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feat: add solutions to lc problems: No.2907,2921 (#1920)

* No.2907.Maximum Profitable Triplets With Increasing Prices I * No.2921.Maximum Profitable Triplets With Increasing Prices II

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`59`	`59`
`60`	`60`	`因此,对于当前节点,如果其左子节点不为空,我们找到左子树的最右节点,作为前驱节点,然后将当前节点的右子节点赋给前驱节点的右子节点。然后将当前节点的左子节点赋给当前节点的右子节点,并将当前节点的左子节点置为空。然后将当前节点的右子节点作为下一个节点,继续处理,直至所有节点处理完毕。`
`61`	`61`
`62`		`-时间复杂度 $O(n),ドル空间复杂度 O(1)$。其中 $n$ 是树中节点的个数。`
	`62`	`+时间复杂度 $O(n),ドル其中 $n$ 是树中节点的个数。空间复杂度 O(1)$。`
`63`	`63`
`64`	`64`	`<!-- tabs:start -->`
`65`	`65`

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`@@ -44,6 +44,32 @@ As shown below, there are 5 ways you can generate "bag" from s.`
`44`	`44`
`45`	`45`	`## Solutions`
`46`	`46`
	`47`	`+Solution 1: Dynamic Programming`
	`48`	`+`
	`49`	`+We define $f[i][j]$ as the number of schemes where the first $i$ characters of string $s$ form the first $j$ characters of string $t$. Initially, $f[i][0]=1$ for all $i \in [0,m]$.`
	`50`	`+`
	`51`	`+When $i > 0,ドル we consider the calculation of $f[i][j]$:`
	`52`	`+`
	`53`	`+- When $s[i-1] \ne t[j-1],ドル we cannot select $s[i-1],ドル so $f[i][j]=f[i-1][j]$;`
	`54`	`+- Otherwise, we can select $s[i-1],ドル so $f[i][j]=f[i-1][j-1]$.`
	`55`	`+`
	`56`	`+Therefore, we have the following state transition equation:`
	`57`	`+`
	`58`	`+$$`
	`59`	`+f[i][j]=\left\{`
	`60`	`+\begin{aligned}`
	`61`	`+&f[i-1][j], &s[i-1] \ne t[j-1] \\`
	`62`	`+&f[i-1][j-1]+f[i-1][j], &s[i-1]=t[j-1]`
	`63`	`+\end{aligned}`
	`64`	`+\right.`
	`65`	`+$$`
	`66`	`+`
	`67`	`+The final answer is $f[m][n],ドル where $m$ and $n$ are the lengths of strings $s$ and $t$ respectively.`
	`68`	`+`
	`69`	`+The time complexity is $O(m \times n),ドル and the space complexity is $O(m \times n)$.`
	`70`	`+`
	`71`	`+We notice that the calculation of $f[i][j]$ is only related to $f[i-1][..]$. Therefore, we can optimize the first dimension, reducing the space complexity to $O(n)$.`
	`72`	`+`
`47`	`73`	`<!-- tabs:start -->`
`48`	`74`
`49`	`75`	`### Python3`

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