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

Author: yanglbme

solution/2800-2899/2809.Minimum Time to Make Array Sum At Most x/README.md

@@ -79,8 +79,6 @@ $$
 
 时间复杂度 $O(n^2)$，空间复杂度 $O(n^2)$。其中 $n$ 是数组的长度。
 
-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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 ```python
@@ -238,7 +236,7 @@ function minimumTime(nums1: number[], nums2: number[], x: number): number {
 
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-### 方法二
+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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solution/2800-2899/2809.Minimum Time to Make Array Sum At Most x/README_EN.md

@@ -76,8 +76,6 @@ Finally, we enumerate $j$ and find the smallest $j$ that satisfies $s_1 + s_2 \t
 
 The time complexity is $O(n^2)$, and the space complexity is $O(n^2)$, where $n$ is the length of the array.
 
-We notice that the state $f[i][j]$ is only related to $f[i-1][j]$ and $f[i-1][j-1]$, so we can optimize the first dimension and reduce the space complexity to $O(n)$.
-
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 ```python
@@ -235,7 +233,7 @@ function minimumTime(nums1: number[], nums2: number[], x: number): number {
 
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-### Solution 2
+We notice that the state $f[i][j]$ is only related to $f[i-1][j]$ and $f[i-1][j-1]$, so we can optimize the first dimension and reduce the space complexity to $O(n)$.
 
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