[pull] main from doocs:main

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 $O(1)$。
 
 <!-- tabs:start -->
 
@@ -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;
+};
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

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

@@ -33,20 +33,20 @@ tags:
 <pre>
 <strong>Input:</strong> word = &quot;aba&quot;
 <strong>Output:</strong> 6
-<strong>Explanation:</strong> 
+<strong>Explanation:</strong>
 All possible substrings are: &quot;a&quot;, &quot;ab&quot;, &quot;aba&quot;, &quot;b&quot;, &quot;ba&quot;, and &quot;a&quot;.
 - &quot;b&quot; has 0 vowels in it
 - &quot;a&quot;, &quot;ab&quot;, &quot;ba&quot;, and &quot;a&quot; have 1 vowel each
 - &quot;aba&quot; 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.
 </pre>
 
 <p><strong class="example">Example 2:</strong></p>
 
 <pre>
 <strong>Input:</strong> word = &quot;abc&quot;
 <strong>Output:</strong> 3
-<strong>Explanation:</strong> 
+<strong>Explanation:</strong>
 All possible substrings are: &quot;a&quot;, &quot;ab&quot;, &quot;abc&quot;, &quot;b&quot;, &quot;bc&quot;, and &quot;c&quot;.
 - &quot;a&quot;, &quot;ab&quot;, and &quot;abc&quot; have 1 vowel each
 - &quot;b&quot;, &quot;bc&quot;, and &quot;c&quot; have 0 vowels each
@@ -75,7 +75,11 @@ Hence, the total sum of vowels = 1 + 1 + 1 + 0 + 0 + 0 = 3.
 
 <!-- solution:start -->
 
-### 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 $O(1)$.
 
 <!-- tabs:start -->
 
@@ -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;
+};
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

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;
+};

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()
+    }
+}

solution/2000-2099/2080.Range Frequency Queries/README.md

@@ -68,7 +68,7 @@ rangeFreqQuery.query(0, 11, 33); // 返回 2 。33 在整个子数组中出现 2
 
 <!-- solution:start -->
 
-### 方法一：哈希表
+### 方法一：哈希表 + 二分查找
 
 我们用一个哈希表 $g$ 来存储每个值对应的下标数组。在构造函数中，我们遍历数组 $\textit{arr}$，将每个值对应的下标加入到哈希表中。
 
@@ -215,20 +215,8 @@ class RangeFreqQuery {
         if (!idx) {
             return 0;
         }
-        const search = (x: number): number => {
-            let [l, r] = [0, idx.length];
-            while (l < r) {
-                const mid = (l + r) >> 1;
-                if (idx[mid] >= x) {
-                    r = mid;
-                } else {
-                    l = mid + 1;
-                }
-            }
-            return l;
-        };
-        const l = search(left);
-        const r = search(right + 1);
+        const l = _.sortedIndex(idx, left);
+        const r = _.sortedIndex(idx, right + 1);
         return r - l;
     }
 }
@@ -240,6 +228,111 @@ class RangeFreqQuery {
  */
 ```
 
+#### Rust
+
+```rust
+use std::collections::HashMap;
+
+struct RangeFreqQuery {
+    g: HashMap<i32, Vec<usize>>,
+}
+
+impl RangeFreqQuery {
+    fn new(arr: Vec<i32>) -> Self {
+        let mut g = HashMap::new();
+        for (i, &value) in arr.iter().enumerate() {
+            g.entry(value).or_insert_with(Vec::new).push(i);
+        }
+        RangeFreqQuery { g }
+    }
+
+    fn query(&self, left: i32, right: i32, value: i32) -> i32 {
+        if let Some(idx) = self.g.get(&value) {
+            let l = idx.partition_point(|&x| x < left as usize);
+            let r = idx.partition_point(|&x| x <= right as usize);
+            return (r - l) as i32;
+        }
+        0
+    }
+}
+```
+
+#### JavaScript
+
+```js
+/**
+ * @param {number[]} arr
+ */
+var RangeFreqQuery = function (arr) {
+    this.g = new Map();
+
+    for (let i = 0; i < arr.length; ++i) {
+        if (!this.g.has(arr[i])) {
+            this.g.set(arr[i], []);
+        }
+        this.g.get(arr[i]).push(i);
+    }
+};
+
+/**
+ * @param {number} left
+ * @param {number} right
+ * @param {number} value
+ * @return {number}
+ */
+RangeFreqQuery.prototype.query = function (left, right, value) {
+    const idx = this.g.get(value);
+    if (!idx) {
+        return 0;
+    }
+    const l = _.sortedIndex(idx, left);
+    const r = _.sortedIndex(idx, right + 1);
+    return r - l;
+};
+
+/**
+ * Your RangeFreqQuery object will be instantiated and called as such:
+ * var obj = new RangeFreqQuery(arr)
+ * var param_1 = obj.query(left,right,value)
+ */
+```
+
+#### C#
+
+```cs
+public class RangeFreqQuery {
+    private Dictionary<int, List<int>> g;
+
+    public RangeFreqQuery(int[] arr) {
+        g = new Dictionary<int, List<int>>();
+        for (int i = 0; i < arr.Length; ++i) {
+            if (!g.ContainsKey(arr[i])) {
+                g[arr[i]] = new List<int>();
+            }
+            g[arr[i]].Add(i);
+        }
+    }
+
+    public int Query(int left, int right, int value) {
+        if (g.ContainsKey(value)) {
+            var idx = g[value];
+            int l = idx.BinarySearch(left);
+            int r = idx.BinarySearch(right + 1);
+            l = l < 0 ? -l - 1 : l;
+            r = r < 0 ? -r - 1 : r;
+            return r - l;
+        }
+        return 0;
+    }
+}
+
+/**
+ * Your RangeFreqQuery object will be instantiated and called as such:
+ * RangeFreqQuery obj = new RangeFreqQuery(arr);
+ * int param_1 = obj.Query(left, right, value);
+ */
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->