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feat: add solutions to lc problems: No.0712,0713
* No.0712.Minimum ASCII Delete Sum for Two Strings * No.0713.Subarray Product Less Than K
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‎solution/0700-0799/0712.Minimum ASCII Delete Sum for Two Strings/README.md

+119-35
Original file line numberDiff line numberDiff line change
@@ -46,11 +46,27 @@
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**方法一：动态规划**
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类似[1143. 最长公共子序列](/solution/1100-1199/1143.Longest%20Common%20Subsequence/README.md)。
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我们定义 $f[i][j]$ 表示使得 $s_1$ 的前 $i$ 个字符和 $s_2$ 的前 $j$ 个字符相等所需删除字符的 ASCII 值的最小和。那么答案就是 $f[m][n]$。
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定义 `dp[i][j]` 表示使得 `s1[0:i-1]` 和 `s2[0:j-1]` 两个字符串相等所需删除的字符的 ASCII 值的最小值。
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如果 $s_1[i-1] = s_2[j-1]$，那么 $f[i][j] = f[i-1][j-1]$。否则，我们可以删除 $s_1[i-1]$ 或者 $s_2[j-1]$ 中的一个，使得 $f[i][j]$ 达到最小。因此，状态转移方程如下：
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53-
时间复杂度 O(mn)。
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$$
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f[i][j]=
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\begin{cases}
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f[i-1][j-1], & s_1[i-1] = s_2[j-1] \\
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min(f[i-1][j] + s_1[i-1], f[i][j-1] + s_2[j-1]), & s_1[i-1] \neq s_2[j-1]
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\end{cases}
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$$
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初始状态为 $f[0][j] = f[0][j-1] + s_2[j-1]$，$f[i][0] = f[i-1][0] + s_1[i-1]$。
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最后返回 $f[m][n]$ 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 $s_1$ 和 $s_2$ 的长度。
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相似题目：
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- [1143. 最长公共子序列](/solution/1100-1199/1143.Longest%20Common%20Subsequence/README.md)
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<!-- tabs:start -->
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@@ -62,20 +78,20 @@
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class Solution:
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def minimumDeleteSum(self, s1: str, s2: str) -> int:
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m, n = len(s1), len(s2)
65-
dp = [[0] * (n + 1) for _ in range(m + 1)]
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f = [[0] * (n + 1) for _ in range(m + 1)]
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for i in range(1, m + 1):
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dp[i][0] = dp[i - 1][0] + ord(s1[i - 1])
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f[i][0] = f[i - 1][0] + ord(s1[i - 1])
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for j in range(1, n + 1):
69-
dp[0][j] = dp[0][j - 1] + ord(s2[j - 1])
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f[0][j] = f[0][j - 1] + ord(s2[j - 1])
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for i in range(1, m + 1):
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for j in range(1, n + 1):
7288
if s1[i - 1] == s2[j - 1]:
73-
dp[i][j] = dp[i - 1][j - 1]
89+
f[i][j] = f[i - 1][j - 1]
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else:
75-
dp[i][j] = min(
76-
dp[i - 1][j] + ord(s1[i - 1]), dp[i][j - 1] + ord(s2[j - 1])
91+
f[i][j] = min(
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f[i - 1][j] + ord(s1[i - 1]), f[i][j - 1] + ord(s2[j - 1])
7793
)
78-
return dp[-1][-1]
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return f[m][n]
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```
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### **Java**
@@ -86,24 +102,23 @@ class Solution:
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class Solution {
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public int minimumDeleteSum(String s1, String s2) {
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int m = s1.length(), n = s2.length();
89-
int[][] dp = new int[m + 1][n + 1];
105+
int[][] f = new int[m + 1][n + 1];
90106
for (int i = 1; i <= m; ++i) {
91-
dp[i][0] = dp[i - 1][0] + s1.codePointAt(i - 1);
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f[i][0] = f[i - 1][0] + s1.charAt(i - 1);
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}
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for (int j = 1; j <= n; ++j) {
94-
dp[0][j] = dp[0][j - 1] + s2.codePointAt(j - 1);
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f[0][j] = f[0][j - 1] + s2.charAt(j - 1);
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}
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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)) {
99-
dp[i][j] = dp[i - 1][j - 1];
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f[i][j] = f[i - 1][j - 1];
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} else {
101-
dp[i][j] = Math.min(
102-
dp[i - 1][j] + s1.codePointAt(i - 1), dp[i][j - 1] + s2.codePointAt(j - 1));
117+
f[i][j] = Math.min(f[i - 1][j] + s1.charAt(i - 1), f[i][j - 1] + s2.charAt(j - 1));
103118
}
104119
}
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}
106-
return dp[m][n];
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return f[m][n];
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}
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}
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```
@@ -115,18 +130,24 @@ class Solution {
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public:
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int minimumDeleteSum(string s1, string s2) {
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int m = s1.size(), n = s2.size();
118-
vector<vector<int>> dp(m + 1, vector<int>(n + 1));
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for (int i = 1; i <= m; ++i) dp[i][0] = dp[i - 1][0] + s1[i - 1];
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for (int j = 1; j <= n; ++j) dp[0][j] = dp[0][j - 1] + s2[j - 1];
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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) {
136+
f[i][0] = f[i - 1][0] + s1[i - 1];
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}
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for (int j = 1; j <= n; ++j) {
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f[0][j] = f[0][j - 1] + s2[j - 1];
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}
121141
for (int i = 1; i <= m; ++i) {
122142
for (int j = 1; j <= n; ++j) {
123-
if (s1[i - 1] == s2[j - 1])
124-
dp[i][j] = dp[i - 1][j - 1];
125-
else
126-
dp[i][j] = min(dp[i - 1][j] + s1[i - 1], dp[i][j - 1] + s2[j - 1]);
143+
if (s1[i - 1] == s2[j - 1]) {
144+
f[i][j] = f[i - 1][j - 1];
145+
} else {
146+
f[i][j] = min(f[i - 1][j] + s1[i - 1], f[i][j - 1] + s2[j - 1]);
147+
}
127148
}
128149
}
129-
return dp[m][n];
150+
return f[m][n];
130151
}
131152
};
132153
```
@@ -136,26 +157,26 @@ public:
136157
```go
137158
func minimumDeleteSum(s1 string, s2 string) int {
138159
m, n := len(s1), len(s2)
139-
dp := make([][]int, m+1)
140-
for i := range dp {
141-
dp[i] = make([]int, n+1)
160+
f := make([][]int, m+1)
161+
for i := range f {
162+
f[i] = make([]int, n+1)
142163
}
143-
for i := 1; i <= m; i++ {
144-
dp[i][0] = dp[i-1][0] + int(s1[i-1])
164+
for i, c := range s1 {
165+
f[i+1][0] = f[i][0] + int(c)
145166
}
146-
for j := 1; j <= n; j++ {
147-
dp[0][j] = dp[0][j-1] + int(s2[j-1])
167+
for j, c := range s2 {
168+
f[0][j+1] = f[0][j] + int(c)
148169
}
149170
for i := 1; i <= m; i++ {
150171
for j := 1; j <= n; j++ {
151172
if s1[i-1] == s2[j-1] {
152-
dp[i][j] = dp[i-1][j-1]
173+
f[i][j] = f[i-1][j-1]
153174
} else {
154-
dp[i][j] = min(dp[i-1][j]+int(s1[i-1]), dp[i][j-1]+int(s2[j-1]))
175+
f[i][j] = min(f[i-1][j]+int(s1[i-1]), f[i][j-1]+int(s2[j-1]))
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}
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}
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}
158-
return dp[m][n]
179+
return f[m][n]
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}
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func min(a, b int) int {
@@ -166,6 +187,69 @@ func min(a, b int) int {
166187
}
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```
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### **TypeScript**
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```ts
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function minimumDeleteSum(s1: string, s2: string): number {
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const m = s1.length;
195+
const n = s2.length;
196+
const f = Array.from({ length: m + 1 }, () => Array(n + 1).fill(0));
197+
for (let i = 1; i <= m; ++i) {
198+
f[i][0] = f[i - 1][0] + s1[i - 1].charCodeAt(0);
199+
}
200+
for (let j = 1; j <= n; ++j) {
201+
f[0][j] = f[0][j - 1] + s2[j - 1].charCodeAt(0);
202+
}
203+
for (let i = 1; i <= m; ++i) {
204+
for (let j = 1; j <= n; ++j) {
205+
if (s1[i - 1] === s2[j - 1]) {
206+
f[i][j] = f[i - 1][j - 1];
207+
} else {
208+
f[i][j] = Math.min(
209+
f[i - 1][j] + s1[i - 1].charCodeAt(0),
210+
f[i][j - 1] + s2[j - 1].charCodeAt(0),
211+
);
212+
}
213+
}
214+
}
215+
return f[m][n];
216+
}
217+
```
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### **JavaScript**
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```js
222+
/**
223+
* @param {string} s1
224+
* @param {string} s2
225+
* @return {number}
226+
*/
227+
var minimumDeleteSum = function (s1, s2) {
228+
const m = s1.length;
229+
const n = s2.length;
230+
const f = Array.from({ length: m + 1 }, () => Array(n + 1).fill(0));
231+
for (let i = 1; i <= m; ++i) {
232+
f[i][0] = f[i - 1][0] + s1[i - 1].charCodeAt(0);
233+
}
234+
for (let j = 1; j <= n; ++j) {
235+
f[0][j] = f[0][j - 1] + s2[j - 1].charCodeAt(0);
236+
}
237+
for (let i = 1; i <= m; ++i) {
238+
for (let j = 1; j <= n; ++j) {
239+
if (s1[i - 1] === s2[j - 1]) {
240+
f[i][j] = f[i - 1][j - 1];
241+
} else {
242+
f[i][j] = Math.min(
243+
f[i - 1][j] + s1[i - 1].charCodeAt(0),
244+
f[i][j - 1] + s2[j - 1].charCodeAt(0),
245+
);
246+
}
247+
}
248+
}
249+
return f[m][n];
250+
};
251+
```
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### **...**
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171255
```

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