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Commit 8264fe9

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feat: add solutions to lc problem: No.1967 (doocs#3943)
No.1967.Number of Strings That Appear as Substrings in Word
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6 files changed

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‎solution/1600-1699/1647.Minimum Deletions to Make Character Frequencies Unique/README.md‎

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### 方法一:数组 + 排序
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我们先用一个长度为 26ドル$ 的数组 `cnt` 统计字符串 $s$ 中每个字母出现的次数。
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我们先用一个长度为 26ドル$ 的数组 $\textit{cnt}$ 统计字符串 $s$ 中每个字母出现的次数。
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然后我们对数组 `cnt` 进行倒序排序。定义一个变量 `pre` 记录当前字母的出现次数。
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然后我们对数组 $\textit{cnt}$ 进行倒序排序。定义一个变量 $\textit{pre}$ 记录当前字母的出现次数。
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接下来,遍历数组 `cnt` 每个元素 $v,ドル如果当前 `pre` 等于 0ドル,ドル我们直接将答案加上 $v$;否则,如果 $v \geq pre,ドル我们将答案加上 $v-pre+1,ドル并且将 `pre` 减去 1ドル,ドル否则,我们直接将 `pre` 更新为 $v$。然后继续遍历下个元素。
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接下来,遍历数组 $\textit{cnt}$ 每个元素 $v,ドル如果当前 $\textit{pre}$ 等于 0ドル,ドル我们直接将答案加上 $v$;否则,如果 $v \geq \textit{pre},ドル我们将答案加上 $v-\textit{pre}+1,ドル并且将 $\textit{pre}$ 减去 1ドル,ドル否则,我们直接将 $\textit{pre}$ 更新为 $v$。然后继续遍历下个元素。
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遍历结束,返回答案即可。
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时间复杂度 $O(n + C \times \log C),ドル空间复杂度 $O(C)$。其中 $n$ 是字符串 $s$ 的长度,而 $C$ 为字母集的大小。本题中 $C=26$。
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时间复杂度 $O(n + |\Sigma| \times \log |\Sigma|),ドル空间复杂度 $O(|\Sigma|)$。其中 $n$ 是字符串 $s$ 的长度,而 $|\Sigma|$ 为字母集的大小。本题中 $|\Sigma|=26$。
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<!-- tabs:start -->
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‎solution/1600-1699/1647.Minimum Deletions to Make Character Frequencies Unique/README_EN.md‎

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<!-- solution:start -->
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### Solution 1
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### Solution 1: Array + Sorting
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First, we use an array $\textit{cnt}$ of length 26ドル$ to count the occurrences of each letter in the string $s$.
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Then, we sort the array $\textit{cnt}$ in descending order. We define a variable $\textit{pre}$ to record the current number of occurrences of the letter.
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Next, we traverse each element $v$ in the array $\textit{cnt}$. If the current $\textit{pre}$ is 0ドル,ドル we directly add $v$ to the answer. Otherwise, if $v \geq \textit{pre},ドル we add $v - \textit{pre} + 1$ to the answer and decrement $\textit{pre}$ by 1ドル$. Otherwise, we directly update $\textit{pre}$ to $v$. Then, we continue to the next element.
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After traversing, we return the answer.
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The time complexity is $O(n + |\Sigma| \times \log |\Sigma|),ドル and the space complexity is $O(|\Sigma|)$. Here, $n$ is the length of the string $s,ドル and $|\Sigma|$ is the size of the alphabet. In this problem, $|\Sigma| = 26$.
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<!-- tabs:start -->
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‎solution/1900-1999/1967.Number of Strings That Appear as Substrings in Word/README.md‎

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### 方法一:模拟
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遍历字符串数组 $patterns$ 中的每个字符串 $p,ドル判断其是否为 $word$ 的子字符串,如果是,答案加一。
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遍历字符串数组 $\textit{patterns}$ 中的每个字符串 $p,ドル判断其是否为 $\textit{word}$ 的子字符串,如果是,答案加一。
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遍历结束后,返回答案。
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时间复杂度 $O(n \times m),ドル空间复杂度 $O(1)$。其中 $n$ 和 $m$ 分别为 $patterns$ 和 $word$ 的长度。
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时间复杂度 $O(n \times m),ドル空间复杂度 $O(1)$。其中 $n$ 和 $m$ 分别为 $\textit{patterns}$ 和 $\textit{word}$ 的长度。
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<!-- tabs:start -->
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```ts
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function numOfStrings(patterns: string[], word: string): number {
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let ans = 0;
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for (const p of patterns) {
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if (word.includes(p)) {
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++ans;
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}
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return patterns.filter(p => word.includes(p)).length;
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}
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```
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#### Rust
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```rust
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impl Solution {
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pub fn num_of_strings(patterns: Vec<String>, word: String) -> i32 {
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patterns.iter().filter(|p| word.contains(&**p)).count() as i32
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}
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return ans;
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}
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```
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‎solution/1900-1999/1967.Number of Strings That Appear as Substrings in Word/README_EN.md‎

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<!-- solution:start -->
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### Solution 1
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### Solution 1: Simulation
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Traverse each string $p$ in the array $\textit{patterns}$ and check if it is a substring of $\textit{word}$. If it is, increment the answer by one.
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After traversing, return the answer.
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The time complexity is $O(n \times m),ドル and the space complexity is $O(1)$. Here, $n$ and $m$ are the lengths of $\textit{patterns}$ and $\textit{word},ドル respectively.
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<!-- tabs:start -->
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```ts
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function numOfStrings(patterns: string[], word: string): number {
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let ans = 0;
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for (const p of patterns) {
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if (word.includes(p)) {
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++ans;
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}
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return patterns.filter(p => word.includes(p)).length;
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}
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```
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#### Rust
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```rust
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impl Solution {
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pub fn num_of_strings(patterns: Vec<String>, word: String) -> i32 {
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patterns.iter().filter(|p| word.contains(&**p)).count() as i32
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}
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return ans;
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}
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```
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impl Solution {
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pub fn num_of_strings(patterns: Vec<String>, word: String) -> i32 {
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patterns.iter().filter(|p| word.contains(&**p)).count() as i32
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}
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}
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function numOfStrings(patterns: string[], word: string): number {
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let ans = 0;
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for (const p of patterns) {
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if (word.includes(p)) {
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++ans;
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}
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}
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return ans;
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return patterns.filter(p => word.includes(p)).length;
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}

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