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Commit 68eed49

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feat: add solutions to lc problem: No.2063 (doocs#4075)
No.2063.Vowels of All Substrings
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‎solution/2000-2099/2063.Vowels of All Substrings/README.md‎

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@@ -85,9 +85,9 @@ tags:
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### 方法一:枚举贡献
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我们可以枚举字符串的每个字符 $word[i],ドル如果 $word[i]$ 是元音字母,那么 $word[i]$ 一共在 $(i + 1) \times (n - i)$ 个子字符串中出现,将这些子字符串的个数累加即可。
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我们可以枚举字符串的每个字符 $\textit{word}[i],ドル如果 $\textit{word}[i]$ 是元音字母,那么 $\textit{word}[i]$ 一共在 $(i + 1) \times (n - i)$ 个子字符串中出现,将这些子字符串的个数累加即可。
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时间复杂度 $O(n),ドル空间复杂度 $O(1)$。其中 $n$ 为字符串 $word$ 的长度。
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时间复杂度 $O(n),ドル其中 $n$ 为字符串 $\textit{word}$ 的长度。空间复杂度 $O(1)$。
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<!-- tabs:start -->
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@@ -163,6 +163,40 @@ function countVowels(word: string): number {
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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 count_vowels(word: String) -> i64 {
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let n = word.len() as i64;
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word.chars()
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.enumerate()
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.filter(|(_, c)| "aeiou".contains(*c))
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.map(|(i, _)| (i as i64 + 1) * (n - i as i64))
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.sum()
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}
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}
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```
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#### JavaScript
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```js
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/**
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* @param {string} word
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* @return {number}
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*/
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var countVowels = function (word) {
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const n = word.length;
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let ans = 0;
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for (let i = 0; i < n; ++i) {
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if (['a', 'e', 'i', 'o', 'u'].includes(word[i])) {
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ans += (i + 1) * (n - i);
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}
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}
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return ans;
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};
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```
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<!-- tabs:end -->
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<!-- solution:end -->

‎solution/2000-2099/2063.Vowels of All Substrings/README_EN.md‎

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<pre>
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<strong>Input:</strong> word = &quot;aba&quot;
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<strong>Output:</strong> 6
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<strong>Explanation:</strong>
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<strong>Explanation:</strong>
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All possible substrings are: &quot;a&quot;, &quot;ab&quot;, &quot;aba&quot;, &quot;b&quot;, &quot;ba&quot;, and &quot;a&quot;.
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- &quot;b&quot; has 0 vowels in it
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- &quot;a&quot;, &quot;ab&quot;, &quot;ba&quot;, and &quot;a&quot; have 1 vowel each
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- &quot;aba&quot; has 2 vowels in it
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Hence, the total sum of vowels = 0 + 1 +たす 1 +たす 1 +たす 1 +たす 2 =わ 6.
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Hence, the total sum of vowels = 0 + 1 +たす 1 +たす 1 +たす 1 +たす 2 =わ 6.
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> word = &quot;abc&quot;
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<strong>Output:</strong> 3
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<strong>Explanation:</strong>
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<strong>Explanation:</strong>
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All possible substrings are: &quot;a&quot;, &quot;ab&quot;, &quot;abc&quot;, &quot;b&quot;, &quot;bc&quot;, and &quot;c&quot;.
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- &quot;a&quot;, &quot;ab&quot;, and &quot;abc&quot; have 1 vowel each
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- &quot;b&quot;, &quot;bc&quot;, and &quot;c&quot; have 0 vowels each
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<!-- solution:start -->
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### Solution 1
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### Solution 1: Enumerate Contribution
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We can enumerate each character $\textit{word}[i]$ in the string. If $\textit{word}[i]$ is a vowel, then $\textit{word}[i]$ appears in $(i + 1) \times (n - i)$ substrings. We sum up the counts of these substrings.
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The time complexity is $O(n),ドル where $n$ is the length of the string $\textit{word}$. The space complexity is $O(1)$.
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<!-- tabs:start -->
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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 count_vowels(word: String) -> i64 {
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let n = word.len() as i64;
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word.chars()
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.enumerate()
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.filter(|(_, c)| "aeiou".contains(*c))
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.map(|(i, _)| (i as i64 + 1) * (n - i as i64))
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.sum()
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}
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}
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```
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#### JavaScript
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```js
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/**
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* @param {string} word
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* @return {number}
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*/
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var countVowels = function (word) {
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const n = word.length;
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let ans = 0;
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for (let i = 0; i < n; ++i) {
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if (['a', 'e', 'i', 'o', 'u'].includes(word[i])) {
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ans += (i + 1) * (n - i);
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}
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}
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return ans;
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};
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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/**
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* @param {string} word
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* @return {number}
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*/
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var countVowels = function (word) {
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const n = word.length;
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let ans = 0;
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for (let i = 0; i < n; ++i) {
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if (['a', 'e', 'i', 'o', 'u'].includes(word[i])) {
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ans += (i + 1) * (n - i);
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}
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}
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return ans;
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};
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impl Solution {
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pub fn count_vowels(word: String) -> i64 {
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let n = word.len() as i64;
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word.chars()
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.enumerate()
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.filter(|(_, c)| "aeiou".contains(*c))
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.map(|(i, _)| (i as i64 + 1) * (n - i as i64))
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.sum()
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}
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}

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