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docs: update solution to lc problem: No.2809 (doocs#2241)

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solution/2800-2899/2809.Minimum Time to Make Array Sum At Most x
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`80`	`80`	`时间复杂度 $O(n^2),ドル空间复杂度 $O(n^2)$。其中 $n$ 是数组的长度。`
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`82`		`-我们注意到,状态 $f[i][j]$ 只与 $f[i-1][j]$ 和 $f[i-1][j-1]$ 有关,因此我们可以优化掉第一维,将空间复杂度降低到 $O(n)$。`
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`84`	`82`	`<!-- tabs:start -->`
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`86`	`84`	```python
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`239`	`237`	`<!-- tabs:end -->`
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`241`		`-### 方法二`
	`239`	`+我们注意到,状态 $f[i][j]$ 只与 $f[i-1][j]$ 和 $f[i-1][j-1]$ 有关,因此我们可以优化掉第一维,将空间复杂度降低到 $O(n)$。`
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`@@ -76,8 +76,6 @@ Finally, we enumerate $j$ and find the smallest $j$ that satisfies $s_1 + s_2 \t`
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`77`	`77`	`The time complexity is $O(n^2),ドル and the space complexity is $O(n^2),ドル where $n$ is the length of the array.`
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`79`		`-We notice that the state $f[i][j]$ is only related to $f[i-1][j]$ and $f[i-1][j-1],ドル so we can optimize the first dimension and reduce the space complexity to $O(n)$.`
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`83`	`81`	```python
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`236`	`234`	`<!-- tabs:end -->`
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`238`		`-### Solution 2`
	`236`	`+We notice that the state $f[i][j]$ is only related to $f[i-1][j]$ and $f[i-1][j-1],ドル so we can optimize the first dimension and reduce the space complexity to $O(n)$.`
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