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Commit 1ea1077

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fix: update solutions to lc problem: No.1955 (doocs#4257)
close doocs#4256 
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‎solution/1900-1999/1955.Count Number of Special Subsequences/README.md‎

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如果 $nums[i] = 0$:如果我们不选择 $nums[i],ドル则 $f[i][0] = f[i-1][0]$;如果我们选择 $nums[i],ドル那么 $f[i][0]=f[i-1][0]+1,ドル因为我们可以在任何一个以 0ドル$ 结尾的特殊子序列后面加上一个 0ドル$ 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],ドル则 $f[i][1] = f[i-1][1]$;如果我们选择 $nums[i],ドル那么 $f[i][1]=f[i-1][1]+f[i-1][0],ドル因为我们可以在任何一个以 0ドル$ 或 1ドル$ 结尾的特殊子序列后面加上一个 1ドル$ 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 $nums[i] = 1$:如果我们不选择 $nums[i],ドル则 $f[i][1] = f[i-1][1]$;如果我们选择 $nums[i],ドル那么 $f[i][1]=f[i-1][1]+f[i-1][0],ドル因为我们可以在任何一个以 0ドル$ 或 1ドル$ 结尾的特殊子序列后面加上一个 1ドル$ 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。
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如果 $nums[i] = 2$:如果我们不选择 $nums[i],ドル则 $f[i][2] = f[i-1][2]$;如果我们选择 $nums[i],ドル那么 $f[i][2]=f[i-1][2]+f[i-1][1],ドル因为我们可以在任何一个以 1ドル$ 或 2ドル$ 结尾的特殊子序列后面加上一个 2ドル$ 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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如果 $nums[i] = 2$:如果我们不选择 $nums[i],ドル则 $f[i][2] = f[i-1][2]$;如果我们选择 $nums[i],ドル那么 $f[i][2]=f[i-1][2]+f[i-1][1],ドル因为我们可以在任何一个以 1ドル$ 或 2ドル$ 结尾的特殊子序列后面加上一个 2ドル$ 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。
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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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$$
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\begin{aligned}
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f[i][0] &= 2 \times f[i - 1][0] + 1, \quad nums[i] = 0 \\
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f[i][1] &= f[i-1][1] + 2 \times f[i - 1][0], \quad nums[i] = 1 \\
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f[i][2] &= f[i-1][2] + 2 \times f[i - 1][1], \quad nums[i] = 2 \\
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f[i][1] &= f[i-1][0] + 2 \times f[i - 1][1], \quad nums[i] = 1 \\
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f[i][2] &= f[i-1][1] + 2 \times f[i - 1][2], \quad nums[i] = 2 \\
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f[i][j] &= f[i-1][j], \quad nums[i] \neq j
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\end{aligned}
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$$
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时间复杂度 $O(n),ドル空间复杂度 $O(n)$。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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#### Python3
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<!-- solution:start -->
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### 方法二
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### 方法二:动态规划(空间优化)
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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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‎solution/1900-1999/1955.Count Number of Special Subsequences/README_EN.md‎

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<!-- solution:start -->
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### Solution 1
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### Solution 1: Dynamic Programming
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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,ドル and if $nums[0]=0,ドル then $f[0][0]=1$.
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For $i \gt 0,ドル we consider the value of $nums[i]$:
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If $nums[i] = 0$: If we do not choose $nums[i],ドル then $f[i][0] = f[i-1][0]$; if we choose $nums[i],ドル then $f[i][0]=f[i-1][0]+1,ドル because we can add a 0ドル$ to the end of any special subsequence ending with 0ドル$ 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 $f[i-1][j]$.
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If $nums[i] = 1$: If we do not choose $nums[i],ドル then $f[i][1] = f[i-1][1]$; if we choose $nums[i],ドル then $f[i][1]=f[i-1][1]+f[i-1][0],ドル because we can add a 1ドル$ to the end of any special subsequence ending with 0ドル$ or 1ドル$ 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 $f[i-1][j]$.
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If $nums[i] = 2$: If we do not choose $nums[i],ドル then $f[i][2] = f[i-1][2]$; if we choose $nums[i],ドル then $f[i][2]=f[i-1][2]+f[i-1][1],ドル because we can add a 2ドル$ to the end of any special subsequence ending with 1ドル$ or 2ドル$ 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 $f[i-1][j]$.
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In summary, we can get the following state transition equations:
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$$
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\begin{aligned}
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f[i][0] &= 2 \times f[i - 1][0] + 1, \quad nums[i] = 0 \\
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f[i][1] &= f[i-1][0] + 2 \times f[i - 1][1], \quad nums[i] = 1 \\
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f[i][2] &= f[i-1][1] + 2 \times f[i - 1][2], \quad nums[i] = 2 \\
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f[i][j] &= f[i-1][j], \quad nums[i] \neq j
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\end{aligned}
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$$
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The final answer is $f[n-1][2]$.
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The time complexity is $O(n),ドル and the space complexity is $O(n)$. Where $n$ is the length of the array $nums$.
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Similar code found with 1 license type
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<!-- tabs:start -->
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<!-- solution:start -->
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### Solution 2
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### Solution 2: Dynamic Programming (Space Optimization)
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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 $O(1)$.
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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.
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The time complexity is $O(n),ドル and the space complexity is $O(1)$. Where $n$ is the length of the array $nums$.
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<!-- tabs:start -->
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