From a73ed8886caaf96bb53c0ca9058630c79fe572c7 Mon Sep 17 00:00:00 2001 From: yanglbme Date: Tue, 18 Feb 2025 13:03:31 +0800 Subject: [PATCH] feat: add solutions to lc problem: NO.2063 No.2063.Vowels of All Substrings --- .../2063.Vowels of All Substrings/README.md | 38 ++++++++++++++- .../README_EN.md | 46 +++++++++++++++++-- .../2063.Vowels of All Substrings/Solution.js | 14 ++++++ .../2063.Vowels of All Substrings/Solution.rs | 10 ++++ 4 files changed, 102 insertions(+), 6 deletions(-) create mode 100644 solution/2000-2099/2063.Vowels of All Substrings/Solution.js create mode 100644 solution/2000-2099/2063.Vowels of All Substrings/Solution.rs diff --git a/solution/2000-2099/2063.Vowels of All Substrings/README.md b/solution/2000-2099/2063.Vowels of All Substrings/README.md index 2653b85def8e6..36b9dd813a5eb 100644 --- a/solution/2000-2099/2063.Vowels of All Substrings/README.md +++ b/solution/2000-2099/2063.Vowels of All Substrings/README.md @@ -85,9 +85,9 @@ tags: ### 方法一：枚举贡献 -我们可以枚举字符串的每个字符 $word[i]$，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。 +我们可以枚举字符串的每个字符 $\textit{word}[i]$，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。 -时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为字符串 $word$ 的长度。 +时间复杂度 $O(n)$，其中 $n$ 为字符串 $\textit{word}$ 的长度。空间复杂度 $O(1)$。 @@ -163,6 +163,40 @@ function countVowels(word: string): number { } ``` +#### Rust + +```rust +impl Solution { + pub fn count_vowels(word: String) -> i64 { + let n = word.len() as i64; + word.chars() + .enumerate() + .filter(|(_, c)| "aeiou".contains(*c)) + .map(|(i, _)| (i as i64 + 1) * (n - i as i64)) + .sum() + } +} +``` + +#### JavaScript + +```js +/** + * @param {string} word + * @return {number} + */ +var countVowels = function (word) { + const n = word.length; + let ans = 0; + for (let i = 0; i < n; ++i) { + if (['a', 'e', 'i', 'o', 'u'].includes(word[i])) { + ans += (i + 1) * (n - i); + } + } + return ans; +}; +``` + diff --git a/solution/2000-2099/2063.Vowels of All Substrings/README_EN.md b/solution/2000-2099/2063.Vowels of All Substrings/README_EN.md index f7c07bc373390..1a5b421d1cba6 100644 --- a/solution/2000-2099/2063.Vowels of All Substrings/README_EN.md +++ b/solution/2000-2099/2063.Vowels of All Substrings/README_EN.md @@ -33,12 +33,12 @@ tags: 
 Input: word = "aba"
 Output: 6
-Explanation: 
+Explanation:
 All possible substrings are: "a", "ab", "aba", "b", "ba", and "a".
 - "b" has 0 vowels in it
 - "a", "ab", "ba", and "a" have 1 vowel each
 - "aba" has 2 vowels in it
-Hence, the total sum of vowels = 0 + 1 + 1 + 1 + 1 + 2 = 6. 
+Hence, the total sum of vowels = 0 + 1 + 1 + 1 + 1 + 2 = 6.

Example 2:

@@ -46,7 +46,7 @@ Hence, the total sum of vowels = 0 + 1 + 1 + 1 + 1 + 2 = 6. 
 Input: word = "abc"
 Output: 3
-Explanation: 
+Explanation:
 All possible substrings are: "a", "ab", "abc", "b", "bc", and "c".
 - "a", "ab", and "abc" have 1 vowel each
 - "b", "bc", and "c" have 0 vowels each
@@ -75,7 +75,11 @@ Hence, the total sum of vowels = 1 + 1 + 1 + 0 + 0 + 0 = 3.
 
 
 
-### Solution 1
+### Solution 1: Enumerate Contribution
+
+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.
+
+The time complexity is $O(n)$, where $n$ is the length of the string $\textit{word}$. The space complexity is $O(1)$.
 
 
 
@@ -151,6 +155,40 @@ function countVowels(word: string): number {
 }
 ```
 
+#### Rust
+
+```rust
+impl Solution {
+    pub fn count_vowels(word: String) -> i64 {
+        let n = word.len() as i64;
+        word.chars()
+            .enumerate()
+            .filter(|(_, c)| "aeiou".contains(*c))
+            .map(|(i, _)| (i as i64 + 1) * (n - i as i64))
+            .sum()
+    }
+}
+```
+
+#### JavaScript
+
+```js
+/**
+ * @param {string} word
+ * @return {number}
+ */
+var countVowels = function (word) {
+    const n = word.length;
+    let ans = 0;
+    for (let i = 0; i < n; ++i) {
+        if (['a', 'e', 'i', 'o', 'u'].includes(word[i])) {
+            ans += (i + 1) * (n - i);
+        }
+    }
+    return ans;
+};
+```
+
 
 
 
diff --git a/solution/2000-2099/2063.Vowels of All Substrings/Solution.js b/solution/2000-2099/2063.Vowels of All Substrings/Solution.js
new file mode 100644
index 0000000000000..1cb57ec60dda5
--- /dev/null
+++ b/solution/2000-2099/2063.Vowels of All Substrings/Solution.js	
@@ -0,0 +1,14 @@
+/**
+ * @param {string} word
+ * @return {number}
+ */
+var countVowels = function (word) {
+    const n = word.length;
+    let ans = 0;
+    for (let i = 0; i < n; ++i) {
+        if (['a', 'e', 'i', 'o', 'u'].includes(word[i])) {
+            ans += (i + 1) * (n - i);
+        }
+    }
+    return ans;
+};
diff --git a/solution/2000-2099/2063.Vowels of All Substrings/Solution.rs b/solution/2000-2099/2063.Vowels of All Substrings/Solution.rs
new file mode 100644
index 0000000000000..55243247dd45f
--- /dev/null
+++ b/solution/2000-2099/2063.Vowels of All Substrings/Solution.rs	
@@ -0,0 +1,10 @@
+impl Solution {
+    pub fn count_vowels(word: String) -> i64 {
+        let n = word.len() as i64;
+        word.chars()
+            .enumerate()
+            .filter(|(_, c)| "aeiou".contains(*c))
+            .map(|(i, _)| (i as i64 + 1) * (n - i as i64))
+            .sum()
+    }
+}