feat: add solutions to lc problems: No.2608,2609

* No.2608.Shortest Cycle in a Graph
* No.2609.Find the Longest Balanced Substring of a Binary String

Author: yanglbme

solution/2600-2699/2608.Shortest Cycle in a Graph/README_EN.md

@@ -41,6 +41,22 @@
 
 ## Solutions
 
+**方法一：Enumerate edges + BFS**
+
+We first construct the adjacency list $g$ of the graph according to the array $edges$, where $g[u]$ represents all the adjacent vertices of vertex $u$.
+
+Then we enumerate the two-directional edge $(u, v)$, if the path from vertex $u$ to vertex $v$ still exists after deleting this edge, then the length of the shortest cycle containing this edge is $dist[v] + 1$, where $dist[v]$ represents the shortest path length from vertex $u$ to vertex $v$. We take the minimum of all these cycles.
+
+The time complexity is $O(m^2)$ and the space complexity is $O(m + n)$, where $m$ and $n$ are the length of the array $edges$ and the number of vertices.
+
+**Approach 2: Enumerate points + BFS**
+
+Similar to Approach 1, we first construct the adjacency list $g$ of the graph according to the array $edges$, where $g[u]$ represents all the adjacent vertices of vertex $u$.
+
+Then we enumerate the vertex $u$, if there are two paths from vertex $u$ to vertex $v$, then we currently find a cycle, the length is the sum of the length of the two paths. We take the minimum of all these cycles.
+
+The time complexity is $O(m \times n)$ and the space complexity is $O(m + n)$, where $m$ and $n$ are the length of the array $edges$ and the number of vertices.
+
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 ### **Python3**

solution/2600-2699/2609.Find the Longest Balanced Substring of a Binary String/README_EN.md

@@ -47,6 +47,25 @@
 
 ## Solutions
 
+**Approach 1: Brute force**
+
+Since the range of $n$ is small, we can enumerate all substrings $s[i..j]$ to check if it is a balanced string. If so, update the answer.
+
+The time complexity is $O(n^3)$, and the space complexity is $O(1)$. Where $n$ is the length of string $s$.
+
+**Approach 2: Enumeration optimization**
+
+We use variables $zero$ and $one$ to record the number of continuous $0$ and $1$.
+
+Traverse the string $s$, for the current character $c$:
+
+-   If the current character is `'0'`, we check if $one$ is greater than $0$, if so, we reset $zero$ and $one$ to $0$, and then add $1$ to $zero$.
+-   If the current character is `'1'`, we add $1$ to $one$, and update the answer to $ans = max(ans, 2 \times min(one, zero))$.
+
+After the traversal is complete, we can get the length of the longest balanced substring.
+
+The time complexity is $O(n)$, and the space complexity is $O(1)$. Where $n$ is the length of string $s$.
+
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 ### **Python3**