From d24325a61317e69e4fed285d8ca2ff295c5be9d3 Mon Sep 17 00:00:00 2001 From: Libin YANG Date: Tue, 18 Feb 2025 12:41:37 +0800 Subject: [PATCH 1/2] feat: add solutions to lc problem: No.2080 (#4074) No.2080.Range Frequency Queries --- .../2080.Range Frequency Queries/README.md | 123 +++++++++++++++--- .../2080.Range Frequency Queries/README_EN.md | 123 +++++++++++++++--- .../2080.Range Frequency Queries/Solution.cs | 31 +++++ .../2080.Range Frequency Queries/Solution.js | 35 +++++ .../2080.Range Frequency Queries/Solution.rs | 24 ++++ .../2080.Range Frequency Queries/Solution.ts | 16 +-- 6 files changed, 308 insertions(+), 44 deletions(-) create mode 100644 solution/2000-2099/2080.Range Frequency Queries/Solution.cs create mode 100644 solution/2000-2099/2080.Range Frequency Queries/Solution.js create mode 100644 solution/2000-2099/2080.Range Frequency Queries/Solution.rs 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; } } From 68eed4908ec0b022d3e1111aaf2f93f491779739 Mon Sep 17 00:00:00 2001 From: Libin YANG Date: Tue, 18 Feb 2025 14:01:07 +0800 Subject: [PATCH 2/2] feat: add solutions to lc problem: No.2063 (#4075) 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()
+    }
+}