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()
+ }
+}
diff --git a/solution/2000-2099/2080.Range Frequency Queries/README.md b/solution/2000-2099/2080.Range Frequency Queries/README.md
index 3e5f7b21a2f5c..4e064fa154100 100644
--- a/solution/2000-2099/2080.Range Frequency Queries/README.md
+++ b/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
-### 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>,
+}
+
+impl RangeFreqQuery {
+ fn new(arr: Vec) -> 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> g;
+
+ public RangeFreqQuery(int[] arr) {
+ g = new Dictionary>();
+ for (int i = 0; i < arr.Length; ++i) {
+ if (!g.ContainsKey(arr[i])) {
+ g[arr[i]] = new List();
+ }
+ 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);
+ */
+```
+
diff --git a/solution/2000-2099/2080.Range Frequency Queries/README_EN.md b/solution/2000-2099/2080.Range Frequency Queries/README_EN.md
index 7945cfe3ed4ee..2cd33bf0b4d26 100644
--- a/solution/2000-2099/2080.Range Frequency Queries/README_EN.md
+++ b/solution/2000-2099/2080.Range Frequency Queries/README_EN.md
@@ -67,7 +67,7 @@ rangeFreqQuery.query(0, 11, 33); // return 2. The value 33 occurs 2 times in the
-### Solution 1: Hash Table
+### Solution 1: Hash Table + Binary Search
We use a hash table $g$ to store the array of indices corresponding to each value. In the constructor, we traverse the array $\textit{arr}$, adding the index corresponding to each value to the hash table.
@@ -214,20 +214,8 @@ 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;
}
}
@@ -239,6 +227,111 @@ class RangeFreqQuery {
*/
```
+#### Rust
+
+```rust
+use std::collections::HashMap;
+
+struct RangeFreqQuery {
+ g: HashMap>,
+}
+
+impl RangeFreqQuery {
+ fn new(arr: Vec) -> 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> g;
+
+ public RangeFreqQuery(int[] arr) {
+ g = new Dictionary>();
+ for (int i = 0; i < arr.Length; ++i) {
+ if (!g.ContainsKey(arr[i])) {
+ g[arr[i]] = new List();
+ }
+ 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);
+ */
+```
+
diff --git a/solution/2000-2099/2080.Range Frequency Queries/Solution.cs b/solution/2000-2099/2080.Range Frequency Queries/Solution.cs
new file mode 100644
index 0000000000000..0f675a7a0c951
--- /dev/null
+++ b/solution/2000-2099/2080.Range Frequency Queries/Solution.cs
@@ -0,0 +1,31 @@
+public class RangeFreqQuery {
+ private Dictionary> g;
+
+ public RangeFreqQuery(int[] arr) {
+ g = new Dictionary>();
+ for (int i = 0; i < arr.Length; ++i) {
+ if (!g.ContainsKey(arr[i])) {
+ g[arr[i]] = new List();
+ }
+ 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);
+ */
diff --git a/solution/2000-2099/2080.Range Frequency Queries/Solution.js b/solution/2000-2099/2080.Range Frequency Queries/Solution.js
new file mode 100644
index 0000000000000..14b374d0ec9ee
--- /dev/null
+++ b/solution/2000-2099/2080.Range Frequency Queries/Solution.js
@@ -0,0 +1,35 @@
+/**
+ * @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)
+ */
diff --git a/solution/2000-2099/2080.Range Frequency Queries/Solution.rs b/solution/2000-2099/2080.Range Frequency Queries/Solution.rs
new file mode 100644
index 0000000000000..d9d360c422398
--- /dev/null
+++ b/solution/2000-2099/2080.Range Frequency Queries/Solution.rs
@@ -0,0 +1,24 @@
+use std::collections::HashMap;
+
+struct RangeFreqQuery {
+ g: HashMap>,
+}
+
+impl RangeFreqQuery {
+ fn new(arr: Vec) -> 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
+ }
+}
diff --git a/solution/2000-2099/2080.Range Frequency Queries/Solution.ts b/solution/2000-2099/2080.Range Frequency Queries/Solution.ts
index 2ff602ed7618b..fb92babdfb118 100644
--- a/solution/2000-2099/2080.Range Frequency Queries/Solution.ts
+++ b/solution/2000-2099/2080.Range Frequency Queries/Solution.ts
@@ -15,20 +15,8 @@ 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;
}
}