fix: update

Author: yanglbme

solution/1300-1399/1326.Minimum Number of Taps to Open to Water a Garden/README.md

@@ -75,15 +75,15 @@ tags:
 
 我们注意到，对于所有能覆盖某个左端点的水龙头，选择能覆盖最远右端点的那个水龙头是最优的。
 
-因此，我们可以先预处理数组 $ranges$，对于第 $i$ 个水龙头，它能覆盖的左端点 $l = max(0, i - ranges[i])$，右端点 $r = i + ranges[i]$，我们算出所有能覆盖左端点 $l$ 的水龙头中，右端点最大的那个位置，记录在数组 $last[i]$ 中。
+因此，我们可以先预处理数组 $ranges$，对于第 $i$ 个水龙头，它能覆盖的左端点 $l = \max(0, i - ranges[i])$，右端点 $r = i + ranges[i]$，我们算出所有能覆盖左端点 $l$ 的水龙头中，右端点最大的那个位置，记录在数组 $last[i]$ 中。
 
 然后我们定义以下三个变量，其中：
 
 -   变量 $ans$ 表示最终答案，即最少水龙头数目；
 -   变量 $mx$ 表示当前能覆盖的最远右端点；
 -   变量 $pre$ 表示上一个水龙头覆盖的最远右端点。
 
-我们在 $[0,...n-1]$ 的范围内遍历所有位置，对于当前位置 $i$，我们用 $last[i]$ 更新 $mx$，即 $mx = max(mx, last[i])$。
+我们在 $[0,...n-1]$ 的范围内遍历所有位置，对于当前位置 $i$，我们用 $last[i]$ 更新 $mx$，即 $mx = \max(mx, last[i])$。
 
 -   如果 $mx \leq i$，说明无法覆盖下一个位置，返回 $-1$。
 -   如果 $pre = i$，说明需要使用一个新的子区间，因此我们将 $ans$ 加 $1$，并且更新 $pre = mx$。

solution/1300-1399/1326.Minimum Number of Taps to Open to Water a Garden/README_EN.md

@@ -66,7 +66,32 @@ Opening Only the second tap will water the whole garden [0,5]
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Greedy
+
+We note that for all taps that can cover a certain left endpoint, choosing the tap that can cover the farthest right endpoint is optimal.
+
+Therefore, we can preprocess the array $ranges$. For the $i$-th tap, it can cover the left endpoint $l = \max(0, i - ranges[i])$ and the right endpoint $r = i + ranges[i]$. We calculate the position of the tap that can cover the left endpoint $l$ with the farthest right endpoint and record it in the array $last[i]$.
+
+Then we define the following three variables:
+
+-   Variable $ans$ represents the final answer, i.e., the minimum number of taps;
+-   Variable $mx$ represents the farthest right endpoint that can currently be covered;
+-   Variable $pre$ represents the farthest right endpoint covered by the previous tap.
+
+We traverse all positions in the range $[0, \ldots, n-1]$. For the current position $i$, we use $last[i]$ to update $mx$, i.e., $mx = \max(mx, last[i])$.
+
+-   If $mx \leq i$, it means the next position cannot be covered, so we return $-1$.
+-   If $pre = i$, it means a new subinterval needs to be used, so we increment $ans$ by $1$ and update $pre = mx$.
+
+After the traversal, we return $ans$.
+
+The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the garden.
+
+Similar problems:
+
+-   [45. Jump Game II](https://github.com/doocs/leetcode/blob/main/solution/0000-0099/0045.Jump%20Game%20II/README.md)
+-   [55. Jump Game](https://github.com/doocs/leetcode/blob/main/solution/0000-0099/0055.Jump%20Game/README.md)
+-   [1024. Video Stitching](https://github.com/doocs/leetcode/blob/main/solution/1000-1099/1024.Video%20Stitching/README.md)
 
 <!-- tabs:start -->
 

solution/1300-1399/1328.Break a Palindrome/README_EN.md

@@ -57,7 +57,13 @@ Of all the ways, "aaccba" is the lexicographically smallest.
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Greedy
+
+First, we check if the length of the string is $1$. If it is, we directly return an empty string.
+
+Otherwise, we traverse the first half of the string from left to right, find the first character that is not `'a'`, and change it to `'a'`. If no such character exists, we change the last character to `'b'`.
+
+The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the string.
 
 <!-- tabs:start -->