fix: update solutions to lc problem: No.1955 (#4257)

close #4256

Author: yanglbme

solution/1900-1999/1955.Count Number of Special Subsequences/README.md

@@ -86,17 +86,17 @@ tags:
 
 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。
 
-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。
+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 $f[i][1] = f[i-1][0] + 2 \times f[i - 1][1]$。其余的 $f[i][j]$ 与 $f[i-1][j]$ 相等。
 
-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。
+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 $f[i][2] = f[i-1][1] + 2 \times f[i - 1][2]$。其余的 $f[i][j]$ 与 $f[i-1][j]$ 相等。
 
 综上，我们可以得到如下的状态转移方程：
 
 $$
 \begin{aligned}
 f[i][0] &= 2 \times f[i - 1][0] + 1, \quad nums[i] = 0 \\
-f[i][1] &= f[i-1][1] + 2 \times f[i - 1][0], \quad nums[i] = 1 \\
-f[i][2] &= f[i-1][2] + 2 \times f[i - 1][1], \quad nums[i] = 2 \\
+f[i][1] &= f[i-1][0] + 2 \times f[i - 1][1], \quad nums[i] = 1 \\
+f[i][2] &= f[i-1][1] + 2 \times f[i - 1][2], \quad nums[i] = 2 \\
 f[i][j] &= f[i-1][j], \quad nums[i] \neq j
 \end{aligned}
 $$
@@ -105,8 +105,6 @@ $$
 
 时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是数组 $nums$ 的长度。
 
-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 $O(1)$。
-
 <!-- tabs:start -->
 
 #### Python3
@@ -258,7 +256,9 @@ function countSpecialSubsequences(nums: number[]): number {
 
 <!-- solution:start -->
 
-### 方法二
+### 方法二：动态规划（空间优化）
+
+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 $O(1)$。
 
 <!-- tabs:start -->
 

solution/1900-1999/1955.Count Number of Special Subsequences/README_EN.md

@@ -76,7 +76,34 @@ tags:
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Dynamic Programming
+
+We define $f[i][j]$ to represent the number of special subsequences ending with $j$ among the first $i+1$ elements. Initially, $f[i][j]=0$, and if $nums[0]=0$, then $f[0][0]=1$.
+
+For $i \gt 0$, we consider the value of $nums[i]$:
+
+If $nums[i] = 0$: If we do not choose $nums[i]$, then $f[i][0] = f[i-1][0]$; if we choose $nums[i]$, then $f[i][0]=f[i-1][0]+1$, because we can add a $0$ to the end of any special subsequence ending with $0$ to get a new special subsequence, or we can use $nums[i]$ alone as a special subsequence. Therefore, $f[i][0] = 2 \times f[i - 1][0] + 1$. The rest of $f[i][j]$ is equal to $f[i-1][j]$.
+
+If $nums[i] = 1$: If we do not choose $nums[i]$, then $f[i][1] = f[i-1][1]$; if we choose $nums[i]$, then $f[i][1]=f[i-1][1]+f[i-1][0]$, because we can add a $1$ to the end of any special subsequence ending with $0$ or $1$ to get a new special subsequence. Therefore, $f[i][1] = f[i-1][0] + 2 \times f[i - 1][1]$. The rest of $f[i][j]$ is equal to $f[i-1][j]$.
+
+If $nums[i] = 2$: If we do not choose $nums[i]$, then $f[i][2] = f[i-1][2]$; if we choose $nums[i]$, then $f[i][2]=f[i-1][2]+f[i-1][1]$, because we can add a $2$ to the end of any special subsequence ending with $1$ or $2$ to get a new special subsequence. Therefore, $f[i][2] = f[i-1][1] + 2 \times f[i - 1][2]$. The rest of $f[i][j]$ is equal to $f[i-1][j]$.
+
+In summary, we can get the following state transition equations:
+
+$$
+\begin{aligned}
+f[i][0] &= 2 \times f[i - 1][0] + 1, \quad nums[i] = 0 \\
+f[i][1] &= f[i-1][0] + 2 \times f[i - 1][1], \quad nums[i] = 1 \\
+f[i][2] &= f[i-1][1] + 2 \times f[i - 1][2], \quad nums[i] = 2 \\
+f[i][j] &= f[i-1][j], \quad nums[i] \neq j
+\end{aligned}
+$$
+
+The final answer is $f[n-1][2]$.
+
+The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the length of the array $nums$.
+
+Similar code found with 1 license type
 
 <!-- tabs:start -->
 
@@ -229,7 +256,13 @@ function countSpecialSubsequences(nums: number[]): number {
 
 <!-- solution:start -->
 
-### Solution 2
+### Solution 2: Dynamic Programming (Space Optimization)
+
+We notice that in the above state transition equations, the value of $f[i][j]$ is only related to $f[i-1][j]$. Therefore, we can remove the first dimension and optimize the space complexity to $O(1)$.
+
+We can use an array $f$ of length 3 to represent the number of special subsequences ending with 0, 1, and 2, respectively. For each element in the array, we update the array $f$ according to the value of the current element.
+
+The time complexity is $O(n)$, and the space complexity is $O(1)$. Where $n$ is the length of the array $nums$.
 
 <!-- tabs:start -->