeros in a Cornered Path
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‎solution/2200-2299/2245.Maximum Trailing Zeros in a Cornered Path/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：前缀和 + 枚举拐点**
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首先我们要明确，对于一个乘积，尾随零的个数取决于因子中 $2$ 和 $5$ 的个数的较小值。另外，每一条转角路径应该覆盖尽可能多的数，因此，它一定是从某个边界出发，到达某个拐点，再到达另一个边界。
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因此，我们可以创建四个二维数组 $r2$, $c2$, $r5$, $c5$ 来记录每一行和每一列中 $2$ 和 $5$ 的个数。其中：
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- `r2[i][j]` 表示第 $i$ 行中从第 $1$ 列到第 $j$ 列的 $2$ 的个数；
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- `c2[i][j]` 表示第 $j$ 列中从第 $1$ 行到第 $i$ 行的 $2$ 的个数；
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- `r5[i][j]` 表示第 $i$ 行中从第 $1$ 列到第 $j$ 列的 $5$ 的个数；
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- `c5[i][j]` 表示第 $j$ 列中从第 $1$ 行到第 $i$ 行的 $5$ 的个数。
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接下来，我们遍历二维数组 `grid`，对于每个数，我们计算它的 $2$ 和 $5$ 的个数，然后更新四个二维数组。
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然后，我们枚举拐点 $(i, j)$，对于每个拐点，我们计算四个值，其中：
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- `a` 表示从 $(i, 1)$ 右移到 $(i, j)$，再从 $(i, j)$ 拐头向上移动到 $(1, j)$ 的路径中 $2$ 的个数和 $5$ 的个数的较小值；
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- `b` 表示从 $(i, 1)$ 右移到 $(i, j)$，再从 $(i, j)$ 拐头向下移动到 $(m, j)$ 的路径中 $2$ 的个数和 $5$ 的个数的较小值；
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- `c` 表示从 $(i, n)$ 左移到 $(i, j)$，再从 $(i, j)$ 拐头向上移动到 $(1, j)$ 的路径中 $2$ 的个数和 $5$ 的个数的较小值；
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- `d` 表示从 $(i, n)$ 左移到 $(i, j)$，再从 $(i, j)$ 拐头向下移动到 $(m, j)$ 的路径中 $2$ 的个数和 $5$ 的个数的较小值。
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每一次枚举，我们取这四个值的最大值，然后更新答案。
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最后，我们返回答案即可。
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时间复杂度 $O(m \times n)$，空间复杂度 $O(m \times n)$。其中 $m$ 和 $n$ 分别是二维数组 `grid` 的行数和列数。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def maxTrailingZeros(self, grid: List[List[int]]) -> int:
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m, n = len(grid), len(grid[0])
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r2 = [[0] * (n + 1) for _ in range(m + 1)]
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c2 = [[0] * (n + 1) for _ in range(m + 1)]
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r5 = [[0] * (n + 1) for _ in range(m + 1)]
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c5 = [[0] * (n + 1) for _ in range(m + 1)]
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for i, row in enumerate(grid, 1):
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for j, x in enumerate(row, 1):
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s2 = s5 = 0
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while x % 2 == 0:
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x //= 2
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s2 += 1
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while x % 5 == 0:
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x //= 5
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s5 += 1
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r2[i][j] = r2[i][j - 1] + s2
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c2[i][j] = c2[i - 1][j] + s2
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r5[i][j] = r5[i][j - 1] + s5
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c5[i][j] = c5[i - 1][j] + s5
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ans = 0
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for i in range(1, m + 1):
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for j in range(1, n + 1):
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a = min(r2[i][j] + c2[i - 1][j], r5[i][j] + c5[i - 1][j])
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b = min(r2[i][j] + c2[m][j] - c2[i][j], r5[i][j] + c5[m][j] - c5[i][j])
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c = min(r2[i][n] - r2[i][j] + c2[i][j], r5[i][n] - r5[i][j] + c5[i][j])
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d = min(
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r2[i][n] - r2[i][j - 1] + c2[m][j] - c2[i][j],
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r5[i][n] - r5[i][j - 1] + c5[m][j] - c5[i][j],
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)
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ans = max(ans, a, b, c, d)
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return ans
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int maxTrailingZeros(int[][] grid) {
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int m = grid.length, n = grid[0].length;
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int[][] r2 = new int[m + 1][n + 1];
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int[][] c2 = new int[m + 1][n + 1];
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int[][] r5 = new int[m + 1][n + 1];
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int[][] c5 = new int[m + 1][n + 1];
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for (int i = 1; i <= m; ++i) {
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for (int j = 1; j <= n; ++j) {
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int x = grid[i - 1][j - 1];
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int s2 = 0, s5 = 0;
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for (; x % 2 == 0; x /= 2) {
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++s2;
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}
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for (; x % 5 == 0; x /= 5) {
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++s5;
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}
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r2[i][j] = r2[i][j - 1] + s2;
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c2[i][j] = c2[i - 1][j] + s2;
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r5[i][j] = r5[i][j - 1] + s5;
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c5[i][j] = c5[i - 1][j] + s5;
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}
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}
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int ans = 0;
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for (int i = 1; i <= m; ++i) {
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for (int j = 1; j <= n; ++j) {
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int a = Math.min(r2[i][j] + c2[i - 1][j], r5[i][j] + c5[i - 1][j]);
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int b = Math.min(r2[i][j] + c2[m][j] - c2[i][j], r5[i][j] + c5[m][j] - c5[i][j]);
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int c = Math.min(r2[i][n] - r2[i][j] + c2[i][j], r5[i][n] - r5[i][j] + c5[i][j]);
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int d = Math.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]);
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ans = Math.max(ans, Math.max(a, Math.max(b, Math.max(c, d))));
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}
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int maxTrailingZeros(vector<vector<int>>& grid) {
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int m = grid.size(), n = grid[0].size();
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vector<vector<int>> r2(m + 1, vector<int>(n + 1));
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vector<vector<int>> c2(m + 1, vector<int>(n + 1));
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vector<vector<int>> r5(m + 1, vector<int>(n + 1));
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vector<vector<int>> c5(m + 1, vector<int>(n + 1));
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for (int i = 1; i <= m; ++i) {
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for (int j = 1; j <= n; ++j) {
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int x = grid[i - 1][j - 1];
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int s2 = 0, s5 = 0;
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for (; x % 2 == 0; x /= 2) {
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++s2;
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}
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for (; x % 5 == 0; x /= 5) {
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++s5;
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}
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r2[i][j] = r2[i][j - 1] + s2;
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c2[i][j] = c2[i - 1][j] + s2;
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r5[i][j] = r5[i][j - 1] + s5;
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c5[i][j] = c5[i - 1][j] + s5;
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}
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}
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int ans = 0;
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for (int i = 1; i <= m; ++i) {
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for (int j = 1; j <= n; ++j) {
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int a = min(r2[i][j] + c2[i - 1][j], r5[i][j] + c5[i - 1][j]);
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int b = min(r2[i][j] + c2[m][j] - c2[i][j], r5[i][j] + c5[m][j] - c5[i][j]);
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int c = min(r2[i][n] - r2[i][j] + c2[i][j], r5[i][n] - r5[i][j] + c5[i][j]);
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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]);
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ans = max({ans, a, b, c, d});
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}
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func maxTrailingZeros(grid [][]int) (ans int) {
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m, n := len(grid), len(grid[0])
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r2 := get(m+1, n+1)
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c2 := get(m+1, n+1)
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r5 := get(m+1, n+1)
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c5 := get(m+1, n+1)
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for i := 1; i <= m; i++ {
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for j := 1; j <= n; j++ {
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x := grid[i-1][j-1]
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s2, s5 := 0, 0
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for ; x%2 == 0; x /= 2 {
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s2++
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}
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for ; x%5 == 0; x /= 5 {
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s5++
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}
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r2[i][j] = r2[i][j-1] + s2
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c2[i][j] = c2[i-1][j] + s2
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r5[i][j] = r5[i][j-1] + s5
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c5[i][j] = c5[i-1][j] + s5
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}
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}
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for i := 1; i <= m; i++ {
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for j := 1; j <= n; j++ {
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a := min(r2[i][j]+c2[i-1][j], r5[i][j]+c5[i-1][j])
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b := min(r2[i][j]+c2[m][j]-c2[i][j], r5[i][j]+c5[m][j]-c5[i][j])
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c := min(r2[i][n]-r2[i][j]+c2[i][j], r5[i][n]-r5[i][j]+c5[i][j])
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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])
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ans = max(ans, max(a, max(b, max(c, d))))
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}
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}
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return
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}
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func get(m, n int) [][]int {
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f := make([][]int, m)
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for i := range f {
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f[i] = make([]int, n)
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}
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return f
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}
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```
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### **TypeScript**
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```ts
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function maxTrailingZeros(grid: number[][]): number {
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let m = grid.length,
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n = grid[0].length;
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let r2 = Array.from({ length: m + 1 }, v => new Array(n + 1).fill(0)),
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c2 = Array.from({ length: m + 1 }, v => new Array(n + 1).fill(0)),
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r5 = Array.from({ length: m + 1 }, v => new Array(n + 1).fill(0)),
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c5 = Array.from({ length: m + 1 }, v => new Array(n + 1).fill(0));
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for (let i = 1; i <= m; i++) {
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for (let j = 1; j <= n; j++) {
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let cur = grid[i - 1][j - 1];
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let two = 0,
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five = 0;
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while (cur % 2 == 0) two++, (cur /= 2);
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while (cur % 5 == 0) five++, (cur /= 5);
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r2[i][j] = r2[i - 1][j] + two;
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c2[i][j] = c2[i][j - 1] + two;
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r5[i][j] = r5[i - 1][j] + five;
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c5[i][j] = c5[i][j - 1] + five;
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const m = grid.length;
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const n = grid[0].length;
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const r2 = Array.from({ length: m + 1 }, () => new Array(n + 1).fill(0));
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const c2 = Array.from({ length: m + 1 }, () => new Array(n + 1).fill(0));
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const r5 = Array.from({ length: m + 1 }, () => new Array(n + 1).fill(0));
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const c5 = Array.from({ length: m + 1 }, () => new Array(n + 1).fill(0));
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for (let i = 1; i <= m; ++i) {
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for (let j = 1; j <= n; ++j) {
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let x = grid[i - 1][j - 1];
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let s2 = 0;
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let s5 = 0;
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for (; x % 2 == 0; x = Math.floor(x / 2)) {
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++s2;
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}
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for (; x % 5 == 0; x = Math.floor(x / 5)) {
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++s5;
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}
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r2[i][j] = r2[i][j - 1] + s2;
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c2[i][j] = c2[i - 1][j] + s2;
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r5[i][j] = r5[i][j - 1] + s5;
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c5[i][j] = c5[i - 1][j] + s5;
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}
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}
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let ans = 0;
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function getMin(i0, j0, i1, j1, i3, j3, i4, j4): number {
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// 横向开始、结束，竖向开始、结束
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const two = c2[i1][j1] - c2[i0][j0] + r2[i4][j4] - r2[i3][j3];
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const five = c5[i1][j1] - c5[i0][j0] + r5[i4][j4] - r5[i3][j3];
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return Math.min(two, five);
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}
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for (let i = 1; i <= m; i++) {
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for (let j = 1; j <= n; j++) {
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const leftToTop = getMin(i, 0, i, j, 0, j, i - 1, j),
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leftToBotton = getMin(i, 0, i, j, i, j, m, j),
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rightToTop = getMin(i, j, i, n, 0, j, i, j),
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rightToBotton = getMin(i, j, i, n, i - 1, j, m, j);
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ans = Math.max(
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leftToTop,
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leftToBotton,
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rightToTop,
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rightToBotton,
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ans,
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for (let i = 1; i <= m; ++i) {
309+
for (let j = 1; j <= n; ++j) {
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const a = Math.min(
311+
r2[i][j] + c2[i - 1][j],
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r5[i][j] + c5[i - 1][j],
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);
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const b = Math.min(
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r2[i][j] + c2[m][j] - c2[i][j],
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r5[i][j] + c5[m][j] - c5[i][j],
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);
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const c = Math.min(
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r2[i][n] - r2[i][j] + c2[i][j],
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r5[i][n] - r5[i][j] + c5[i][j],
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);
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const d = Math.min(
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r2[i][n] - r2[i][j - 1] + c2[m][j] - c2[i][j],
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r5[i][n] - r5[i][j - 1] + c5[m][j] - c5[i][j],
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);
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ans = Math.max(ans, a, b, c, d);
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}
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}
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return ans;

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