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feat: add solutions to lc problem: No.1643
No.1643.Kth Smallest Instructions
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‎solution/1600-1699/1641.Count Sorted Vowel Strings/README.md

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@@ -64,9 +64,9 @@
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**方法二：动态规划 + 前缀和**
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定义 $f[i][j]$ 表示当前已经选了 $i$ 个元音字母，且最后一个元音字母是 $j$ 的方案数。初始时 $f[0][j]=1$。答案是 $\sum_{j = 0}^4 f[n - 1][j]$。
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定义 $f[i][j]$ 表示当前已经选了 $i$ 个元音字母，且最后一个元音字母是 $j$ 的方案数。初始时 $f[1][j]=1$。答案是 $\sum_{j = 0}^4 f[n][j]$。
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我们可以发现，$f[i][j]$ 只与 $f[i - 1][j]$ 有关，因此我们可以使用滚动数组优化空间复杂度。
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我们可以发现 $f[i][j]$ 只与 $f[i - 1][j]$ 有关，因此可以将二维数组压缩为一维数组，从而优化空间复杂度。
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时间复杂度 $O(n \times C)$，空间复杂度 $O(C)$。其中 $n$ 为题目给定的整数，而 $C$ 是元音字母的数量，本题中 $C = 5$。
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‎solution/1600-1699/1643.Kth Smallest Instructions/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：组合计数**
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根据题目描述我们可以知道，最终的路径是由 $destination[0]$ 个 `'V'` 和 $destination[1]$ 个 `'H'` 组成的，且按字典序排列后的第 $k$ 条最小指令。
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我们首先考虑字典序的最高位，即最左边的字符。如果高位字符是 `'V'`，那么所有以 `'H'` 开头的路径的字典序都比它小，而以 `'H'` 开头的路径总数为 $x = C_{v+h-1}^{h-1}$。
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如果 $k \lt x$，那么高位字符一定是 `'V'`，我们将 $k$ 减去 $x$，并将 $v$ 减 $1$，然后继续考虑下一位字符；否则，高位字符一定是 `'H'`，我们将 $h$ 减 $1$，然后继续考虑下一位字符。
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注意，如果 $h = 0$，那么高位字符一定是 `'V'`，因为剩下的字符都是 `'V'`。
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问题可以转换为求解 $C_{n}^{k}$，我们可以通过公式 $C_{n}^{k} = C_{n-1}^{k-1} + C_{n-1}^{k}$ 递推求解。
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时间复杂度 $O((h + v) \times h)$，空间复杂度 $O((h + v) \times h)$。其中 $h$ 和 $v$ 分别为 $destination[1]$ 和 $destination[0]$。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def kthSmallestPath(self, destination: List[int], k: int) -> str:
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v, h = destination
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ans = []
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for _ in range(h + v):
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if h == 0:
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ans.append("V")
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else:
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x = comb(h + v - 1, h - 1)
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if k > x:
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ans.append("V")
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v -= 1
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k -= x
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else:
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ans.append("H")
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h -= 1
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return "".join(ans)
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public String kthSmallestPath(int[] destination, int k) {
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int v = destination[0], h = destination[1];
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int n = v + h;
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int[][] c = new int[n + 1][h + 1];
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c[0][0] = 1;
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for (int i = 1; i <= n; ++i) {
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c[i][0] = 1;
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for (int j = 1; j <= h; ++j) {
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c[i][j] = c[i - 1][j] + c[i - 1][j - 1];
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}
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}
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StringBuilder ans = new StringBuilder();
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for (int i = n; i > 0; --i) {
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if (h == 0) {
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ans.append('V');
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} else {
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int x = c[v + h - 1][h - 1];
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if (k > x) {
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ans.append('V');
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k -= x;
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--v;
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} else {
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ans.append('H');
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--h;
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}
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}
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}
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return ans.toString();
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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string kthSmallestPath(vector<int>& destination, int k) {
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int v = destination[0], h = destination[1];
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int n = v + h;
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int c[n + 1][h + 1];
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memset(c, 0, sizeof(c));
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c[0][0] = 1;
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for (int i = 1; i <= n; ++i) {
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c[i][0] = 1;
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for (int j = 1; j <= h; ++j) {
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c[i][j] = c[i - 1][j] + c[i - 1][j - 1];
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}
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}
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string ans;
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for (int i = 0; i < n; ++i) {
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if (h == 0) {
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ans.push_back('V');
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} else {
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int x = c[v + h - 1][h - 1];
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if (k > x) {
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ans.push_back('V');
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--v;
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k -= x;
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} else {
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ans.push_back('H');
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--h;
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}
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}
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func kthSmallestPath(destination []int, k int) string {
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v, h := destination[0], destination[1]
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n := v + h
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c := make([][]int, n+1)
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for i := range c {
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c[i] = make([]int, h+1)
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c[i][0] = 1
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}
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for i := 1; i <= n; i++ {
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for j := 1; j <= h; j++ {
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c[i][j] = c[i-1][j] + c[i-1][j-1]
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}
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}
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ans := []byte{}
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for i := 0; i < n; i++ {
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if h == 0 {
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ans = append(ans, 'V')
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} else {
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x := c[v+h-1][h-1]
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if k > x {
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ans = append(ans, 'V')
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k -= x
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v--
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} else {
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ans = append(ans, 'H')
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h--
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}
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}
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}
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return string(ans)
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}
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```
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### **...**

‎solution/1600-1699/1643.Kth Smallest Instructions/README_EN.md

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### **Python3**
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```python
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class Solution:
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def kthSmallestPath(self, destination: List[int], k: int) -> str:
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v, h = destination
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ans = []
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for _ in range(h + v):
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if h == 0:
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ans.append("V")
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else:
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x = comb(h + v - 1, h - 1)
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if k > x:
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ans.append("V")
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v -= 1
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k -= x
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else:
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ans.append("H")
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h -= 1
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return "".join(ans)
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```
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### **Java**
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```java
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class Solution {
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public String kthSmallestPath(int[] destination, int k) {
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int v = destination[0], h = destination[1];
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int n = v + h;
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int[][] c = new int[n + 1][h + 1];
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c[0][0] = 1;
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for (int i = 1; i <= n; ++i) {
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c[i][0] = 1;
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for (int j = 1; j <= h; ++j) {
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c[i][j] = c[i - 1][j] + c[i - 1][j - 1];
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}
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}
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StringBuilder ans = new StringBuilder();
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for (int i = n; i > 0; --i) {
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if (h == 0) {
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ans.append('V');
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} else {
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int x = c[v + h - 1][h - 1];
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if (k > x) {
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ans.append('V');
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k -= x;
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--v;
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} else {
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ans.append('H');
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--h;
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}
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}
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}
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return ans.toString();
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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string kthSmallestPath(vector<int>& destination, int k) {
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int v = destination[0], h = destination[1];
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int n = v + h;
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int c[n + 1][h + 1];
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memset(c, 0, sizeof(c));
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c[0][0] = 1;
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for (int i = 1; i <= n; ++i) {
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c[i][0] = 1;
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for (int j = 1; j <= h; ++j) {
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c[i][j] = c[i - 1][j] + c[i - 1][j - 1];
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}
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}
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string ans;
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for (int i = 0; i < n; ++i) {
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if (h == 0) {
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ans.push_back('V');
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} else {
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int x = c[v + h - 1][h - 1];
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if (k > x) {
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ans.push_back('V');
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--v;
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k -= x;
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} else {
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ans.push_back('H');
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--h;
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}
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}
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func kthSmallestPath(destination []int, k int) string {
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v, h := destination[0], destination[1]
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n := v + h
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c := make([][]int, n+1)
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for i := range c {
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c[i] = make([]int, h+1)
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c[i][0] = 1
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}
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for i := 1; i <= n; i++ {
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for j := 1; j <= h; j++ {
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c[i][j] = c[i-1][j] + c[i-1][j-1]
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}
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}
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ans := []byte{}
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for i := 0; i < n; i++ {
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if h == 0 {
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ans = append(ans, 'V')
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} else {
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x := c[v+h-1][h-1]
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if k > x {
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ans = append(ans, 'V')
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k -= x
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v--
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} else {
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ans = append(ans, 'H')
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h--
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}
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}
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}
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return string(ans)
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}
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```
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### **...**

‎solution/1600-1699/1643.Kth Smallest Instructions/Solution.cpp

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class Solution {
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public:
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string kthSmallestPath(vector<int>& destination, int k) {
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int v = destination[0], h = destination[1];
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int n = v + h;
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int c[n + 1][h + 1];
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memset(c, 0, sizeof(c));
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c[0][0] = 1;
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for (int i = 1; i <= n; ++i) {
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c[i][0] = 1;
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for (int j = 1; j <= h; ++j) {
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c[i][j] = c[i - 1][j] + c[i - 1][j - 1];
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}
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}
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string ans;
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for (int i = 0; i < n; ++i) {
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if (h == 0) {
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ans.push_back('V');
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} else {
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int x = c[v + h - 1][h - 1];
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if (k > x) {
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ans.push_back('V');
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--v;
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k -= x;
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} else {
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ans.push_back('H');
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--h;
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}
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}
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}
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return ans;
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}
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};

‎solution/1600-1699/1643.Kth Smallest Instructions/Solution.go

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func kthSmallestPath(destination []int, k int) string {
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v, h := destination[0], destination[1]
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n := v + h
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c := make([][]int, n+1)
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for i := range c {
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c[i] = make([]int, h+1)
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c[i][0] = 1
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}
9+
for i := 1; i <= n; i++ {
10+
for j := 1; j <= h; j++ {
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c[i][j] = c[i-1][j] + c[i-1][j-1]
12+
}
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}
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ans := []byte{}
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for i := 0; i < n; i++ {
16+
if h == 0 {
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ans = append(ans, 'V')
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} else {
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x := c[v+h-1][h-1]
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if k > x {
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ans = append(ans, 'V')
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k -= x
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v--
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} else {
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ans = append(ans, 'H')
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h--
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}
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}
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}
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return string(ans)
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}

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