feat: add solutions to lc problem: No.0440 (#4469)

No.0440.K-th Smallest in Lexicographical Order

Author: yanglbme

solution/0400-0499/0440.K-th Smallest in Lexicographical Order/README.md

@@ -49,7 +49,37 @@ tags:
 
 <!-- solution:start -->
 
-### 方法一
+### 方法一：字典树计数 + 贪心构造
+
+本题要求在区间 $[1, n]$ 中，按**字典序**排序后，找到第 $k$ 小的数字。由于 $n$ 的范围非常大（最多可达 $10^9$），我们无法直接枚举所有数字后排序。因此我们采用**贪心 + 字典树模拟**的策略。
+
+我们将 $[1, n]$ 看作一棵 **十叉字典树（Trie）**：
+
+-   每个节点是一个前缀，根节点为空串；
+-   节点的子节点是当前前缀拼接上 $0 \sim 9$；
+-   例如前缀 $1$ 会有子节点 $10, 11, \ldots, 19$，而 $10$ 会有 $100, 101, \ldots, 109$；
+-   这种结构天然符合字典序遍历。
+
+```
+根
+├── 1
+│   ├── 10
+│   ├── 11
+│   ├── ...
+├── 2
+├── ...
+```
+
+我们使用变量 $\textit{curr}$ 表示当前前缀，初始为 $1$。每次我们尝试向下扩展前缀，直到找到第 $k$ 小的数字。
+
+每次我们计算当前前缀下有多少个合法数字（即以 $\textit{curr}$ 为前缀、且不超过 $n$ 的整数个数），记作 $\textit{count}(\text{curr})$：
+
+-   如果 $k \ge \text{count}(\text{curr})$：说明目标不在这棵子树中，跳过整棵子树，前缀右移：$\textit{curr} \leftarrow \text{curr} + 1$，并更新 $k \leftarrow k - \text{count}(\text{curr})$；
+-   否则：说明目标在当前前缀的子树中，进入下一层：$\textit{curr} \leftarrow \text{curr} \times 10$，并消耗一个前缀：$k \leftarrow k - 1$。
+
+每一层我们将当前区间扩大 $10$ 倍，向下延伸到更长的前缀，直到超出 $n$。
+
+时间复杂度 $O(\log^2 n)$，空间复杂度 $O(1)$。
 
 <!-- tabs:start -->
 
@@ -181,6 +211,75 @@ func findKthNumber(n int, k int) int {
 }
 ```
 
+#### TypeScript
+
+```ts
+function findKthNumber(n: number, k: number): number {
+    function count(curr: number): number {
+        let next = curr + 1;
+        let cnt = 0;
+        while (curr <= n) {
+            cnt += Math.min(n - curr + 1, next - curr);
+            curr *= 10;
+            next *= 10;
+        }
+        return cnt;
+    }
+
+    let curr = 1;
+    k--;
+
+    while (k > 0) {
+        const cnt = count(curr);
+        if (k >= cnt) {
+            k -= cnt;
+            curr += 1;
+        } else {
+            k -= 1;
+            curr *= 10;
+        }
+    }
+
+    return curr;
+}
+```
+
+#### Rust
+
+```rust
+impl Solution {
+    pub fn find_kth_number(n: i32, k: i32) -> i32 {
+        fn count(mut curr: i64, n: i32) -> i32 {
+            let mut next = curr + 1;
+            let mut total = 0;
+            let n = n as i64;
+            while curr <= n {
+                total += std::cmp::min(n - curr + 1, next - curr);
+                curr *= 10;
+                next *= 10;
+            }
+            total as i32
+        }
+
+        let mut curr = 1;
+        let mut k = k - 1;
+
+        while k > 0 {
+            let cnt = count(curr as i64, n);
+            if k >= cnt {
+                k -= cnt;
+                curr += 1;
+            } else {
+                k -= 1;
+                curr *= 10;
+            }
+        }
+
+        curr
+    }
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/0400-0499/0440.K-th Smallest in Lexicographical Order/README_EN.md

@@ -47,7 +47,46 @@ tags:
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Trie-Based Counting + Greedy Construction
+
+The problem asks for the \$k\$-th smallest number in the range $[1, n]$ when all numbers are sorted in **lexicographical order**. Since $n$ can be as large as $10^9$, we cannot afford to generate and sort all the numbers explicitly. Instead, we adopt a strategy based on **greedy traversal over a conceptual Trie**.
+
+We treat the range $[1, n]$ as a **10-ary prefix tree (Trie)**:
+
+-   Each node represents a numeric prefix, starting from an empty root;
+-   Each node has 10 children, corresponding to appending digits $0 \sim 9$;
+-   For example, prefix $1$ has children $10, 11, \ldots, 19$, and node $10$ has children $100, 101, \ldots, 109$;
+-   This tree naturally reflects lexicographical order traversal.
+
+```
+root
+├── 1
+│   ├── 10
+│   ├── 11
+│   ├── ...
+├── 2
+├── ...
+```
+
+We use a variable $\textit{curr}$ to denote the current prefix, initialized as $1$. At each step, we try to expand or skip prefixes until we find the \$k\$-th smallest number.
+
+At each step, we calculate how many valid numbers (i.e., numbers $\le n$ with prefix $\textit{curr}$) exist under this prefix subtree. Let this count be $\textit{count}(\text{curr})$:
+
+-   If $k \ge \text{count}(\text{curr})$: the target number is not in this subtree. We skip the entire subtree by moving to the next sibling:
+
+    $$
+    \textit{curr} \leftarrow \textit{curr} + 1,\quad k \leftarrow k - \text{count}(\text{curr})
+    $$
+
+-   Otherwise: the target is within this subtree. We go one level deeper:
+
+    $$
+    \textit{curr} \leftarrow \textit{curr} \times 10,\quad k \leftarrow k - 1
+    $$
+
+At each level, we enlarge the current range by multiplying by 10 and continue descending until we exceed $n$.
+
+The time complexity is $O(\log^2 n)$, as we perform logarithmic operations for counting and traversing the Trie structure. The space complexity is $O(1)$ since we only use a few variables to track the current prefix and count.
 
 <!-- tabs:start -->
 
@@ -179,6 +218,75 @@ func findKthNumber(n int, k int) int {
 }
 ```
 
+#### TypeScript
+
+```ts
+function findKthNumber(n: number, k: number): number {
+    function count(curr: number): number {
+        let next = curr + 1;
+        let cnt = 0;
+        while (curr <= n) {
+            cnt += Math.min(n - curr + 1, next - curr);
+            curr *= 10;
+            next *= 10;
+        }
+        return cnt;
+    }
+
+    let curr = 1;
+    k--;
+
+    while (k > 0) {
+        const cnt = count(curr);
+        if (k >= cnt) {
+            k -= cnt;
+            curr += 1;
+        } else {
+            k -= 1;
+            curr *= 10;
+        }
+    }
+
+    return curr;
+}
+```
+
+#### Rust
+
+```rust
+impl Solution {
+    pub fn find_kth_number(n: i32, k: i32) -> i32 {
+        fn count(mut curr: i64, n: i32) -> i32 {
+            let mut next = curr + 1;
+            let mut total = 0;
+            let n = n as i64;
+            while curr <= n {
+                total += std::cmp::min(n - curr + 1, next - curr);
+                curr *= 10;
+                next *= 10;
+            }
+            total as i32
+        }
+
+        let mut curr = 1;
+        let mut k = k - 1;
+
+        while k > 0 {
+            let cnt = count(curr as i64, n);
+            if k >= cnt {
+                k -= cnt;
+                curr += 1;
+            } else {
+                k -= 1;
+                curr *= 10;
+            }
+        }
+
+        curr
+    }
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/0400-0499/0440.K-th Smallest in Lexicographical Order/Solution.rs

@@ -0,0 +1,31 @@
+impl Solution {
+    pub fn find_kth_number(n: i32, k: i32) -> i32 {
+        fn count(mut curr: i64, n: i32) -> i32 {
+            let mut next = curr + 1;
+            let mut total = 0;
+            let n = n as i64;
+            while curr <= n {
+                total += std::cmp::min(n - curr + 1, next - curr);
+                curr *= 10;
+                next *= 10;
+            }
+            total as i32
+        }
+
+        let mut curr = 1;
+        let mut k = k - 1;
+
+        while k > 0 {
+            let cnt = count(curr as i64, n);
+            if k >= cnt {
+                k -= cnt;
+                curr += 1;
+            } else {
+                k -= 1;
+                curr *= 10;
+            }
+        }
+
+        curr
+    }
+}

solution/0400-0499/0440.K-th Smallest in Lexicographical Order/Solution.ts

@@ -0,0 +1,28 @@
+function findKthNumber(n: number, k: number): number {
+    function count(curr: number): number {
+        let next = curr + 1;
+        let cnt = 0;
+        while (curr <= n) {
+            cnt += Math.min(n - curr + 1, next - curr);
+            curr *= 10;
+            next *= 10;
+        }
+        return cnt;
+    }
+
+    let curr = 1;
+    k--;
+
+    while (k > 0) {
+        const cnt = count(curr);
+        if (k >= cnt) {
+            k -= cnt;
+            curr += 1;
+        } else {
+            k -= 1;
+            curr *= 10;
+        }
+    }
+
+    return curr;
+}