diff --git a/solution/0400-0499/0440.K-th Smallest in Lexicographical Order/README.md b/solution/0400-0499/0440.K-th Smallest in Lexicographical Order/README.md index 79d1666a074ef..04f5e0e21142b 100644 --- a/solution/0400-0499/0440.K-th Smallest in Lexicographical Order/README.md +++ b/solution/0400-0499/0440.K-th Smallest in Lexicographical Order/README.md @@ -49,7 +49,37 @@ tags: -### 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 $O(1)$。 @@ -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 + } +} +``` + diff --git a/solution/0400-0499/0440.K-th Smallest in Lexicographical Order/README_EN.md b/solution/0400-0499/0440.K-th Smallest in Lexicographical Order/README_EN.md index f8c7bc38a68b3..abdae7d0cad00 100644 --- a/solution/0400-0499/0440.K-th Smallest in Lexicographical Order/README_EN.md +++ b/solution/0400-0499/0440.K-th Smallest in Lexicographical Order/README_EN.md @@ -47,7 +47,46 @@ tags: -### 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. @@ -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 + } +} +``` + diff --git a/solution/0400-0499/0440.K-th Smallest in Lexicographical Order/Solution.rs b/solution/0400-0499/0440.K-th Smallest in Lexicographical Order/Solution.rs new file mode 100644 index 0000000000000..52351a5f8ff07 --- /dev/null +++ b/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 + } +} diff --git a/solution/0400-0499/0440.K-th Smallest in Lexicographical Order/Solution.ts b/solution/0400-0499/0440.K-th Smallest in Lexicographical Order/Solution.ts new file mode 100644 index 0000000000000..7e54456e1b47a --- /dev/null +++ b/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; +}