[pull] main from doocs:main

solution/1700-1799/1761.Minimum Degree of a Connected Trio in a Graph/README.md

@@ -68,9 +68,9 @@ tags:
 
 ### 方法一：暴力枚举
 
-我们先将所有边存入邻接矩阵 $g$ 中，再将每个节点的度数存入数组 $deg$ 中。初始化答案 $ans=+\infty$。
+我们先将所有边存入邻接矩阵 $\textit{g}$ 中，再将每个节点的度数存入数组 $\textit{deg}$ 中。初始化答案 $\textit{ans}=+\infty$。
 
-然后枚举所有的三元组 $(i, j, k)$，其中 $i \lt j \lt k$，如果 $g[i][j] = g[j][k] = g[i][k] = 1$，则说明这三个节点构成了一个连通三元组，此时更新答案为 $ans = \min(ans, deg[i] + deg[j] + deg[k] - 6)$。
+然后枚举所有的三元组 $(i, j, k)$，其中 $i \lt j \lt k$，如果 $\textit{g}[i][j] = \textit{g}[j][k] = \textit{g}[i][k] = 1$，则说明这三个节点构成了一个连通三元组，此时更新答案为 $\textit{ans} = \min(\textit{ans}, \textit{deg}[i] + \textit{deg}[j] + \textit{deg}[k] - 6)$。
 
 枚举完所有的三元组后，如果答案仍然为 $+\infty$，说明图中不存在连通三元组，返回 $-1$，否则返回答案。
 
@@ -81,6 +81,10 @@ tags:
 #### Python3
 
 ```python
+def min(a: int, b: int) -> int:
+    return a if a < b else b
+
+
 class Solution:
     def minTrioDegree(self, n: int, edges: List[List[int]]) -> int:
         g = [[False] * n for _ in range(n)]

solution/1700-1799/1761.Minimum Degree of a Connected Trio in a Graph/README_EN.md

@@ -64,13 +64,25 @@ tags:
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Brute Force Enumeration
+
+We first store all edges in the adjacency matrix $\textit{g}$, and then store the degree of each node in the array $\textit{deg}$. Initialize the answer $\textit{ans} = +\infty$.
+
+Then enumerate all triplets $(i, j, k)$, where $i \lt j \lt k$. If $\textit{g}[i][j] = \textit{g}[j][k] = \textit{g}[i][k] = 1$, it means these three nodes form a connected trio. In this case, update the answer to $\textit{ans} = \min(\textit{ans}, \textit{deg}[i] + \textit{deg}[j] + \textit{deg}[k] - 6)$.
+
+After enumerating all triplets, if the answer is still $+\infty$, it means there is no connected trio in the graph, return $-1$. Otherwise, return the answer.
+
+The time complexity is $O(n^3)$, and the space complexity is $O(n^2)$. Here, $n$ is the number of nodes.
 
 <!-- tabs:start -->
 
 #### Python3
 
 ```python
+def min(a: int, b: int) -> int:
+    return a if a < b else b
+
+
 class Solution:
     def minTrioDegree(self, n: int, edges: List[List[int]]) -> int:
         g = [[False] * n for _ in range(n)]

solution/1700-1799/1761.Minimum Degree of a Connected Trio in a Graph/Solution.py

@@ -1,3 +1,7 @@
+def min(a: int, b: int) -> int:
+    return a if a < b else b
+
+
 class Solution:
     def minTrioDegree(self, n: int, edges: List[List[int]]) -> int:
         g = [[False] * n for _ in range(n)]

solution/1700-1799/1762.Buildings With an Ocean View/README.md

@@ -72,9 +72,9 @@ tags:
 
 ### 方法一：逆序遍历求右侧最大值
 
-我们逆序遍历数组 $height$ 每个元素 $v$，判断 $v$ 与右侧最大元素 $mx$ 的大小关系，若 $mx \lt v$，说明右侧所有元素都比当前元素小，当前位置能看到海景，加入结果数组 $ans$。然后我们更新 $mx$ 为 $v$。
+我们逆序遍历数组 $\textit{height}$ 每个元素 $v$，判断 $v$ 与右侧最大元素 $mx$ 的大小关系，若 $mx \lt v$，说明右侧所有元素都比当前元素小，当前位置能看到海景，加入结果数组 $\textit{ans}$。然后我们更新 $mx$ 为 $v$。
 
-遍历结束后，逆序返回 $ans$ 即可。
+遍历结束后，逆序返回 $\textit{ans}$ 即可。
 
 时间复杂度 $O(n)$，其中 $n$ 为数组长度。忽略答案数组的空间消耗，空间复杂度 $O(1)$。
 

solution/1700-1799/1762.Buildings With an Ocean View/README_EN.md

@@ -63,7 +63,13 @@ tags:
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Reverse Traversal to Find the Maximum on the Right
+
+We traverse the array $\textit{height}$ in reverse order for each element $v$, comparing $v$ with the maximum element $mx$ on the right. If $mx \lt v$, it means all elements to the right are smaller than the current element, so the current position can see the ocean and is added to the result array $\textit{ans}$. Then we update $mx$ to $v$.
+
+After the traversal, return $\textit{ans}$ in reverse order.
+
+The time complexity is $O(n)$, where $n$ is the length of the array. Ignoring the space consumption of the answer array, the space complexity is $O(1)$.
 
 <!-- tabs:start -->