Commit 085570c

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feat: add solutions to lc problems: No.1690,1727,1730,1733 (doocs#2302)

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`@@ -304,4 +304,6 @@ function stoneGameVII(stones: number[]): number {`
`304`	`304`	`}`
`305`	`305`	```
`306`	`306`
	`307`	`+<!-- tabs:end -->`
	`308`	`+`
`307`	`309`	`<!-- end -->`

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`@@ -71,7 +71,7 @@`
`71`	`71`
`72`	`72`	`对于矩阵的某一行,我们记第 $k$ 大元素的值为 $val_k,ドル其中 $k \geq 1,ドル那么该行至少有 $k$ 个元素不小于 $val_k,ドル组成的全 1ドル$ 子矩阵面积为 $val_k \times k$。从大到小遍历矩阵该行的每个元素,取 $val_k \times k$ 的最大值,更新答案。`
`73`	`73`
`74`		`-时间复杂度 $O(m\times n\times \log n)$。其中 $m$ 和 $n$ 分别为矩阵的行数和列数。`
	`74`	`+时间复杂度 $O(m\times n\times \log n)$。其中 $m$ 和 $n$ 分别为矩阵的行数和列数。`
`75`	`75`
`76`	`76`	`<!-- tabs:start -->`
`77`	`77`

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`@@ -47,7 +47,17 @@ The largest submatrix of 1s, in bold, has an area of 3.`
`47`	`47`
`48`	`48`	`## Solutions`
`49`	`49`
`50`		`-### Solution 1`
	`50`	`+### Solution 1: Preprocessing + Sorting`
	`51`	`+`
	`52`	`+Since the matrix is rearranged by columns according to the problem, we can first preprocess each column of the matrix.`
	`53`	`+`
	`54`	`+For each element with a value of 1ドル,ドル we update its value to the maximum consecutive number of 1ドル$s above it, that is, $matrix[i][j] = matrix[i-1][j] + 1$.`
	`55`	`+`
	`56`	`+Next, we can sort each row of the updated matrix. Then traverse each row, calculate the area of the largest sub-matrix full of 1ドル$s with this row as the bottom edge. The specific calculation logic is as follows:`
	`57`	`+`
	`58`	`+For a row of the matrix, we denote the value of the $k$-th largest element as $val_k,ドル where $k \geq 1,ドル then there are at least $k$ elements in this row that are not less than $val_k,ドル forming a sub-matrix full of 1ドル$s with an area of $val_k \times k$. Traverse each element of this row from large to small, take the maximum value of $val_k \times k,ドル and update the answer.`
	`59`	`+`
	`60`	`+The time complexity is $O(m \times n \times \log n)$. Here, $m$ and $n$ are the number of rows and columns of the matrix, respectively.`
`51`	`61`
`52`	`62`	`<!-- tabs:start -->`
`53`	`63`

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`@@ -56,7 +56,15 @@`
`56`	`56`
`57`	`57`	`## Solutions`
`58`	`58`
`59`		`-### Solution 1`
	`59`	`+### Solution 1: BFS (Breadth-First Search)`
	`60`	`+`
	`61`	+According to the problem, we need to start from `*`, find the nearest `#`, and return the shortest path length.
	`62`	`+`
	`63`	+First, we traverse the entire two-dimensional array to find the position of `*`, which will be the starting point for BFS, and put it into the queue.
	`64`	`+`
	`65`	+Then, we start BFS, traversing the elements in the queue. Each time we traverse an element, we add the elements in the four directions (up, down, left, and right) of it into the queue, until we encounter `#`, and return the current layer number.
	`66`	`+`
	`67`	`+The time complexity is $O(m \times n),ドル and the space complexity is $O(1)$. Here, $m$ and $n$ are the number of rows and columns of the two-dimensional array, respectively.`
`60`	`68`
`61`	`69`	`<!-- tabs:start -->`
`62`	`70`

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`@@ -50,7 +50,13 @@ Note that friendships are not transitive, meaning if <code>x</code> is a friend`
`50`	`50`
`51`	`51`	`## Solutions`
`52`	`52`
`53`		`-### Solution 1`
	`53`	`+### Solution 1: Simulation + Statistics`
	`54`	`+`
	`55`	`+For each friendship, if the sets of languages known by the two people do not intersect, then a language needs to be taught so that the two people can communicate with each other. We put these people into a hash set $s$.`
	`56`	`+`
	`57`	+Then in this set $s,ドル we count the number of people who know each language, and get the maximum number, which we denote as $mx$. So the answer is `len(s) - mx`.
	`58`	`+`
	`59`	`+The time complexity is $O(m^2 \times k)$. Here, $m$ is the number of languages, and $k$ is the number of friendships.`
`54`	`60`
`55`	`61`	`<!-- tabs:start -->`
`56`	`62`

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