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Jan 19, 2024
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Update README.md #2241

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bfc7ffc
Update README.md
yanglbme Jan 19, 2024
7c84360
Update README_EN.md
yanglbme Jan 19, 2024
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4 changes: 1 addition & 3 deletions solution/2800-2899/2809.Minimum Time to Make Array Sum At Most x/README.md
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Expand Up @@ -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
Expand Down Expand Up @@ -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 $O(n)$。

<!-- tabs:start -->

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4 changes: 1 addition & 3 deletions solution/2800-2899/2809.Minimum Time to Make Array Sum At Most x/README_EN.md
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Expand Up @@ -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
Expand Down Expand Up @@ -235,7 +233,7 @@ function minimumTime(nums1: number[], nums2: number[], x: number): number {

<!-- tabs:end -->

### 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 $O(n)$.

<!-- tabs:start -->

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