diff --git a/solution/2200-2299/2209.Minimum White Tiles After Covering With Carpets/README.md b/solution/2200-2299/2209.Minimum White Tiles After Covering With Carpets/README.md index bb4eabb2ad096..1e7c2bee47fbc 100644 --- a/solution/2200-2299/2209.Minimum White Tiles After Covering With Carpets/README.md +++ b/solution/2200-2299/2209.Minimum White Tiles After Covering With Carpets/README.md @@ -73,18 +73,19 @@ tags: ### 方法一：记忆化搜索 -我们设计一个函数 $dfs(i, j)$ 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 $dfs(0, numCarpets)$。 +我们设计一个函数 $\textit{dfs}(i, j)$ 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 $\textit{dfs}(0, \textit{numCarpets})$。 对于下标 $i$，我们分情况讨论： - 如果 $i \ge n$，说明已经覆盖完所有砖块，返回 $0$； -- 如果 $floor[i] = 0$，则不需要使用地毯，直接跳过即可，即 $dfs(i, j) = dfs(i + 1, j)$； -- 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 $dfs(i, j) = s[n] - s[i]$； -- 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 $dfs(i, j) = min(dfs(i + 1, j), dfs(i + carpetLen, j - 1))$。 +- 如果 $\textit{floor}[i] = 0$，则不需要使用地毯，直接跳过即可，即 $\textit{dfs}(i, j) = \textit{dfs}(i + 1, j)$； +- 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 $\textit{dfs}(i, j) = s[n] - s[i]$； +- 如果 $\textit{floor}[i] = 1$，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 $\textit{dfs}(i, j) = \min(\textit{dfs}(i + 1, + j), \textit{dfs}(i + \textit{carpetLen}, j - 1))$。 记忆化搜索即可。 -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。 +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 $\textit{floor}$ 的长度和 $\textit{numCarpets}$ 的值。 @@ -94,10 +95,10 @@ tags: class Solution: def minimumWhiteTiles(self, floor: str, numCarpets: int, carpetLen: int) -> int: @cache - def dfs(i, j): + def dfs(i: int, j: int) -> int: if i >= n: return 0 - if floor[i] == '0': + if floor[i] == "0": return dfs(i + 1, j) if j == 0: return s[-1] - s[i] @@ -106,7 +107,7 @@ class Solution: n = len(floor) s = [0] * (n + 1) for i, c in enumerate(floor): - s[i + 1] = s[i] + int(c == '1') + s[i + 1] = s[i] + int(c == "1") ans = dfs(0, numCarpets) dfs.cache_clear() return ans @@ -116,17 +117,14 @@ class Solution: ```java class Solution { - private int[][] f; + private Integer[][] f; private int[] s; private int n; private int k; public int minimumWhiteTiles(String floor, int numCarpets, int carpetLen) { n = floor.length(); - f = new int[n][numCarpets + 1]; - for (var e : f) { - Arrays.fill(e, -1); - } + f = new Integer[n][numCarpets + 1]; s = new int[n + 1]; for (int i = 0; i < n; ++i) { s[i + 1] = s[i] + (floor.charAt(i) == '1' ? 1 : 0); @@ -142,7 +140,7 @@ class Solution { if (j == 0) { return s[n] - s[i]; } - if (f[i][j] != -1) { + if (f[i][j] != null) { return f[i][j]; } if (s[i + 1] == s[i]) { @@ -167,12 +165,19 @@ public: for (int i = 0; i < n; ++i) { s[i + 1] = s[i] + (floor[i] == '1'); } - function dfs; - dfs = [&](int i, int j) { - if (i >= n) return 0; - if (j == 0) return s[n] - s[i]; - if (f[i][j] != -1) return f[i][j]; - if (s[i + 1] == s[i]) return dfs(i + 1, j); + auto dfs = [&](this auto&& dfs, int i, int j) -> int { + if (i >= n) { + return 0; + } + if (j == 0) { + return s[n] - s[i]; + } + if (f[i][j] != -1) { + return f[i][j]; + } + if (s[i + 1] == s[i]) { + return dfs(i + 1, j); + } int ans = min(1 + dfs(i + 1, j), dfs(i + carpetLen, j - 1)); f[i][j] = ans; return ans; @@ -220,6 +225,37 @@ func minimumWhiteTiles(floor string, numCarpets int, carpetLen int) int { } ``` +#### TypeScript + +```ts +function minimumWhiteTiles(floor: string, numCarpets: number, carpetLen: number): number { + const n = floor.length; + const f: number[][] = Array.from({ length: n }, () => Array(numCarpets + 1).fill(-1)); + const s: number[] = Array(n + 1).fill(0); + for (let i = 0; i < n; ++i) { + s[i + 1] = s[i] + (floor[i] === '1' ? 1 : 0); + } + const dfs = (i: number, j: number): number => { + if (i >= n) { + return 0; + } + if (j === 0) { + return s[n] - s[i]; + } + if (f[i][j] !== -1) { + return f[i][j]; + } + if (s[i + 1] === s[i]) { + return dfs(i + 1, j); + } + const ans = Math.min(1 + dfs(i + 1, j), dfs(i + carpetLen, j - 1)); + f[i][j] = ans; + return ans; + }; + return dfs(0, numCarpets); +} +``` + diff --git a/solution/2200-2299/2209.Minimum White Tiles After Covering With Carpets/README_EN.md b/solution/2200-2299/2209.Minimum White Tiles After Covering With Carpets/README_EN.md index 0d9cd9ff602aa..a6d4c6fcd9c7f 100644 --- a/solution/2200-2299/2209.Minimum White Tiles After Covering With Carpets/README_EN.md +++ b/solution/2200-2299/2209.Minimum White Tiles After Covering With Carpets/README_EN.md @@ -37,7 +37,7 @@ tags: 
 Input: floor = "10110101", numCarpets = 2, carpetLen = 2
 Output: 2
-Explanation: 
+Explanation:
 The figure above shows one way of covering the tiles with the carpets such that only 2 white tiles are visible.
 No other way of covering the tiles with the carpets can leave less than 2 white tiles visible.
@@ -47,7 +47,7 @@ No other way of covering the tiles with the carpets can leave less than 2 white 
 Input: floor = "11111", numCarpets = 2, carpetLen = 3
 Output: 0
-Explanation: 
+Explanation:
 The figure above shows one way of covering the tiles with the carpets such that no white tiles are visible.
 Note that the carpets are able to overlap one another.
@@ -69,18 +69,18 @@ Note that the carpets are able to overlap one another. ### Solution 1: Memoization Search -Design a function $dfs(i, j)$ to represent the minimum number of white bricks that are not covered starting from index $i$ using $j$ carpets. The answer is $dfs(0, numCarpets)$. +We design a function $\textit{dfs}(i, j)$ to represent the minimum number of white tiles that are not covered starting from index $i$ using $j$ carpets. The answer is $\textit{dfs}(0, \textit{numCarpets})$. -For index $i$, we discuss different cases: +For index $i$, we discuss the following cases: -- If $i \ge n$, it means that all bricks have been covered, return $0$; -- If $floor[i] = 0$, there is no need to use a carpet, just skip it, that is, $dfs(i, j) = dfs(i + 1, j)$; -- If $j = 0$, we can directly calculate the number of remaining white bricks that have not been covered using the prefix sum array $s$, that is, $dfs(i, j) = s[n] - s[i]$; -- If $floor[i] = 1$, we can choose to use a carpet to cover it, or choose not to use a carpet to cover it, and take the minimum of the two, that is, $dfs(i, j) = min(dfs(i + 1, j), dfs(i + carpetLen, j - 1))$. +- If $i \ge n$, it means all tiles have been covered, return $0$; +- If $\textit{floor}[i] = 0$, then we do not need to use a carpet, just skip it, i.e., $\textit{dfs}(i, j) = \textit{dfs}(i + 1, j)$; +- If $j = 0$, then we can directly use the prefix sum array $s$ to calculate the number of remaining uncovered white tiles, i.e., $\textit{dfs}(i, j) = s[n] - s[i]$; +- If $\textit{floor}[i] = 1$, then we can choose to use a carpet or not, and take the minimum of the two, i.e., $\textit{dfs}(i, j) = \min(\textit{dfs}(i + 1, j), \textit{dfs}(i + \textit{carpetLen}, j - 1))$. -Use memoization search. +We use memoization search to solve this problem. -The time complexity is $O(n\times m)$, and the space complexity is $O(n\times m)$. Where $n$ and $m$ are the lengths of the string $floor$ and the value of $numCarpets$ respectively. +The time complexity is $O(n \times m)$, and the space complexity is $O(n \times m)$. Here, $n$ and $m$ are the length of the string $\textit{floor}$ and the value of $\textit{numCarpets}$, respectively. @@ -90,10 +90,10 @@ The time complexity is $O(n\times m)$, and the space complexity is $O(n\times m) class Solution: def minimumWhiteTiles(self, floor: str, numCarpets: int, carpetLen: int) -> int: @cache - def dfs(i, j): + def dfs(i: int, j: int) -> int: if i >= n: return 0 - if floor[i] == '0': + if floor[i] == "0": return dfs(i + 1, j) if j == 0: return s[-1] - s[i] @@ -102,7 +102,7 @@ class Solution: n = len(floor) s = [0] * (n + 1) for i, c in enumerate(floor): - s[i + 1] = s[i] + int(c == '1') + s[i + 1] = s[i] + int(c == "1") ans = dfs(0, numCarpets) dfs.cache_clear() return ans @@ -112,17 +112,14 @@ class Solution: ```java class Solution { - private int[][] f; + private Integer[][] f; private int[] s; private int n; private int k; public int minimumWhiteTiles(String floor, int numCarpets, int carpetLen) { n = floor.length(); - f = new int[n][numCarpets + 1]; - for (var e : f) { - Arrays.fill(e, -1); - } + f = new Integer[n][numCarpets + 1]; s = new int[n + 1]; for (int i = 0; i < n; ++i) { s[i + 1] = s[i] + (floor.charAt(i) == '1' ? 1 : 0); @@ -138,7 +135,7 @@ class Solution { if (j == 0) { return s[n] - s[i]; } - if (f[i][j] != -1) { + if (f[i][j] != null) { return f[i][j]; } if (s[i + 1] == s[i]) { @@ -163,12 +160,19 @@ public: for (int i = 0; i < n; ++i) { s[i + 1] = s[i] + (floor[i] == '1'); } - function dfs; - dfs = [&](int i, int j) { - if (i >= n) return 0; - if (j == 0) return s[n] - s[i]; - if (f[i][j] != -1) return f[i][j]; - if (s[i + 1] == s[i]) return dfs(i + 1, j); + auto dfs = [&](this auto&& dfs, int i, int j) -> int { + if (i >= n) { + return 0; + } + if (j == 0) { + return s[n] - s[i]; + } + if (f[i][j] != -1) { + return f[i][j]; + } + if (s[i + 1] == s[i]) { + return dfs(i + 1, j); + } int ans = min(1 + dfs(i + 1, j), dfs(i + carpetLen, j - 1)); f[i][j] = ans; return ans; @@ -216,6 +220,37 @@ func minimumWhiteTiles(floor string, numCarpets int, carpetLen int) int { } ``` +#### TypeScript + +```ts +function minimumWhiteTiles(floor: string, numCarpets: number, carpetLen: number): number { + const n = floor.length; + const f: number[][] = Array.from({ length: n }, () => Array(numCarpets + 1).fill(-1)); + const s: number[] = Array(n + 1).fill(0); + for (let i = 0; i < n; ++i) { + s[i + 1] = s[i] + (floor[i] === '1' ? 1 : 0); + } + const dfs = (i: number, j: number): number => { + if (i >= n) { + return 0; + } + if (j === 0) { + return s[n] - s[i]; + } + if (f[i][j] !== -1) { + return f[i][j]; + } + if (s[i + 1] === s[i]) { + return dfs(i + 1, j); + } + const ans = Math.min(1 + dfs(i + 1, j), dfs(i + carpetLen, j - 1)); + f[i][j] = ans; + return ans; + }; + return dfs(0, numCarpets); +} +``` + diff --git a/solution/2200-2299/2209.Minimum White Tiles After Covering With Carpets/Solution.cpp b/solution/2200-2299/2209.Minimum White Tiles After Covering With Carpets/Solution.cpp index 4cf3b8c1c9102..0cb2d3cff4d8e 100644 --- a/solution/2200-2299/2209.Minimum White Tiles After Covering With Carpets/Solution.cpp +++ b/solution/2200-2299/2209.Minimum White Tiles After Covering With Carpets/Solution.cpp @@ -7,16 +7,23 @@ class Solution { for (int i = 0; i < n; ++i) { s[i + 1] = s[i] + (floor[i] == '1'); } - function dfs; - dfs = [&](int i, int j) { - if (i >= n) return 0; - if (j == 0) return s[n] - s[i]; - if (f[i][j] != -1) return f[i][j]; - if (s[i + 1] == s[i]) return dfs(i + 1, j); + auto dfs = [&](this auto&& dfs, int i, int j) -> int { + if (i >= n) { + return 0; + } + if (j == 0) { + return s[n] - s[i]; + } + if (f[i][j] != -1) { + return f[i][j]; + } + if (s[i + 1] == s[i]) { + return dfs(i + 1, j); + } int ans = min(1 + dfs(i + 1, j), dfs(i + carpetLen, j - 1)); f[i][j] = ans; return ans; }; return dfs(0, numCarpets); } -}; \ No newline at end of file +}; diff --git a/solution/2200-2299/2209.Minimum White Tiles After Covering With Carpets/Solution.java b/solution/2200-2299/2209.Minimum White Tiles After Covering With Carpets/Solution.java index b3032cf4c8bee..06550c907dfdf 100644 --- a/solution/2200-2299/2209.Minimum White Tiles After Covering With Carpets/Solution.java +++ b/solution/2200-2299/2209.Minimum White Tiles After Covering With Carpets/Solution.java @@ -1,15 +1,12 @@ class Solution { - private int[][] f; + private Integer[][] f; private int[] s; private int n; private int k; public int minimumWhiteTiles(String floor, int numCarpets, int carpetLen) { n = floor.length(); - f = new int[n][numCarpets + 1]; - for (var e : f) { - Arrays.fill(e, -1); - } + f = new Integer[n][numCarpets + 1]; s = new int[n + 1]; for (int i = 0; i < n; ++i) { s[i + 1] = s[i] + (floor.charAt(i) == '1' ? 1 : 0); @@ -25,7 +22,7 @@ private int dfs(int i, int j) { if (j == 0) { return s[n] - s[i]; } - if (f[i][j] != -1) { + if (f[i][j] != null) { return f[i][j]; } if (s[i + 1] == s[i]) { @@ -35,4 +32,4 @@ private int dfs(int i, int j) { f[i][j] = ans; return ans; } -} \ No newline at end of file +} diff --git a/solution/2200-2299/2209.Minimum White Tiles After Covering With Carpets/Solution.py b/solution/2200-2299/2209.Minimum White Tiles After Covering With Carpets/Solution.py index 0e158a0899c7d..a358b8bf399b2 100644 --- a/solution/2200-2299/2209.Minimum White Tiles After Covering With Carpets/Solution.py +++ b/solution/2200-2299/2209.Minimum White Tiles After Covering With Carpets/Solution.py @@ -1,10 +1,10 @@ class Solution: def minimumWhiteTiles(self, floor: str, numCarpets: int, carpetLen: int) -> int: @cache - def dfs(i, j): + def dfs(i: int, j: int) -> int: if i >= n: return 0 - if floor[i] == '0': + if floor[i] == "0": return dfs(i + 1, j) if j == 0: return s[-1] - s[i] @@ -13,7 +13,7 @@ def dfs(i, j): n = len(floor) s = [0] * (n + 1) for i, c in enumerate(floor): - s[i + 1] = s[i] + int(c == '1') + s[i + 1] = s[i] + int(c == "1") ans = dfs(0, numCarpets) dfs.cache_clear() return ans diff --git a/solution/2200-2299/2209.Minimum White Tiles After Covering With Carpets/Solution.ts b/solution/2200-2299/2209.Minimum White Tiles After Covering With Carpets/Solution.ts new file mode 100644 index 0000000000000..23e3b636cbc21 --- /dev/null +++ b/solution/2200-2299/2209.Minimum White Tiles After Covering With Carpets/Solution.ts @@ -0,0 +1,26 @@ +function minimumWhiteTiles(floor: string, numCarpets: number, carpetLen: number): number { + const n = floor.length; + const f: number[][] = Array.from({ length: n }, () => Array(numCarpets + 1).fill(-1)); + const s: number[] = Array(n + 1).fill(0); + for (let i = 0; i < n; ++i) { + s[i + 1] = s[i] + (floor[i] === '1' ? 1 : 0); + } + const dfs = (i: number, j: number): number => { + if (i >= n) { + return 0; + } + if (j === 0) { + return s[n] - s[i]; + } + if (f[i][j] !== -1) { + return f[i][j]; + } + if (s[i + 1] === s[i]) { + return dfs(i + 1, j); + } + const ans = Math.min(1 + dfs(i + 1, j), dfs(i + carpetLen, j - 1)); + f[i][j] = ans; + return ans; + }; + return dfs(0, numCarpets); +}