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feat: add solutions to lc problem: No.2209
No.2209.Minimum White Tiles After Covering With Carpets
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‎solution/2200-2299/2209.Minimum White Tiles After Covering With Carpets/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：记忆化搜索**
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设计函数 $dfs(i, j)$ 表示从下标 $i$ 开始，使用 $j$ 条地毯，最少有多少个白色砖块没有被覆盖。答案即为 $dfs(0, numCarpets)$。
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对于下标 $i$，我们分情况讨论：
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- 如果 $i \ge n$，说明已经覆盖完所有砖块，返回 $0$；
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- 如果 $floor[i] = 0$，则不需要使用地毯，直接跳过即可，即 $dfs(i, j) = dfs(i + 1, j)$；
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- 如果 $j = 0$，那么我们可以直接利用前缀和数组 $s$ 计算出剩余未被覆盖的白色砖块的数目，即 $dfs(i, j) = s[n] - s[i]$；
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- 如果 $floor[i] = 1$，那么我们可以选择使用地毯覆盖，也可以选择不使用地毯覆盖，取两者的最小值即可，即 $dfs(i, j) = min(dfs(i + 1, j), dfs(i + carpetLen, j - 1))$。
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记忆化搜索即可。
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时间复杂度 $O(n\times m)$，空间复杂度 $O(n\times m)$。其中 $n$ 和 $m$ 分别为字符串 $floor$ 的长度和 $numCarpets$ 的值。
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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 minimumWhiteTiles(self, floor: str, numCarpets: int, carpetLen: int) -> int:
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@cache
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def dfs(i, j):
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if i >= n:
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return 0
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if floor[i] == '0':
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return dfs(i + 1, j)
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if j == 0:
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return s[-1] - s[i]
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return min(1 + dfs(i + 1, j), dfs(i + carpetLen, j - 1))
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n = len(floor)
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s = [0] * (n + 1)
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for i, c in enumerate(floor):
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s[i + 1] = s[i] + int(c == '1')
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ans = dfs(0, numCarpets)
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dfs.cache_clear()
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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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private int[][] f;
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private int[] s;
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private int n;
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private int k;
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public int minimumWhiteTiles(String floor, int numCarpets, int carpetLen) {
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n = floor.length();
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f = new int[n][numCarpets + 1];
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for (var e : f) {
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Arrays.fill(e, -1);
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}
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s = new int[n + 1];
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for (int i = 0; i < n; ++i) {
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s[i + 1] = s[i] + (floor.charAt(i) == '1' ? 1 : 0);
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}
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k = carpetLen;
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return dfs(0, numCarpets);
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}
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private int dfs(int i, int j) {
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if (i >= n) {
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return 0;
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}
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if (j == 0) {
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return s[n] - s[i];
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}
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if (f[i][j] != -1) {
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return f[i][j];
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}
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if (s[i + 1] == s[i]) {
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return dfs(i + 1, j);
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}
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int ans = Math.min(1 + dfs(i + 1, j), dfs(i + k, j - 1));
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f[i][j] = ans;
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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 minimumWhiteTiles(string floor, int numCarpets, int carpetLen) {
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int n = floor.size();
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vector<vector<int>> f(n, vector<int>(numCarpets + 1, -1));
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vector<int> s(n + 1);
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for (int i = 0; i < n; ++i) {
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s[i + 1] = s[i] + (floor[i] == '1');
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}
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function<int(int, int)> dfs;
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dfs = [&](int i, int j) {
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if (i >= n) return 0;
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if (j == 0) return s[n] - s[i];
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if (f[i][j] != -1) return f[i][j];
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if (s[i + 1] == s[i]) return dfs(i + 1, j);
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int ans = min(1 + dfs(i + 1, j), dfs(i + carpetLen, j - 1));
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f[i][j] = ans;
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return ans;
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};
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return dfs(0, numCarpets);
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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 minimumWhiteTiles(floor string, numCarpets int, carpetLen int) int {
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n := len(floor)
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f := make([][]int, n)
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for i := range f {
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f[i] = make([]int, numCarpets+1)
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for j := range f[i] {
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f[i][j] = -1
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}
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}
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s := make([]int, n+1)
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for i, c := range floor {
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s[i+1] = s[i] + int(c-'0')
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}
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var dfs func(i, j int) int
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dfs = func(i, j int) int {
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if i >= n {
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return 0
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}
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if j == 0 {
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return s[n] - s[i]
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}
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if f[i][j] != -1 {
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return f[i][j]
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}
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if s[i+1] == s[i] {
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return dfs(i+1, j)
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}
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ans := min(1+dfs(i+1, j), dfs(i+carpetLen, j-1))
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f[i][j] = ans
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return ans
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}
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return dfs(0, numCarpets)
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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**

‎solution/2200-2299/2209.Minimum White Tiles After Covering With Carpets/README_EN.md

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Original file line numberDiff line numberDiff line change
@@ -52,13 +52,140 @@ Note that the carpets are able to overlap one another.
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### **Python3**
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```python
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class Solution:
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def minimumWhiteTiles(self, floor: str, numCarpets: int, carpetLen: int) -> int:
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@cache
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def dfs(i, j):
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if i >= n:
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return 0
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if floor[i] == '0':
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return dfs(i + 1, j)
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if j == 0:
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return s[-1] - s[i]
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return min(1 + dfs(i + 1, j), dfs(i + carpetLen, j - 1))
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n = len(floor)
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s = [0] * (n + 1)
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for i, c in enumerate(floor):
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s[i + 1] = s[i] + int(c == '1')
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ans = dfs(0, numCarpets)
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dfs.cache_clear()
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return ans
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```
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### **Java**
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```java
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class Solution {
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private int[][] f;
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private int[] s;
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private int n;
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private int k;
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public int minimumWhiteTiles(String floor, int numCarpets, int carpetLen) {
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n = floor.length();
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f = new int[n][numCarpets + 1];
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for (var e : f) {
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Arrays.fill(e, -1);
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}
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s = new int[n + 1];
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for (int i = 0; i < n; ++i) {
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s[i + 1] = s[i] + (floor.charAt(i) == '1' ? 1 : 0);
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}
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k = carpetLen;
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return dfs(0, numCarpets);
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}
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private int dfs(int i, int j) {
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if (i >= n) {
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return 0;
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}
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if (j == 0) {
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return s[n] - s[i];
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}
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if (f[i][j] != -1) {
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return f[i][j];
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}
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if (s[i + 1] == s[i]) {
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return dfs(i + 1, j);
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}
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int ans = Math.min(1 + dfs(i + 1, j), dfs(i + k, j - 1));
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f[i][j] = ans;
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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 minimumWhiteTiles(string floor, int numCarpets, int carpetLen) {
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int n = floor.size();
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vector<vector<int>> f(n, vector<int>(numCarpets + 1, -1));
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vector<int> s(n + 1);
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for (int i = 0; i < n; ++i) {
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s[i + 1] = s[i] + (floor[i] == '1');
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}
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function<int(int, int)> dfs;
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dfs = [&](int i, int j) {
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if (i >= n) return 0;
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if (j == 0) return s[n] - s[i];
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if (f[i][j] != -1) return f[i][j];
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if (s[i + 1] == s[i]) return dfs(i + 1, j);
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int ans = min(1 + dfs(i + 1, j), dfs(i + carpetLen, j - 1));
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f[i][j] = ans;
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return ans;
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};
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return dfs(0, numCarpets);
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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 minimumWhiteTiles(floor string, numCarpets int, carpetLen int) int {
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n := len(floor)
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f := make([][]int, n)
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for i := range f {
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f[i] = make([]int, numCarpets+1)
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for j := range f[i] {
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f[i][j] = -1
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}
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}
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s := make([]int, n+1)
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for i, c := range floor {
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s[i+1] = s[i] + int(c-'0')
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}
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var dfs func(i, j int) int
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dfs = func(i, j int) int {
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if i >= n {
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return 0
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}
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if j == 0 {
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return s[n] - s[i]
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}
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if f[i][j] != -1 {
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return f[i][j]
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}
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if s[i+1] == s[i] {
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return dfs(i+1, j)
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}
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ans := min(1+dfs(i+1, j), dfs(i+carpetLen, j-1))
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f[i][j] = ans
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return ans
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}
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return dfs(0, numCarpets)
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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**

‎solution/2200-2299/2209.Minimum White Tiles After Covering With Carpets/Solution.cpp

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class Solution {
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public:
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int minimumWhiteTiles(string floor, int numCarpets, int carpetLen) {
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int n = floor.size();
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vector<vector<int>> f(n, vector<int>(numCarpets + 1, -1));
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vector<int> s(n + 1);
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for (int i = 0; i < n; ++i) {
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s[i + 1] = s[i] + (floor[i] == '1');
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}
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function<int(int, int)> dfs;
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dfs = [&](int i, int j) {
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if (i >= n) return 0;
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if (j == 0) return s[n] - s[i];
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if (f[i][j] != -1) return f[i][j];
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if (s[i + 1] == s[i]) return dfs(i + 1, j);
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int ans = min(1 + dfs(i + 1, j), dfs(i + carpetLen, j - 1));
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f[i][j] = ans;
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return ans;
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};
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return dfs(0, numCarpets);
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}
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};

‎solution/2200-2299/2209.Minimum White Tiles After Covering With Carpets/Solution.go

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func minimumWhiteTiles(floor string, numCarpets int, carpetLen int) int {
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n := len(floor)
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f := make([][]int, n)
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for i := range f {
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f[i] = make([]int, numCarpets+1)
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for j := range f[i] {
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f[i][j] = -1
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}
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}
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s := make([]int, n+1)
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for i, c := range floor {
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s[i+1] = s[i] + int(c-'0')
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}
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var dfs func(i, j int) int
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dfs = func(i, j int) int {
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if i >= n {
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return 0
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}
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if j == 0 {
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return s[n] - s[i]
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}
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if f[i][j] != -1 {
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return f[i][j]
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}
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if s[i+1] == s[i] {
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return dfs(i+1, j)
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}
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ans := min(1+dfs(i+1, j), dfs(i+carpetLen, j-1))
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f[i][j] = ans
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return ans
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}
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return dfs(0, numCarpets)
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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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