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feat: add solutions to lc problem: No.0727 (doocs#1432)
No.0727.Minimum Window Subsequence
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‎solution/0700-0799/0727.Minimum Window Subsequence/README.md

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Original file line numberDiff line numberDiff line change
@@ -35,22 +35,187 @@ S = "abcdebdde", T = "bde"
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：动态规划**
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我们定义 $f[i][j]$ 表示字符串 $s1$ 的前 $i$ 个字符包含字符串 $s2$ 的前 $j$ 个字符时的最短子串的起始位置，如果不存在则为 $0$。
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我们可以得到状态转移方程：
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$$
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f[i][j] = \begin{cases}
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i, & j = 1 \text{ and } s1[i-1] = s2[j] \\
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f[i - 1][j - 1], & j > 1 \text{ and } s1[i-1] = s2[j-1] \\
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f[i - 1][j], & s1[i-1] \ne s2[j-1]
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\end{cases}
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$$
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接下来我们只需要遍历 $s1$，如果 $f[i][n] \gt 0$，则更新最短子串的起始位置和长度。最后返回最短子串即可。
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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 $n$ 分别为字符串 $s1$ 和 $s2$ 的长度。
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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 minWindow(self, s1: str, s2: str) -> str:
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m, n = len(s1), len(s2)
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f = [[0] * (n + 1) for _ in range(m + 1)]
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for i, a in enumerate(s1, 1):
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for j, b in enumerate(s2, 1):
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if a == b:
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f[i][j] = i if j == 1 else f[i - 1][j - 1]
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else:
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f[i][j] = f[i - 1][j]
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p, k = 0, m + 1
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for i, a in enumerate(s1, 1):
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if a == s2[n - 1] and f[i][n]:
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j = f[i][n] - 1
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if i - j < k:
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k = i - j
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p = j
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return "" if k > m else s1[p : p + k]
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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 String minWindow(String s1, String s2) {
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int m = s1.length(), n = s2.length();
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int[][] f = 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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if (s1.charAt(i - 1) == s2.charAt(j - 1)) {
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f[i][j] = j == 1 ? i : f[i - 1][j - 1];
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} else {
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f[i][j] = f[i - 1][j];
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}
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}
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}
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int p = 0, k = m + 1;
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for (int i = 1; i <= m; ++i) {
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if (s1.charAt(i - 1) == s2.charAt(n - 1) && f[i][n] > 0) {
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int j = f[i][n] - 1;
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if (i - j < k) {
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k = i - j;
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p = j;
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}
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}
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}
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return k > m ? "" : s1.substring(p, p + k);
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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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string minWindow(string s1, string s2) {
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int m = s1.size(), n = s2.size();
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int f[m + 1][n + 1];
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memset(f, 0, sizeof(f));
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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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if (s1[i - 1] == s2[j - 1]) {
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f[i][j] = j == 1 ? i : f[i - 1][j - 1];
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} else {
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f[i][j] = f[i - 1][j];
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}
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}
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}
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int p = 0, k = m + 1;
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for (int i = 1; i <= m; ++i) {
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if (s1[i - 1] == s2[n - 1] && f[i][n]) {
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int j = f[i][n] - 1;
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if (i - j < k) {
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k = i - j;
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p = j;
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}
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}
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}
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return k > m ? "" : s1.substr(p, k);
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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 minWindow(s1 string, s2 string) string {
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m, n := len(s1), len(s2)
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f := make([][]int, m+1)
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for i := range f {
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f[i] = make([]int, n+1)
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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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if s1[i-1] == s2[j-1] {
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if j == 1 {
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f[i][j] = i
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} else {
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f[i][j] = f[i-1][j-1]
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}
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} else {
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f[i][j] = f[i-1][j]
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}
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}
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}
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p, k := 0, m+1
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for i := 1; i <= m; i++ {
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if s1[i-1] == s2[n-1] && f[i][n] > 0 {
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j := f[i][n] - 1
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if i-j < k {
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k = i - j
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p = j
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}
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}
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}
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if k > m {
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return ""
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}
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return s1[p : p+k]
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}
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```
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### **TypeScript**
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```ts
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function minWindow(s1: string, s2: string): string {
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const m = s1.length;
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const n = s2.length;
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const f: number[][] = Array(m + 1)
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.fill(0)
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.map(() => 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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if (s1[i - 1] === s2[j - 1]) {
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f[i][j] = j === 1 ? i : f[i - 1][j - 1];
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} else {
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f[i][j] = f[i - 1][j];
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}
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}
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}
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let p = 0;
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let k = m + 1;
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for (let i = 1; i <= m; ++i) {
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if (s1[i - 1] === s2[n - 1] && f[i][n]) {
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const j = f[i][n] - 1;
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if (i - j < k) {
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k = i - j;
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p = j;
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}
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}
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}
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return k > m ? '' : s1.slice(p, p + k);
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}
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```
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### **...**

‎solution/0700-0799/0727.Minimum Window Subsequence/README_EN.md

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@@ -42,13 +42,160 @@
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### **Python3**
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```python
45-
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class Solution:
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def minWindow(self, s1: str, s2: str) -> str:
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m, n = len(s1), len(s2)
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f = [[0] * (n + 1) for _ in range(m + 1)]
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for i, a in enumerate(s1, 1):
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for j, b in enumerate(s2, 1):
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if a == b:
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f[i][j] = i if j == 1 else f[i - 1][j - 1]
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else:
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f[i][j] = f[i - 1][j]
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p, k = 0, m + 1
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for i, a in enumerate(s1, 1):
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if a == s2[n - 1] and f[i][n]:
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j = f[i][n] - 1
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if i - j < k:
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k = i - j
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p = j
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return "" if k > m else s1[p : p + k]
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```
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### **Java**
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```java
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class Solution {
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public String minWindow(String s1, String s2) {
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int m = s1.length(), n = s2.length();
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int[][] f = 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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if (s1.charAt(i - 1) == s2.charAt(j - 1)) {
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f[i][j] = j == 1 ? i : f[i - 1][j - 1];
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} else {
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f[i][j] = f[i - 1][j];
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}
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}
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}
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int p = 0, k = m + 1;
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for (int i = 1; i <= m; ++i) {
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if (s1.charAt(i - 1) == s2.charAt(n - 1) && f[i][n] > 0) {
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int j = f[i][n] - 1;
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if (i - j < k) {
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k = i - j;
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p = j;
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}
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}
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}
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return k > m ? "" : s1.substring(p, p + k);
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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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string minWindow(string s1, string s2) {
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int m = s1.size(), n = s2.size();
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int f[m + 1][n + 1];
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memset(f, 0, sizeof(f));
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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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if (s1[i - 1] == s2[j - 1]) {
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f[i][j] = j == 1 ? i : f[i - 1][j - 1];
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} else {
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f[i][j] = f[i - 1][j];
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}
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}
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}
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int p = 0, k = m + 1;
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for (int i = 1; i <= m; ++i) {
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if (s1[i - 1] == s2[n - 1] && f[i][n]) {
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int j = f[i][n] - 1;
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if (i - j < k) {
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k = i - j;
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p = j;
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}
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}
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}
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return k > m ? "" : s1.substr(p, k);
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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 minWindow(s1 string, s2 string) string {
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m, n := len(s1), len(s2)
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f := make([][]int, m+1)
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for i := range f {
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f[i] = make([]int, n+1)
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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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if s1[i-1] == s2[j-1] {
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if j == 1 {
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f[i][j] = i
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} else {
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f[i][j] = f[i-1][j-1]
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}
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} else {
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f[i][j] = f[i-1][j]
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}
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}
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}
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p, k := 0, m+1
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for i := 1; i <= m; i++ {
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if s1[i-1] == s2[n-1] && f[i][n] > 0 {
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j := f[i][n] - 1
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if i-j < k {
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k = i - j
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p = j
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}
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}
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}
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if k > m {
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return ""
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}
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return s1[p : p+k]
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}
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```
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### **TypeScript**
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```ts
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function minWindow(s1: string, s2: string): string {
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const m = s1.length;
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const n = s2.length;
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const f: number[][] = Array(m + 1)
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.fill(0)
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.map(() => 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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if (s1[i - 1] === s2[j - 1]) {
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f[i][j] = j === 1 ? i : f[i - 1][j - 1];
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} else {
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f[i][j] = f[i - 1][j];
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}
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}
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}
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let p = 0;
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let k = m + 1;
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for (let i = 1; i <= m; ++i) {
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if (s1[i - 1] === s2[n - 1] && f[i][n]) {
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const j = f[i][n] - 1;
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if (i - j < k) {
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k = i - j;
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p = j;
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}
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}
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}
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return k > m ? '' : s1.slice(p, p + k);
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}
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```
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### **...**

‎solution/0700-0799/0727.Minimum Window Subsequence/Solution.cpp

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class Solution {
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public:
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string minWindow(string s1, string s2) {
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int m = s1.size(), n = s2.size();
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int f[m + 1][n + 1];
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memset(f, 0, sizeof(f));
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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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if (s1[i - 1] == s2[j - 1]) {
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f[i][j] = j == 1 ? i : f[i - 1][j - 1];
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} else {
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f[i][j] = f[i - 1][j];
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}
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}
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}
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int p = 0, k = m + 1;
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for (int i = 1; i <= m; ++i) {
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if (s1[i - 1] == s2[n - 1] && f[i][n]) {
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int j = f[i][n] - 1;
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if (i - j < k) {
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k = i - j;
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p = j;
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}
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}
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}
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return k > m ? "" : s1.substr(p, k);
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}
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};

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