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feat: add solutions to lc problem: No.1955 (doocs#1359)
No.1955.Count Number of Special Subsequences
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‎solution/1900-1999/1955.Count Number of Special Subsequences/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：动态规划**
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我们定义 $f[i][j]$ 表示前 $i+1$ 个元素中，以 $j$ 结尾的特殊子序列的个数。初始时 $f[i][j]=0$，如果 $nums[0]=0$，则 $f[0][0]=1$。
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对于 $i \gt 0$，我们考虑 $nums[i]$ 的值：
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如果 $nums[i] = 0$：如果我们不选择 $nums[i]$，则 $f[i][0] = f[i-1][0]$；如果我们选择 $nums[i]$，那么 $f[i][0]=f[i-1][0]+1$，因为我们可以在任何一个以 $0$ 结尾的特殊子序列后面加上一个 $0$ 得到一个新的特殊子序列，也可以将 $nums[i]$ 单独作为一个特殊子序列。因此 $f[i][0] = 2 \times f[i - 1][0] + 1$。其余的 $f[i][j]$ 与 $f[i-1][j]$ 相等。
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如果 $nums[i] = 1$：如果我们不选择 $nums[i]$，则 $f[i][1] = f[i-1][1]$；如果我们选择 $nums[i]$，那么 $f[i][1]=f[i-1][1]+f[i-1][0]$，因为我们可以在任何一个以 $0$ 或 $1$ 结尾的特殊子序列后面加上一个 $1$ 得到一个新的特殊子序列。因此 $f[i][1] = f[i-1][1] + 2 \times f[i - 1][0]$。其余的 $f[i][j]$ 与 $f[i-1][j]$ 相等。
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如果 $nums[i] = 2$：如果我们不选择 $nums[i]$，则 $f[i][2] = f[i-1][2]$；如果我们选择 $nums[i]$，那么 $f[i][2]=f[i-1][2]+f[i-1][1]$，因为我们可以在任何一个以 $1$ 或 $2$ 结尾的特殊子序列后面加上一个 $2$ 得到一个新的特殊子序列。因此 $f[i][2] = f[i-1][2] + 2 \times f[i - 1][1]$。其余的 $f[i][j]$ 与 $f[i-1][j]$ 相等。
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综上，我们可以得到如下的状态转移方程：
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$$
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\begin{aligned}
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f[i][0] &= 2 \times f[i - 1][0] + 1, \quad nums[i] = 0 \\
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f[i][1] &= f[i-1][1] + 2 \times f[i - 1][0], \quad nums[i] = 1 \\
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f[i][2] &= f[i-1][2] + 2 \times f[i - 1][1], \quad nums[i] = 2 \\
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f[i][j] &= f[i-1][j], \quad nums[i] \neq j
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\end{aligned}
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$$
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最终的答案即为 $f[n-1][2]$。
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时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是数组 $nums$ 的长度。
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我们注意到，上述的状态转移方程中，$f[i][j]$ 的值仅与 $f[i-1][j]$ 有关，因此我们可以去掉第一维，将空间复杂度优化到 $O(1)$。
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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 countSpecialSubsequences(self, nums: List[int]) -> int:
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mod = 10**9 + 7
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n = len(nums)
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f = [[0] * 3 for _ in range(n)]
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f[0][0] = nums[0] == 0
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for i in range(1, n):
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if nums[i] == 0:
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f[i][0] = (2 * f[i - 1][0] + 1) % mod
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f[i][1] = f[i - 1][1]
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f[i][2] = f[i - 1][2]
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elif nums[i] == 1:
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f[i][0] = f[i - 1][0]
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f[i][1] = (f[i - 1][0] + 2 * f[i - 1][1]) % mod
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f[i][2] = f[i - 1][2]
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else:
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f[i][0] = f[i - 1][0]
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f[i][1] = f[i - 1][1]
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f[i][2] = (f[i - 1][1] + 2 * f[i - 1][2]) % mod
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return f[n - 1][2]
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```
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```python
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class Solution:
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def countSpecialSubsequences(self, nums: List[int]) -> int:
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mod = 10**9 + 7
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n = len(nums)
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f = [0] * 3
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f[0] = nums[0] == 0
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for i in range(1, n):
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if nums[i] == 0:
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f[0] = (2 * f[0] + 1) % mod
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elif nums[i] == 1:
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f[1] = (f[0] + 2 * f[1]) % mod
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else:
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f[2] = (f[1] + 2 * f[2]) % mod
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return f[2]
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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 int countSpecialSubsequences(int[] nums) {
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final int mod = (int) 1e9 + 7;
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int n = nums.length;
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int[][] f = new int[n][3];
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f[0][0] = nums[0] == 0 ? 1 : 0;
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for (int i = 1; i < n; ++i) {
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if (nums[i] == 0) {
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f[i][0] = (2 * f[i - 1][0] % mod + 1) % mod;
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f[i][1] = f[i - 1][1];
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f[i][2] = f[i - 1][2];
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} else if (nums[i] == 1) {
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f[i][0] = f[i - 1][0];
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f[i][1] = (f[i - 1][0] + 2 * f[i - 1][1] % mod) % mod;
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f[i][2] = f[i - 1][2];
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} else {
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f[i][0] = f[i - 1][0];
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f[i][1] = f[i - 1][1];
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f[i][2] = (f[i - 1][1] + 2 * f[i - 1][2] % mod) % mod;
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}
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}
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return f[n - 1][2];
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}
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}
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```
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```java
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class Solution {
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public int countSpecialSubsequences(int[] nums) {
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final int mod = (int) 1e9 + 7;
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int n = nums.length;
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int[] f = new int[3];
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f[0] = nums[0] == 0 ? 1 : 0;
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for (int i = 1; i < n; ++i) {
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if (nums[i] == 0) {
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f[0] = (2 * f[0] % mod + 1) % mod;
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} else if (nums[i] == 1) {
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f[1] = (f[0] + 2 * f[1] % mod) % mod;
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} else {
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f[2] = (f[1] + 2 * f[2] % mod) % mod;
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}
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}
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return f[2];
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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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int countSpecialSubsequences(vector<int>& nums) {
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const int mod = 1e9 + 7;
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int n = nums.size();
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int f[n][3];
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memset(f, 0, sizeof(f));
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f[0][0] = nums[0] == 0;
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for (int i = 1; i < n; ++i) {
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if (nums[i] == 0) {
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f[i][0] = (2 * f[i - 1][0] % mod + 1) % mod;
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f[i][1] = f[i - 1][1];
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f[i][2] = f[i - 1][2];
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} else if (nums[i] == 1) {
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f[i][0] = f[i - 1][0];
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f[i][1] = (f[i - 1][0] + 2 * f[i - 1][1] % mod) % mod;
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f[i][2] = f[i - 1][2];
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} else {
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f[i][0] = f[i - 1][0];
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f[i][1] = f[i - 1][1];
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f[i][2] = (f[i - 1][1] + 2 * f[i - 1][2] % mod) % mod;
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}
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}
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return f[n - 1][2];
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}
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};
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```
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```cpp
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class Solution {
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public:
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int countSpecialSubsequences(vector<int>& nums) {
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const int mod = 1e9 + 7;
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int n = nums.size();
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int f[3]{0};
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f[0] = nums[0] == 0;
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for (int i = 1; i < n; ++i) {
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if (nums[i] == 0) {
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f[0] = (2 * f[0] % mod + 1) % mod;
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} else if (nums[i] == 1) {
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f[1] = (f[0] + 2 * f[1] % mod) % mod;
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} else {
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f[2] = (f[1] + 2 * f[2] % mod) % mod;
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}
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}
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return f[2];
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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 countSpecialSubsequences(nums []int) int {
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const mod = 1e9 + 7
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n := len(nums)
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f := make([][3]int, n)
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if nums[0] == 0 {
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f[0][0] = 1
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}
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for i := 1; i < n; i++ {
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if nums[i] == 0 {
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f[i][0] = (2*f[i-1][0] + 1) % mod
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f[i][1] = f[i-1][1]
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f[i][2] = f[i-1][2]
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} else if nums[i] == 1 {
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f[i][0] = f[i-1][0]
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f[i][1] = (f[i-1][0] + 2*f[i-1][1]) % mod
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f[i][2] = f[i-1][2]
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} else {
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f[i][0] = f[i-1][0]
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f[i][1] = f[i-1][1]
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f[i][2] = (f[i-1][1] + 2*f[i-1][2]) % mod
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}
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}
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return f[n-1][2]
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}
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```
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```go
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func countSpecialSubsequences(nums []int) int {
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const mod = 1e9 + 7
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n := len(nums)
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f := [3]int{}
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if nums[0] == 0 {
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f[0] = 1
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}
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for i := 1; i < n; i++ {
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if nums[i] == 0 {
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f[0] = (2*f[0] + 1) % mod
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} else if nums[i] == 1 {
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f[1] = (f[0] + 2*f[1]) % mod
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} else {
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f[2] = (f[1] + 2*f[2]) % mod
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}
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}
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return f[2]
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}
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```
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### **TypeScript**
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```ts
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function countSpecialSubsequences(nums: number[]): number {
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const mod = 1e9 + 7;
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const n = nums.length;
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const f: number[][] = Array(n)
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.fill(0)
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.map(() => Array(3).fill(0));
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f[0][0] = nums[0] === 0 ? 1 : 0;
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for (let i = 1; i < n; ++i) {
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if (nums[i] === 0) {
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f[i][0] = (((2 * f[i - 1][0]) % mod) + 1) % mod;
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f[i][1] = f[i - 1][1];
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f[i][2] = f[i - 1][2];
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} else if (nums[i] === 1) {
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f[i][0] = f[i - 1][0];
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f[i][1] = (f[i - 1][0] + ((2 * f[i - 1][1]) % mod)) % mod;
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f[i][2] = f[i - 1][2];
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} else {
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f[i][0] = f[i - 1][0];
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f[i][1] = f[i - 1][1];
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f[i][2] = (f[i - 1][1] + ((2 * f[i - 1][2]) % mod)) % mod;
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}
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}
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return f[n - 1][2];
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}
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```
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```ts
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function countSpecialSubsequences(nums: number[]): number {
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const mod = 1e9 + 7;
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const n = nums.length;
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const f: number[] = [0, 0, 0];
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f[0] = nums[0] === 0 ? 1 : 0;
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for (let i = 1; i < n; ++i) {
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if (nums[i] === 0) {
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f[0] = (((2 * f[0]) % mod) + 1) % mod;
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f[1] = f[1];
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f[2] = f[2];
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} else if (nums[i] === 1) {
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f[0] = f[0];
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f[1] = (f[0] + ((2 * f[1]) % mod)) % mod;
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f[2] = f[2];
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} else {
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f[0] = f[0];
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f[1] = f[1];
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f[2] = (f[1] + ((2 * f[2]) % mod)) % mod;
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}
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}
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return f[2];
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}
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```
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### **...**

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