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feat: update lc problems (doocs#4263)

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- 2600-2699/2610.Convert an Array Into a 2D Array With Conditions
- 2800-2899
  - 2873.Maximum Value of an Ordered Triplet I
    - README.md
    - README_EN.md
  - 2874.Maximum Value of an Ordered Triplet II
    - README.md
    - README_EN.md
- 3400-3499
  - 3481.Apply Substitutions
    - README.md
  - 3482.Analyze Organization Hierarchy
    - README.md
  - 3491.Phone Number Prefix
    - README.md
- DATABASE_README.md
- README.md

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`‎solution/2600-2699/2610.Convert an Array Into a 2D Array With Conditions/README.md‎`

Lines changed: 31 additions & 4 deletions

Original file line number	Diff line number	Diff line change
`@@ -68,13 +68,13 @@ nums 中的所有元素都有用到,并且每一行都由不同的整数组成`
`68`	`68`
`69`	`69`	`### 方法一:数组或哈希表`
`70`	`70`
`71`		`-我们先用数组或哈希表 $cnt$ 统计数组 $nums$ 中每个元素出现的次数。`
	`71`	`+我们先用一个数组或者哈希表 $\textit{cnt}$ 统计数组 $\textit{nums}$ 中每个元素出现的次数。`
`72`	`72`
`73`		`-然后遍历 $cnt,ドル对于每个元素 $x,ドル我们将其添加到答案列表中的第 0ドル$ 行,第 1ドル$ 行,第 2ドル$ 行,...,第 $cnt[x]-1$ 行。`
	`73`	`+然后遍历 $\textit{cnt},ドル对于每个元素 $x,ドル我们将其添加到答案列表中的第 0ドル$ 行,第 1ドル$ 行,第 2ドル$ 行,...,第 $\textit{cnt}[x]-1$ 行。`
`74`	`74`
`75`	`75`	`最后返回答案列表即可。`
`76`	`76`
`77`		`-时间复杂度 $O(n),ドル空间复杂度 $O(n)$。其中 $n$ 为数组 $nums$ 的长度。`
	`77`	`+时间复杂度 $O(n),ドル空间复杂度 $O(n)$。其中 $n$ 为数组 $\textit{nums}$ 的长度。`
`78`	`78`
`79`	`79`	`<!-- tabs:start -->`
`80`	`80`
`@@ -171,7 +171,7 @@ func findMatrix(nums []int) (ans [][]int) {`
`171`	`171`	`function findMatrix(nums: number[]): number[][] {`
`172`	`172`	`const ans: number[][] = [];`
`173`	`173`	`const n = nums.length;`
`174`		`- const cnt: number[] = newArray(n + 1).fill(0);`
	`174`	`+ const cnt: number[] = Array(n + 1).fill(0);`
`175`	`175`	`for (const x of nums) {`
`176`	`176`	`++cnt[x];`
`177`	`177`	`}`
`@@ -187,6 +187,33 @@ function findMatrix(nums: number[]): number[][] {`
`187`	`187`	`}`
`188`	`188`	```
`189`	`189`
	`190`	`+#### Rust`
	`191`	`+`
	`192`	+```rust
	`193`	`+impl Solution {`
	`194`	`+ pub fn find_matrix(nums: Vec<i32>) -> Vec<Vec<i32>> {`
	`195`	`+ let n = nums.len();`
	`196`	`+ let mut cnt = vec![0; n + 1];`
	`197`	`+ let mut ans: Vec<Vec<i32>> = Vec::new();`
	`198`	`+`
	`199`	`+ for &x in &nums {`
	`200`	`+ cnt[x as usize] += 1;`
	`201`	`+ }`
	`202`	`+`
	`203`	`+ for x in 1..=n as i32 {`
	`204`	`+ for j in 0..cnt[x as usize] {`
	`205`	`+ if ans.len() <= j {`
	`206`	`+ ans.push(Vec::new());`
	`207`	`+ }`
	`208`	`+ ans[j].push(x);`
	`209`	`+ }`
	`210`	`+ }`
	`211`	`+`
	`212`	`+ ans`
	`213`	`+ }`
	`214`	`+}`
	`215`	+```
	`216`	`+`
`190`	`217`	`<!-- tabs:end -->`
`191`	`218`
`192`	`219`	`<!-- solution:end -->`

`‎solution/2600-2699/2610.Convert an Array Into a 2D Array With Conditions/README_EN.md‎`

Lines changed: 32 additions & 5 deletions

Original file line number	Diff line number	Diff line change
`@@ -68,13 +68,13 @@ It can be shown that we cannot have less than 3 rows in a valid array.</pre>`
`68`	`68`
`69`	`69`	`### Solution 1: Array or Hash Table`
`70`	`70`
`71`		`-We use an array or hash table $cnt$ to count the number of occurrences of each element in the array $nums$.`
	`71`	`+We first use an array or hash table $\textit{cnt}$ to count the frequency of each element in the array $\textit{nums}$.`
`72`	`72`
`73`		`-Then we traverse the $cnt$ array, add $x$to the 0ドル$th row, the 1ドル$st row, the 2ドル$nd row, ..., the ($cnt[x]-1$)th row of the answer list.`
	`73`	`+Then we iterate through $\textit{cnt}$. For each element $x$, we add it to the 0th row, 1st row, 2nd row, ..., and $(cnt[x]-1)$th row of the answer list.`
`74`	`74`
`75`		`-Finally, return the answer list.`
	`75`	`+Finally, we return the answer list.`
`76`	`76`
`77`		`-The time complexity is $O(n)$ and the space complexity is $O(n)$, where $n$ is the length of the array $nums$.`
	`77`	`+The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the length of the array $\textit{nums}$.`
`78`	`78`
`79`	`79`	`<!-- tabs:start -->`
`80`	`80`
`@@ -171,7 +171,7 @@ func findMatrix(nums []int) (ans [][]int) {`
`171`	`171`	`function findMatrix(nums: number[]): number[][] {`
`172`	`172`	`const ans: number[][] = [];`
`173`	`173`	`const n = nums.length;`
`174`		`- const cnt: number[] = newArray(n + 1).fill(0);`
	`174`	`+ const cnt: number[] = Array(n + 1).fill(0);`
`175`	`175`	`for (const x of nums) {`
`176`	`176`	`++cnt[x];`
`177`	`177`	`}`
`@@ -187,6 +187,33 @@ function findMatrix(nums: number[]): number[][] {`
`187`	`187`	`}`
`188`	`188`	```
`189`	`189`
	`190`	`+#### Rust`
	`191`	`+`
	`192`	+```rust
	`193`	`+impl Solution {`
	`194`	`+ pub fn find_matrix(nums: Vec<i32>) -> Vec<Vec<i32>> {`
	`195`	`+ let n = nums.len();`
	`196`	`+ let mut cnt = vec![0; n + 1];`
	`197`	`+ let mut ans: Vec<Vec<i32>> = Vec::new();`
	`198`	`+`
	`199`	`+ for &x in &nums {`
	`200`	`+ cnt[x as usize] += 1;`
	`201`	`+ }`
	`202`	`+`
	`203`	`+ for x in 1..=n as i32 {`
	`204`	`+ for j in 0..cnt[x as usize] {`
	`205`	`+ if ans.len() <= j {`
	`206`	`+ ans.push(Vec::new());`
	`207`	`+ }`
	`208`	`+ ans[j].push(x);`
	`209`	`+ }`
	`210`	`+ }`
	`211`	`+`
	`212`	`+ ans`
	`213`	`+ }`
	`214`	`+}`
	`215`	+```
	`216`	`+`
`190`	`217`	`<!-- tabs:end -->`
`191`	`218`
`192`	`219`	`<!-- solution:end -->`